6 concepts15 worked examples23 exercises3 exam-level6 figures
What are you here for?
05Review and consolidation I: measurement, kinematics, and Newton's laws as one machine
A midterm hands you a 2.00 kg box, a smooth ramp tilted at 30.0 degrees, and one sentence: released from rest, how far has it slid after 1.50 seconds. You know the ramp half and you know the sliding half, and you still lose eleven minutes staring at it, because you met the two halves in different weeks and nobody ever wrote down the single line that joins them.
By the end of this section you can look at any mechanics problem from the first four weeks, name in one sentence which end of it you are given and which end you are asked for, write the joining line, and finish with a number whose units, digits and size you can defend without a calculator.
In 60 seconds
The first four weeks are not four topics but one machine with three stations: forces decide the acceleration, the acceleration decides the motion, and measurement decides how many digits of the answer are real.
The joining line
$$\sum \vec{F} = m\vec{a}\;\longrightarrow\; a \;\longrightarrow\; x,\,v,\,t$$
always, and it is the first thing to write on the page: it says which station you are standing at
Newton's second law, one axis at a time
$$\sum F_x = ma_x,\qquad \sum F_y = ma_y$$
after a exists and the axes have been drawn, never before
and only when the , and therefore the acceleration, stays the same for the whole interval
Block on a smooth incline
$$a = g\sin\theta,\qquad F_N = mg\cos\theta$$
nothing pushes or pulls along the slope except the , and the surface is smooth
Two bodies joined by one string
$$|a_1| = |a_2|,\qquad a = \frac{\sum F_{\text{external}}}{m_1+m_2}$$
the string stays taut and does not stretch, and the pulley is light and turns freely
Three most common mistakes
Reaching for a kinematic equation before the acceleration has been found. All three constant acceleration formulas contain a and none can produce it, so until a has a value they have nothing to work with.
Writing the as mg out of habit. That is its value only on a level surface with zero vertical acceleration and nothing else acting vertically, and it is the most reliable place on a paper to lose marks.
Reading zero velocity as zero force. A ball at the top of its flight and a book on a table both have v = 0 and are opposite cases: the ball carries its full weight, the book has no net force at all.
The published weights are Midterm 1 20 per cent, Midterm 2 20 per cent, quizzes 10 per cent, homework 5 per cent, the final 25 per cent and the laboratory 20 per cent. This week's line is a catch up and review week, so nothing new is examined; what is examined is whether the four weeks behind you can be used together in one question.
How much time do you have?
10 minutes
The one line that tells you where to start on any problem, and the two component equations that turn a picture of forces into arithmetic. If you read nothing else, these stop the blank page.
card, c-acceleration-hinge, c-fbd-to-equations, formula card
45 minutes
Everything that turns into a number on a script: choosing axes and a sign once and keeping them, picking the kinematic equation by the variable that is missing, going from a free body diagram to two equations, handling two bodies on one string, and the four checks that catch a wrong answer before the marker does.
card, c-acceleration-hinge, c-setup-protocol, c-fbd-to-equations, c-two-body-constraints, c-answer-checking, faded ladder, practice B
full read
Adds the part a formula sheet cannot carry: deciding which kind of problem you are looking at before you start writing, seeing the same picture read forwards and backwards, and the interleaved practice that mixes all four weeks so that the decision, not the algebra, is what you are training.
hook, recall first, pretest, c-acceleration-hinge, c-setup-protocol, c-fbd-to-equations, c-two-body-constraints, c-answer-checking, c-triage, method boxes, contrast pair, faded ladder, exam example, practice A to D, mistake ledger, self audit
By the end of this section
Connect a force problem to a motion problem through the acceleration, stating which quantity carries the information across and in which direction the calculation runs.
Set up a problem by declaring an origin, a positive direction and a time zero, and select the constant acceleration equation by naming the variable that does not appear.
Translate a free body diagram into one scalar equation per axis, and compute a normal force or a that is not equal to the weight.
Solve for the acceleration and the tension of two bodies joined by a light string over a light pulley, using the constraint that ties the two accelerations together.
Check a finished answer against its units, its number of significant figures, its order of magnitude and at least one limiting case, and say which check would have caught a given error.
Classify an unseen mechanics problem by its signature phrases, name the first move it needs, and say which of the four weeks supplies each step of the route.
Syllabus coverage
Catch up and Review
The four weeks already taught, used together: measurement and significant figures, one dimensional kinematics, vectors and two dimensional kinematics, and Newton's three laws
The week line names no chapter and no section number, so none is quoted anywhere in this section. The scope is fixed by what has already been taught and by nothing else: every worked example here can be solved with the results of the first four weeks and no others.
covered
Friction, circular motion and drag forces
Surfaces that resist sliding, motion on a curved path, and forces that grow with speed
Deferred to the week immediately after this review. Every surface here is stated to be smooth and every pulley light and free, because the quantity that describes a real surface has not been defined yet.
deferred
Work, energy and momentum methods
Solving the same problems by tracking energy or momentum instead of forces and accelerations
Deferred to the later weeks that build them. Several problems here would be shorter with an energy argument and are done the long way on purpose, because a method you have not been taught cannot be quoted as a reason.
deferred
Laboratory measurement and report writing
Taking readings, estimating uncertainties in the laboratory, and writing the experiment up
The laboratory carries 20 per cent of the grade, but this week's line announces no laboratory content, so nothing is claimed about any particular experiment. What is reviewed is the course wide part taught in the first week: how many digits a result may carry, and how a number is checked before it is written down.
off_syllabus
Recall first
Significant figures and the weakest datum
A product or a quotient keeps as many significant figures as the least precise number that went into it, and a sum keeps as many decimal places as the coarsest one. Guard digits are carried through the middle of a calculation and dropped only at the end, so that rounding twice cannot push the last digit.
Almost every answer in this section is a product of three numbers, and every one of them is quoted to three significant figures. Writing 694.4 N when the data support 694 N is not a rounding preference; it claims a precision the measurement never had.
Displacement, velocity and acceleration on one axis
With one straight axis and a declared positive direction, the displacement is $\Delta x = x - x_0$, the average velocity is $\Delta x/\Delta t$, and the average acceleration is $\Delta v/\Delta t$. The instantaneous versions are the derivatives $v = dx/dt$ and $a = dv/dt$.
The whole of the second half of this section produces an acceleration from forces. That acceleration is worthless unless you remember what it is a rate of change of, and in particular that it is the rate of change of the velocity, not of the position.
The constant acceleration kit
For motion on one axis with a constant acceleration $a$: $v = v_0 + at$, then $x = x_0 + v_0t + \tfrac{1}{2}at^{2}$, then the time free one, $v^{2} = v_0^{2} + 2a(x-x_0)$, and the average form $x = x_0 + \tfrac{1}{2}(v_0+v)t$.
These four are the second half of every problem here. They are also the four that get used out of turn: they are true only while the acceleration does not change, and the moment a force switches on or off in the middle of a motion, the interval has to be split and the kit applied twice.
Free fall near the ground
An object moving freely near the surface of the Earth, with air resistance neglected, has a downward acceleration of magnitude $g = 9.80\ \mathrm{m/s^{2}}$, independent of its mass and independent of how fast it is already moving.
It is the one acceleration you are allowed to write down without computing it from forces, and it is the number every other acceleration in this section gets compared with when you ask whether an answer is plausible.
Resolving a vector and rebuilding it
With $\theta$ measured counterclockwise from the positive $x$ axis, $A_x = A\cos\theta$ and $A_y = A\sin\theta$, both signed. Going back, $A = \sqrt{A_x^{2}+A_y^{2}}$ and $\tan\theta = A_y/A_x$, with a sketch to settle which quadrant the arctangent meant.
A free body diagram is a set of arrows at different angles, and the only way to add arrows is to resolve them. Every force equation in this section is really two equations produced by this rule.
Newton's three laws
First: with no net force a body keeps its velocity, zero or not. Second: $\sum \vec{F} = m\vec{a}$, with the sum running over every force acting on that one body. Third: if A pushes B with $\vec{F}$, then B pushes A with $-\vec{F}$, and the two forces act on different bodies.
The first law is what lets you write zero on the right hand side; the second is the only equation in this section that produces an acceleration; the third is what stops you putting both members of a pair into the same free body diagram, which would cancel them and give zero acceleration for everything.
Weight, normal force and tension
The weight of a body of mass $m$ near the ground is $mg$, directed downward, and it is measured in newtons while the mass is in kilograms. A surface pushes on a body perpendicular to itself with the normal force $F_N$; a taut string pulls along its own length with the tension $F_T$. Neither $F_N$ nor $F_T$ has a formula of its own.
Because they have no formula, they are always unknowns to be solved for, and the whole difficulty of a dynamics problem is that the interesting force is the one you cannot write down at the start. Assuming a value for either is the fastest way to a wrong answer that looks tidy.
Try it yourself first (2 questions)
1§05.1 — net force on a body moving at constant velocity●●○○○
Three quick questions before the section starts, to find out which of the last four weeks are still to hand. Getting them wrong costs nothing and is recorded nowhere; a miss only tells you which recall above to read twice. A crate is sliding across a smooth level floor in a straight line, and a stopwatch and a tape show that it covers equal distances in equal times.
Given
Mass of the crate $m = 24.0\ \mathrm{kg}$
The floor is smooth and level, and the crate slides in a straight line
Successive 1.00 s intervals each show a displacement of $2.40\ \mathrm{m}$
Find
(a) What is the net force on the crate while it slides?
Hint 1/4
The question is not asking what keeps the crate moving. It is asking what the sum of the forces is, and the second law ties that sum to one single quantity, so decide what that quantity is doing here first.
Hint 2/4
Newton's second law reads $\sum \vec{F} = m\vec{a}$, and the first law is the special case of it in which the acceleration is zero and the velocity, zero or not, stays whatever it was.
Hint 3/4
Substitute the data given above: equal displacements of $2.40\ \mathrm{m}$ in equal times of $1.00\ \mathrm{s}$ mean $v = 2.40\ \mathrm{m/s}$ and unchanging, so $a = 0$, and with $m = 24.0\ \mathrm{kg}$ the sum is $24.0 \times 0$.
Hint 4/4
So the net force on the sliding crate is zero, even though it is moving and even though several separate forces act on it.
Show solution
The acceleration is read off the data before any force is named, because the second law runs in that direction here: the motion is known and the force is wanted, which is the reverse of the usual dynamics question.
the second law is an equality, so a zero on one side forces a zero on the other; nothing about the crate's speed enters
Answer $$\boxed{\sum \vec{F} = 0}$$
Check
Check it against the individual forces instead of the sum: the weight is $mg = 24.0 \times 9.80 = 235\ \mathrm{N}$ downward and the floor pushes up with $F_N$. Since the crate does not sink or rise, $F_N = 235\ \mathrm{N}$, and the two cancel, which is the same zero arrived at from the other end.
The trap is a leftover from everyday life, where letting go of something moving makes it stop. On a smooth floor there is nothing to make it stop, and the first law is precisely the statement that no force is needed to keep a velocity.
2§05.3 — the component of the weight along a slope●●○○○
A block rests on a smooth ramp that makes an angle of 25.0 degrees with the horizontal, and it is about to be released. Only one force on it has a component along the slope, and the whole problem turns on which trigonometric function that component uses.
Given
Mass $m = 3.00\ \mathrm{kg}$, so the weight is $mg = 29.4\ \mathrm{N}$
Slope angle $\theta = 25.0^{\circ}$ measured from the horizontal
The ramp surface is smooth, and nothing is attached to the block
Find
(a) What is the component of the weight parallel to the slope, pointing down the slope?
Hint 1/4
Rather than recalling which function it is, decide what the answer must do at the two extremes: a flat ramp, where the block does not slide at all, and a vertical wall, where it falls freely.
Hint 2/4
With the axes turned so that $x$ runs along the slope, the weight splits into $mg\sin\theta$ along the slope and $mg\cos\theta$ into the surface, where $\theta$ is the angle of the ramp with the horizontal.
Hint 3/4
Substitute the data given above, $m = 3.00\ \mathrm{kg}$ and $\theta = 25.0^{\circ}$: the weight is $29.4\ \mathrm{N}$ and $\sin 25.0^{\circ} = 0.4226$.
Hint 4/4
So the component down the slope is $29.4 \times 0.4226 = 12.4\ \mathrm{N}$.
Show solution
The limits are checked before the calculator is used, because the sine and cosine are equally easy to type and only one of them survives the flat ramp test.
Settle the function by its behaviour, not by memory
sine of the ramp angle, with the calculator in degrees, and three figures kept because the data carry three
Answer $$\boxed{mg\sin\theta = 12.4\ \mathrm{N}\ \text{down the slope}}$$
Check
The two components must rebuild the weight: $\sqrt{12.4^{2}+26.6^{2}} = \sqrt{154+708} = \sqrt{862} = 29.4\ \mathrm{N}$, which is the weight, so nothing has been lost or double counted.
The sine and cosine swap places if the angle is measured from the vertical instead of from the horizontal. Reading the angle off the drawing, not off the habit, is the whole of the defence.
Notation
symbol
reads as
means
watch out
$\sum \vec{F}$
the net force, or the sum of the forces
the vector sum of every force acting on one chosen body, and nothing else: forces on other bodies never appear in it
it is a sum over forces on a single body, so the moment two bodies are in the problem there are two of these sums and they are different
$F_N$
the normal force
the push of a surface on a body, always perpendicular to that surface and always a push, never a pull
it has no formula of its own; $F_N = mg$ is a result that holds only on a level surface with nothing else acting vertically
$F_T$
the tension
the pull of a taut string along its own length, the same at both ends when the string is light and the pulley turns freely
a tension is a pull on the body, so it points away from the body along the string, and it is never drawn pointing into the body
$w = mg$
the weight
the downward force the Earth exerts on a body of mass $m$ near the ground, measured in newtons
the mass in kilograms and the weight in newtons are different quantities, and a body taken elsewhere keeps its mass and changes its weight
$a_x,\ a_y$
the x and y components of the acceleration
the two signed numbers the acceleration splits into once the axes are drawn, each obeying its own second law equation
an object can have $a_x = 0$ and $a_y \ne 0$ at the same instant; zero on one axis says nothing about the other
$g$
g
the positive constant $9.80\ \mathrm{m/s^{2}}$, the magnitude of the free fall acceleration near the ground
$g$ is a magnitude and carries no sign; the sign appears once, when you write $a_y = -g$ with up chosen positive
$\theta$
theta
in this section, the angle of a slope with the horizontal, or the angle a rope makes with the horizontal, always stated in the sentence that introduces it
a component is $\cos$ or $\sin$ of it depending on where the angle was measured from, so the reference direction is part of the symbol's meaning
Conventions used here
Signs and axes in this section.
Every problem starts by drawing the axes and saying in words which way is positive, before any equation. On level ground $x$ points along the motion and $y$ points up; on a slope the axes are turned so that $x$ runs along the slope, because that is the direction the acceleration actually has. The choice is written down once and then never changed inside that problem, and a different problem may choose differently.
A sign convention that drifts between two lines of the same solution produces an answer wrong by a factor of minus one that looks perfectly reasonable. Turning the axes on a slope is not decoration either: it makes one of the two acceleration components zero, which is what turns a two unknown problem into a one unknown problem.
The constant g in every calculation here.
The symbol $g$ stands for the positive number $9.80\ \mathrm{m/s^{2}}$ throughout, and the sign is written into the equation instead: with up positive, a freely falling body has $a_y = -g$. These notes never write $g$ with a minus sign attached, and never use $10\ \mathrm{m/s^{2}}$ for it, even in a rough estimate.
Two minus signs in the same term is the commonest way a projectile ends up accelerating upward on a script. Keeping the constant positive leaves exactly one place where the sign can be wrong, and that place is visible.
What is neglected in every problem in this section.
Every surface is smooth, every string is light and does not stretch, every pulley is light and turns freely, and air resistance is absent. Where a problem needs a resisting force, that force is given as a number in newtons rather than derived from a property of the surface.
The quantities that describe a real rough surface or a real air flow are introduced in the week after this review, so using them here would mean quoting a result you have not been given. Stating the idealisation in each problem is also what an examiner expects: an answer that silently assumes a on a rough one is not a small error.
How many significant figures survive.
Data in this section are quoted to three significant figures and answers are given to three. Extra guard digits are carried through intermediate lines and dropped only in the final box, and a number written as $1.20\times 10^{3}$ means three figures while $1200$ on its own is ambiguous and is avoided.
Rounding at every line pushes the last digit around by several units in a three step calculation. Rounding once, at the end, keeps the answer's last figure meaningful, and quoting more figures than the data support claims a precision no measurement had.
What the words smooth, light and ideal mean here.
Smooth means the surface exerts no force along itself. Light, for a string or a pulley, means its mass may be taken as zero, so the tension is the same at both ends of the string and the pulley only changes the direction of the pull. Ideal, for a pulley, adds that it turns without resisting.
These three words are doing real algebraic work, not setting a scene. Each one removes an unknown from the equations, and a problem that omits one of them is a different problem with more unknowns than equations.
Mass and weight are kept apart.
Mass is in kilograms and never appears in a force equation as a force; weight is in newtons and never appears as a mass. A statement such as a 5.00 kg force, or a body of weight 5.00 kg, is treated as an error rather than as loose speech, and the sentence is rewritten before the algebra starts.
The second law multiplies a mass by an acceleration to get a force, so putting a weight where the mass belongs multiplies the answer by 9.80 and putting a mass where the weight belongs divides it by 9.80. Both errors leave a plausible looking number.
5.1Acceleration is the only door between forces and motion
Forces produce an acceleration; the acceleration produces the motion. Nothing else crosses between the two halves.
Four weeks have gone by and they look like four subjects. They are one subject with a joint in the middle, and this section is about the joint.
Solvable with what we have
Given an acceleration, find the distance covered, the speed reached, or the time taken, on one axis or on two.
Given the forces on a body, draw them, resolve them and add them into a single net force.
Given a measurement in the wrong unit, convert it and say how many digits of the result are real.
Given a velocity and a launch angle, split a flight into a horizontal problem and a vertical problem.
Not solvable yet
Find how far a block slides down a ramp in a given time, when the ramp angle is all you are told about the push.
Find the force a lift cable carries while the lift speeds up, when only the change in speed and the time are measured.
Decide, before starting, whether a problem is asking you to run the machine forwards or backwards.
Take the ramp problem and reach straight for the equation that gives distance in a given time. It needs an acceleration, and the only acceleration written anywhere on the page is $9.80\ \mathrm{m/s^{2}}$, so that goes in: $x = \tfrac{1}{2}(9.80)(1.50)^{2} = 11.0\ \mathrm{m}$.
Why it fails
The number $9.80\ \mathrm{m/s^{2}}$ belongs to a body falling freely, with nothing touching it. This block is held up by a surface that cancels most of the weight, so it does not fall, it slides, and the answer is $5.51\ \mathrm{m}$, exactly half the naive one because the ramp is at thirty degrees. The failure is not arithmetic: the acceleration was assumed rather than earned.
RuleRule: the two halves of mechanics meet at the acceleration
Conditions
the force sum runs over every force acting on the one body you have chosen, and over no others
the kinematic formulas below require the acceleration to be constant over the whole interval, which is true here only because the forces do not change during the motion
the mass in the second law is in kilograms, so a weight in newtons has to be divided by $g$ before it can be used there
the acceleration produced on one axis is used only in the equations of that same axis
Add up the forces on one body and divide by its mass, and you have its acceleration. Feed that acceleration into the motion formulas and you have where the body is and how fast it is going. The arrow can be walked in either direction: if the motion is what you measured, the same acceleration comes out of the second half and goes into the first, and out comes a force you never touched.
Why the two halves can be joined at all
The second law is not derived here; it is the experimental statement that the acceleration of a body is proportional to the net force on it and inversely proportional to its mass.
The kinematic formulas are not experimental at all. They follow from the definitions $a = dv/dt$ and $v = dx/dt$ by integrating twice with $a$ held constant, which is why they carry that condition and the second law does not.
So the arrow between them is not a physical law but a bookkeeping fact: the symbol $a$ in the second law and the symbol $a$ in the kinematic formulas denote the same thing, the rate of change of the velocity of the same body along the same axis. That single shared meaning is the whole of the joint.
The block from the opening problem and the diagram that answers it. The axes are turned so that one runs along the slope, the direction the block will accelerate in. The normal force then sits entirely on the other axis and drops out, leaving one force, one unknown and the number 4.90.
Looks like this, but is not
This is the rule working: the forces on the block are fixed for the whole slide, so $a = g\sin\theta$ is one number for the whole slide, and the constant acceleration formulas apply from the first instant to the last.
This looks like the same rule and is not: a parachutist has forces and an acceleration too, and the joint still holds, but the upward force grows as she speeds up, so the acceleration changes at every instant. The second law survives; the kinematic formulas do not, because they were derived with $a$ held constant. The joint is unconditional, the second half of the machine is not.
How far a 2.00 kg block slides down a smooth 30.0 degree ramp in 1.50 s
The block of the opening problem is released from rest at the top of a smooth ramp. Find how far it has slid along the slope after 1.50 s, and how fast it is then moving.
Given
Mass $m = 2.00\ \mathrm{kg}$
Ramp angle $\theta = 30.0^{\circ}$ above the horizontal, surface smooth
Released from rest, so $v_0 = 0$ at $t = 0$
Forces acting: the weight $mg$ downward and the normal force $F_N$ perpendicular to the surface, and nothing else
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The distance travelled along the slope in 1.50 s, and the speed at that instant.
Solution
The axes are turned along the slope rather than left horizontal and vertical. With horizontal axes the block would have two non zero acceleration components and the unknown normal force would appear in both equations; with turned axes the acceleration lies entirely on one axis and the normal force entirely on the other, which is the difference between two coupled equations and one easy one.
Get the acceleration from the forces
$$\sum F_x = mg\sin\theta = ma_x$$
along the slope the normal force has no component at all, so the weight component is the entire net force in that direction
perpendicular to the surface the block neither sinks in nor lifts off, so that acceleration component is zero; this line is not needed for the distance but it is the line that would be needed next week
the speed comes from the same acceleration and the same clock, with no new physics needed
Answer $$\boxed{x = 5.51\ \mathrm{m}\ \text{along the slope},\qquad v = 7.35\ \mathrm{m/s}}$$
Check
Independent route to the same speed, without using the time: $v^{2} = 2a_xx = 2(4.90)(5.51) = 54.0$, so $v = 7.35\ \mathrm{m/s}$. The two answers were produced by different equations from different data, and they agree. The size is also right: 7.35 m/s is a fast jog, which is what a metre and a half of slide at a gentle angle ought to give.
Two lines of force work and two lines of motion work. The whole difficulty was in the first line, where the weight had to be resolved; everything after it is arithmetic.
Notice what the answer does not contain: the mass. It cancelled in the second step and never came back, so the same ramp gives 5.51 m for a 2.00 kg block and for a 200 kg one. That is a fact worth carrying, and it is also a fast way to check that a ramp answer has not been mangled.
What ramp angle gives an acceleration of exactly 2.00 m/s squared
The same smooth ramp is now adjustable, and the same 2.00 kg block is placed on it. Find the angle at which the block would accelerate at 2.00 m/s squared, and the normal force at that angle.
Given
Mass $m = 2.00\ \mathrm{kg}$, weight $mg = 19.6\ \mathrm{N}$
Required acceleration along the slope $a = 2.00\ \mathrm{m/s^{2}}$, down the slope
Smooth surface; the forces are the weight and the normal force only
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The ramp angle, and the normal force at that angle.
Solution
This is the same picture as the previous example with the arrow reversed: the motion is given and a feature of the forces is wanted. Because the relation $a = g\sin\theta$ was already established for this picture, the reversal costs one line of algebra rather than a new diagram.
Invert the relation between angle and acceleration
Two independent checks. First the limits: at $\theta = 0$ the formula gives $a = 0$ and $F_N = 19.6\ \mathrm{N}$, a block sitting still on a level floor with the full weight supported, which is right; at $\theta = 90^{\circ}$ it gives $a = 9.80\ \mathrm{m/s^{2}}$ and $F_N = 0$, free fall down a wall that is no longer touching the block, which is also right. Second the size: 11.8 degrees is a shallow ramp and 2.00 m/s squared is a fifth of free fall, and the two match.
The normal force at 11.8 degrees is 19.2 N while the weight is 19.6 N, close but not equal. On a shallow ramp the difference is small and invisible in a sketch, which is exactly why writing $F_N = mg$ survives so long as a habit: it is nearly right on gentle slopes and badly wrong on steep ones.
Checkpoint
§05.1 — what changing the mass does to a ramp acceleration●●○○○
Half a minute, no calculator. The 2.00 kg block on the smooth 30.0 degree ramp accelerated at 4.90 m/s squared. It is now swapped for a 6.00 kg block on the same ramp, released from rest in the same place.
Given
Same smooth ramp, $\theta = 30.0^{\circ}$
New mass $m = 6.00\ \mathrm{kg}$ in place of $2.00\ \mathrm{kg}$
Released from rest, forces are the weight and the normal force only
Find
(a) What is the acceleration of the heavier block along the slope?
Hint 1/4
Do not recompute. Look at the along slope equation and ask where the mass appears in it, and whether it survives to the end.
Hint 2/4
Along the slope, $mg\sin\theta = ma$, and the same $m$ stands on both sides of the equals sign.
Hint 3/4
Substitute the data from the question, $\theta = 30.0^{\circ}$ and $m = 6.00\ \mathrm{kg}$: dividing both sides by 6.00 leaves $a = g\sin 30.0^{\circ}$, with no mass in it.
Hint 4/4
So the acceleration is unchanged at $4.90\ \mathrm{m/s^{2}}$, while the forces have all tripled.
Show solution
The symbols are kept until the last line instead of substituting the numbers first, because the cancellation is the answer and substituting early hides it inside two numbers that happen to divide.
Write the along slope equation in symbols
$$mg\sin\theta = ma$$
the only force with an along slope component is the weight, whatever the mass is
$$a = g\sin\theta$$
dividing by m removes it from the problem entirely, so no numerical value of the mass is needed
the forces did triple; it is only their ratio to the that stayed put
Answer $$\boxed{a = 4.90\ \mathrm{m/s^{2}},\ \text{independent of the mass}}$$
Check
This is the ramp version of a fact already established for free fall: with only gravity and a frictionless contact acting, the acceleration never depends on the mass. Setting the ramp angle to ninety degrees turns the result into $a = g$, which is the free fall statement, so the two agree at the one angle where they overlap.
Whenever a mass cancels, say so out loud in the solution. It is a strong check on the algebra and it usually means the answer applies to a whole family of problems rather than to the one on the page.
⚠ Using the whole weight as the along slope force
the weight is the only force anyone remembers a formula for, and on a picture with one obvious arrow the temptation is to use all of it
⚠ Starting from a kinematic equation before the acceleration exists
the question asks for a distance and a time, so the formula containing distance and time looks like the right place to begin, and the acceleration slot then gets filled with whatever number is nearby
on every level surface met so far the two were equal, so the equality gets remembered as a property of the normal force rather than as the result of one particular balance
The three stations and the two directions between them. Blue is the way most problems run, from forces to motion; orange is the reverse, for when the motion was measured and a force is wanted. Every question in these four weeks is a walk along this map, and naming the starting station is the setup.
5.2Choosing the origin, the sign and the equation before touching a calculator
Three decisions made in writing before any arithmetic, and one rule for picking which formula to use.
The joint is useless if the two halves are set up in different coordinate systems. So before the joint gets used, three things get written down.
MethodMethod: declare the frame, then choose the equation by what is missing
Conditions
the origin, the positive direction and the instant called $t=0$ are chosen once and never changed inside a single problem
the four formulas below hold only while the acceleration is constant, so an interval in which a force switches on or off must be split in two and each piece treated separately
a quadratic in the time has two roots and both are mathematically valid, so the physical one is selected by an argument, not by taste
all four are one dimensional; in two dimensions they are used once per axis with that axis's own acceleration
Five quantities describe a constant acceleration motion, and each formula is missing exactly one of them. So the choice is mechanical rather than a matter of judgement: list the quantity you were given, the one you want, and the one nobody mentioned, then take the formula that leaves out the one nobody mentioned. It is the only formula that will not force you to find something you were not asked about.
Where the four formulas come from, and why each one is missing a variable
Start from the definition $a = dv/dt$ with $a$ constant and integrate once: $v = v_0 + at$. That is the first formula, and no position appears in it because no position was integrated.
Integrate again, using $v = dx/dt$, to get $x = x_0 + v_0t + \tfrac{1}{2}at^{2}$. The final velocity never entered the integration, so it is the one absent here.
Eliminate $t$ between those two, or equivalently integrate $v\,dv = a\,dx$, and the time free formula $v^{2} = v_0^{2}+2a(x-x_0)$ appears. Averaging the first formula against the definition of the mean velocity gives the fourth, in which the acceleration has cancelled. Four formulas, four different variables left out, and that is the whole basis of the selection rule.
One throw, two legitimate origins, one answer. Zero at the balcony puts the ground at minus eighteen metres; zero at the ground puts the launch at plus eighteen. Different numbers, identical equation: the origin is a bookkeeping decision, and the only fatal thing is to use both inside one solution.
Looks like this, but is not
This is the rule working: a ball is thrown up at $12.0\ \mathrm{m/s}$ and, with up positive and the origin at the hand, the whole flight is covered by one equation with $a = -9.80\ \mathrm{m/s^{2}}$ throughout, because gravity acts the whole time.
This looks like the same situation and is not: a lift starts from rest, speeds up for 3.0 s, runs at a steady speed for 8.0 s, then slows to a stop in 2.0 s. Writing one constant acceleration equation across the thirteen seconds gives a confident and meaningless number, because the acceleration took three different values. The motion has to be cut at the two instants where the forces changed, and the kit applied three times, with the end of each piece becoming the start of the next.
you know
you want
not mentioned anywhere
the formula to use
$v_0$, $a$, $t$
$v$
the position
$v = v_0+at$
$v_0$, $a$, $t$
$x-x_0$
the final velocity
$x-x_0 = v_0t+\tfrac{1}{2}at^{2}$
$v_0$, $a$, $x-x_0$
$v$
the time
$v^{2} = v_0^{2}+2a(x-x_0)$
$v_0$, $v$, $t$
$x-x_0$
the acceleration
$x-x_0 = \tfrac{1}{2}(v_0+v)t$
Read the third column first. Whichever of the five quantities the problem never speaks about is the one that must be missing from your formula, and that single observation picks the row for you without any trial and error. If two rows seem to fit, the problem has given you a redundant datum, and you can use the spare one afterwards as a check.
A ball thrown up from an 18.0 m balcony: when does it land
A ball is thrown vertically upward from a balcony and misses the balcony on the way down, landing on the ground below. Find the time from the throw to the landing, and the speed at impact.
Height of the balcony above the ground: $18.0\ \mathrm{m}$
Air resistance neglected, $g = 9.80\ \mathrm{m/s^{2}}$
The ball is in free flight from the moment it leaves the hand until it reaches the ground
Find
The total time of flight and the impact speed.
Solution
The whole flight is treated as one interval rather than as an upward part and a downward part. Splitting it is not wrong but costs three extra lines, and it is only necessary when something changes at the top, which here it does not: the acceleration is the same going up, at the top, and coming down.
Declare the frame in writing
$$\text{origin at the balcony},\quad +y\ \text{upward},\quad t=0\ \text{at the throw}$$
the origin is placed where the most data are zero, which is the launch point, so that $y_0 = 0$ disappears from the equation
with up positive, the throw is positive and both the acceleration and the landing point are negative; the minus signs are consequences of the declaration, not extra decisions
the discriminant is $b^{2}-4ac$ with $a=4.90$, $b=-12.0$, $c=-18.0$
$$t = 3.50\ \mathrm{s}\quad\text{or}\quad t = -1.05\ \mathrm{s}$$
both roots satisfy the algebra; only one of them happens after the ball was thrown
$$t = 3.50\ \mathrm{s}$$
the negative root describes where the ball would have been launched from the ground to arrive at the balcony with this velocity, which is a different problem
the minus sign says downward, which is the only direction the answer could have had
Answer $$\boxed{t = 3.50\ \mathrm{s},\qquad v = 22.3\ \mathrm{m/s}\ \text{downward}}$$
Check
Independent check on the speed, using the time free formula and therefore none of the work above: $v^{2} = v_0^{2} + 2a\,\Delta y = 144 + 2(-9.80)(-18.0) = 496.8$, giving $v = 22.3\ \mathrm{m/s}$. It also passes the size test: 22.3 m/s is about 80 km/h, which is what six storeys of fall produce and why dropped objects are dangerous.
One quadratic, one root rejected with a stated reason, one substitution. The frame declaration at the top is three symbols long and it is what makes every sign in the rest of the solution automatic.
The upward throw did not need to be treated separately from the downward fall. Whenever the acceleration is the same throughout, one equation spans the whole motion, and the extra work of splitting at the highest point buys nothing.
Checkpoint
§05.2 — picking the constant acceleration formula in one step●●○○○
Thirty seconds. A car starts from rest and accelerates uniformly along a straight road. You are told how hard it accelerates and how far it travels, and you are asked how fast it is going at the end. Nobody has mentioned how long any of it took.
(a) Which single formula gives the final speed without first finding anything else?
Hint 1/4
Do not solve anything. List the five quantities and mark which one the problem is silent about, because that is the one your formula must not contain.
Hint 2/4
Each of the four formulas omits exactly one of $x$, $v$, $t$ and $a$, and the right one omits whichever quantity is neither given nor asked for.
Hint 3/4
Substitute the data from the question: given $v_0 = 0$, $a = 2.40\ \mathrm{m/s^{2}}$ and $x-x_0 = 85.0\ \mathrm{m}$, wanted $v$, never mentioned $t$.
Hint 4/4
So the time free formula is the one, and it gives $v^{2} = 2(2.40)(85.0) = 408$, hence $v = 20.2\ \mathrm{m/s}$.
Show solution
The selection is made from the list of what is absent rather than by trying formulas, which is what makes it a one step problem instead of a two step one.
Select
$$\text{absent: } t \;\Rightarrow\; v^{2} = v_0^{2}+2a(x-x_0)$$
the time is neither given nor asked for, so it must not appear in the formula
starting from rest kills the first term, and the units of the product are metres squared per second squared as they must be
$$v = \sqrt{408} = 20.2\ \mathrm{m/s}$$
the positive root, since the car is moving forwards along the positive axis
Answer $$\boxed{v = 20.2\ \mathrm{m/s}}$$
Check
Cross check by the longer route: $85.0 = \tfrac{1}{2}(2.40)t^{2}$ gives $t = 8.42\ \mathrm{s}$, and then $v = (2.40)(8.42) = 20.2\ \mathrm{m/s}$. Same answer, two extra lines. The size is also sensible, since 20.2 m/s is about 73 km/h, a reasonable speed to reach in 85 m.
The absent variable rule turns formula choice from a memory task into a reading task, and it is worth the ten seconds it takes even when the problem looks obvious.
⚠ Keeping both roots of the flight time, or keeping the wrong one
the quadratic formula hands over two numbers with equal authority, and nothing in the algebra marks one of them as belonging to a time before the throw
right$$y_0 = 0,\ y = -18.0,\ a = -9.80\ \text{used in every line of the same solution}$$
5.3From a diagram of forces to one equation per axis
Draw only the forces on the chosen body, resolve them, and write one scalar equation for each axis.
The joint needs an acceleration, and the acceleration comes from a sum of forces. Producing that sum reliably is a drawing skill before it is an algebra skill.
MethodMethod: one body, one diagram, one equation per axis
Conditions
the diagram carries only the forces acting ON the chosen body, never the forces that body exerts on anything else
the two members of an action and reaction pair act on different bodies, so they never both appear in the same diagram
an acceleration is not a force and is never drawn as one of the arrows; if it is shown at all it is drawn beside the diagram
the axes are chosen so that the acceleration lies along one of them, which makes the other equation a balance
Pick one body. List every force that touches it or acts on it at a distance, and draw each as an arrow leaving the body. Then choose axes and write, for each axis separately, that the sum of the components along that axis equals the mass times the acceleration component along the same axis. The normal force is whatever that arithmetic says it is, and it equals the weight only in the special case where the surface is level and nothing else pulls or pushes vertically.
The same person, the same scale, three different readings. The weight arrow never changes; the acceleration does, and the reading follows it: larger when the acceleration points up, smaller when it points down, equal to the weight only in the middle case. Nothing depends on which way the lift is travelling.
Looks like this, but is not
This is a correct free body diagram: a block resting on a table carries two arrows, its weight pulling down and the table pushing up, and the two are equal because the block is not accelerating.
This looks like the same diagram and is wrong: the block's downward push on the table is also equal and opposite to the weight, and it is tempting to call that the balancing pair. It acts on the table, not on the block, so it does not belong here. The test is mechanical: two forces that balance act on the SAME body and differ in size once that body accelerates, whereas a third law pair acts on two bodies and is equal always.
what the lift is doing
direction of $\vec{v}$
direction of $\vec{a}$
scale reads $F_N$
starting upward
up
up
$694\ \mathrm{N}$
arriving at an upper floor
up
down
$521\ \mathrm{N}$
cruising, or standing still
either or none
zero
$608\ \mathrm{N}$
starting downward
down
down
$521\ \mathrm{N}$
arriving at a lower floor
down
up
$694\ \mathrm{N}$
Read the last column, then look back at the second. Rows two and four have opposite velocities and identical readings, and rows one and five likewise, so the direction of travel predicts nothing. The scale measures which way your velocity is changing, which is why the lurch comes at the start and the end of a lift ride and not in the middle.
What the scale reads for a 62.0 kg person in a lift accelerating at 1.40 m/s squared
A person stands on a bathroom scale on the floor of a lift. Find the scale reading while the lift accelerates upward at 1.40 m/s squared, while it moves at constant velocity, and while it accelerates downward at 1.40 m/s squared.
Given
Mass of the person $m = 62.0\ \mathrm{kg}$
Forces on the person: the weight $mg$ downward and the normal force $F_N$ from the scale upward, and nothing else
Case one: $a = 1.40\ \mathrm{m/s^{2}}$ upward. Case two: $a = 0$. Case three: $a = 1.40\ \mathrm{m/s^{2}}$ downward
$g = 9.80\ \mathrm{m/s^{2}}$; the scale reads the force it pushes up with, in newtons
Find
The three scale readings.
Solution
The person is chosen as the body, not the lift and not the scale. The lift would bring in the cable tension and the lift's own mass, neither of which is given, and the question is about what the scale pushes on the person with, so the person is the body that has that force in its diagram.
Two independent checks. The extreme case: setting $a_y = -g$ gives $F_N = 0$, which is what a scale reads in free fall, and that is the known answer for a falling lift. The size: the readings differ from the weight by 86 N, and $86/608 = 14$ per cent, which is the same as $1.40/9.80 = 14$ per cent, as the formula requires.
One equation, solved once in symbols and evaluated three times. Rewriting the equation for each case would have tripled the chances of a sign error for no gain.
The phrase means this reading and nothing more mysterious: it is the force a support pushes you with, and your actual weight of 608 N was the same in all three cases. Nothing about the person changed; only the acceleration did.
Finding the lift's acceleration from a scale reading of 550 N
The same person watches the same scale and sees it settle at 550 N for a few seconds. Find the acceleration of the lift during that time, and say whether the lift is going up or down.
Given
Mass of the person $m = 62.0\ \mathrm{kg}$, so the weight is $mg = 608\ \mathrm{N}$
Scale reading $F_N = 550\ \mathrm{N}$, steady
Forces on the person: the weight downward and the normal force upward, nothing else
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The acceleration, in magnitude and direction, and whether the direction of travel can be deduced.
Solution
This is the previous example run backwards, so the same single equation is used and solved for a different symbol. Reaching for a kinematic formula here would be a mistake: no distance and no time have been given, and none is needed.
the sign of the acceleration follows directly from the declared positive direction
$$\vec{v}\ \text{unknown}$$
a downward acceleration means either descending and speeding up or ascending and slowing down, and the scale reading cannot distinguish them
Answer $$\boxed{a = 0.929\ \mathrm{m/s^{2}}\ \text{downward; the direction of travel cannot be determined}}$$
Check
Substitute back into the forward version: $F_N = m(g+a_y) = 62.0(9.80-0.929) = 62.0(8.871) = 550\ \mathrm{N}$, which is the reading given. The size is also ordinary, since a real lift accelerates at about one metre per second squared and shifts a scale reading by about a tenth.
The last line is the one worth carrying into an exam. A question that gives you a scale reading, or a tension, or a normal force, has told you about the acceleration only. Anyone who reports the direction of travel from it has answered a question the physics did not ask.
Checkpoint
§05.3 — scale reading in a lift descending at constant speed●●○○○
Thirty seconds. The same 62.0 kg person stands on the same scale, and the lift is now moving steadily downward between two floors, neither speeding up nor slowing down.
Given
$m = 62.0\ \mathrm{kg}$, so $mg = 608\ \mathrm{N}$
The lift descends at a constant speed of $2.10\ \mathrm{m/s}$
Forces on the person: the weight downward and the normal force upward
Find
(a) What does the scale read?
Hint 1/4
The question mentions a speed and a direction. Decide first which of the two quantities in the second law those words actually fix, and whether either of them is the one the equation needs.
Hint 2/4
With up positive, $F_N - mg = ma_y$, and $a_y$ is the rate of change of the velocity, not the velocity.
Hint 3/4
Substitute the data from the question: constant speed of $2.10\ \mathrm{m/s}$ means $a_y = 0$, and with $m = 62.0\ \mathrm{kg}$ this gives $F_N = mg$.
Hint 4/4
So the scale reads $608\ \mathrm{N}$, the same as it would on the ground floor with the doors open.
Show solution
The velocity is written down and then deliberately set aside, because naming the irrelevant datum out loud is what stops it from being used.
constant means unchanging, and the acceleration is defined as the rate of that change
Apply the second law
$$F_N - mg = m(0) = 0$$
with a zero on the right, the two forces must balance exactly
$$F_N = mg = 62.0(9.80) = 608\ \mathrm{N}$$
the special case in which the habit $F_N = mg$ happens to be right, and it is right because of this line and not by default
Answer $$\boxed{F_N = 608\ \mathrm{N}}$$
Check
Check against the table above: rows one and five give 694 N, rows two and four give 521 N, and the cruising row gives 608 N regardless of direction. The answer sits in the row whose acceleration is zero, which is the row this problem describes.
The datum 2.10 m/s was there to be discarded. Exam questions carry spare numbers on purpose, and identifying which ones the equation cannot use is part of the reading.
⚠ Writing the normal force as the weight by default
the first dozen problems anyone meets are on level ground with nothing else acting, so the equality gets stored as a definition instead of as one line of arithmetic
⚠ Putting the acceleration into the diagram as a force arrow
it has a direction and a magnitude and it is what the problem is about, so it looks like it belongs with the other arrows
wrong$$\sum F = F_N - mg - ma = 0$$
right$$\sum F = F_N - mg = ma$$
⚠ Reading the direction of travel from the sign of the acceleration
both are directions attached to the same moving body, and in the commonest examples they happen to agree
wrong$$a_y < 0 \Rightarrow \text{the lift is going down}$$
right$$a_y < 0 \Rightarrow \vec{a}\ \text{is downward};\ \vec{v}\ \text{may be either way}$$
The same recipe on a car. Three forces act: the weight down, the road's push up, and the brakes' backward force. The vertical pair cancels, so the horizontal equation is the whole problem and the acceleration points backwards while the velocity still points forwards. This is the diagram behind the exam question later in the section.
5.4Two bodies, one string: the constraint that makes the problem solvable
Two connected bodies need two diagrams and two equations, joined by the fact that they share one acceleration.
One body needs one diagram. Two bodies tied together need two diagrams and one extra fact, and the extra fact is not a force at all.
RuleRule: separate diagrams, shared acceleration, shared tension
Conditions
the string does not stretch, so the two speeds and the two accelerations have equal magnitudes at every instant
the string is light and the pulley is light and turns freely, so the tension has the same magnitude everywhere along the string
each body gets its own equation with only the forces acting on that body in it, and the tension appears in both, once as a pull one way and once as a pull the other
the system shortcut in the last line holds only when every body of the system has the same acceleration magnitude and the internal forces cancel in pairs
Write one equation for each block, taking the direction each block actually moves as positive for that block. The tension is the same number in both equations and it points in opposite senses, so adding the two equations makes it vanish and leaves the acceleration in terms of the masses alone. Put that acceleration back into either equation and the tension falls out. The hanging weight drives the system and the total mass resists it, which is what the fraction says.
Why adding the two equations is legitimate
The block on the table has the tension pulling it horizontally and nothing else horizontal, so $F_T = m_1a$. Its vertical equation says only $F_N = m_1g$ and is never needed.
The hanging block has its weight pulling down and the tension pulling up, and it accelerates downward, so taking downward as positive for that block gives $m_2g - F_T = m_2a$. Note that positive means a different compass direction for each block, and that is allowed because they are two separate equations; what is not allowed is a different positive direction inside one equation.
Adding the two cancels the tension, because it appears once with a plus and once with a minus, and leaves $m_2g = (m_1+m_2)a$. That is the same statement as treating the pair as a single object of mass $m_1+m_2$ pulled by the external force $m_2g$, which is why the shortcut works and why it stops working the moment the two bodies stop sharing one acceleration.
Two blocks, one string, two diagrams drawn separately. The tension appears in both with the same size and opposite senses, which is why it cancels when the equations are added. The normal force appears only in the first and never affects the answer, and the hanging block's weight is the only force driving anything.
Looks like this, but is not
This is the constraint working: the string is taut and inextensible, so when the hanging block descends 10 cm the table block advances 10 cm, in the same time, and the two accelerations therefore have the same magnitude.
This looks like the same setup and the constraint fails: if the hanging block is given a shove upward, the string goes slack for a moment. During that moment the tension is zero, the hanging block is in free fall at $9.80\ \mathrm{m/s^{2}}$, and the table block is not accelerating at all. The two accelerations are now completely different, and every equation in the box is false. A string can pull and cannot push, and every connected body answer quietly assumes that it is pulling.
Acceleration and tension for a 3.00 kg block pulled by a 2.00 kg hanging block
A block on a smooth horizontal table is joined by a light string, running over a light free turning pulley at the edge of the table, to a second block that hangs beside the table. The system is released from rest. Find the acceleration and the tension.
Given
Block on the table: $m_1 = 3.00\ \mathrm{kg}$; the table is smooth
Hanging block: $m_2 = 2.00\ \mathrm{kg}$
Forces on $m_1$: tension $F_T$ horizontally toward the pulley, weight $m_1g$ down, normal force $F_N$ up
Forces on $m_2$: tension $F_T$ up, weight $m_2g$ down
The string does not stretch and the pulley is light and turns freely; $g = 9.80\ \mathrm{m/s^{2}}$
Find
The magnitude of the acceleration and the tension in the string.
Solution
Two separate diagrams are drawn rather than one diagram of the pair, because the tension is one of the things asked for and an internal force is invisible to a system level equation. The system shortcut is then used at the end as a check rather than as the method.
One equation for the block on the table
$$\text{taking toward the pulley as positive}:\quad F_T = m_1a = 3.00a$$
the tension is the only horizontal force; the weight and the normal force are vertical and cancel each other, so the vertical equation is never used
One equation for the hanging block
$$\text{taking downward as positive for this block}:\quad m_2g - F_T = m_2a$$
the block accelerates downward, so the larger force must be the weight and the tension enters with a minus
$$19.6 - F_T = 2.00a$$
with $m_2g = 2.00 \times 9.80 = 19.6\ \mathrm{N}$; note that the acceleration is the same symbol as in the first equation, which is the constraint doing its work
Add, solve, and substitute back
$$19.6 = 3.00a + 2.00a = 5.00a$$
adding the two equations removes the tension because it appeared once positive and once negative
Substitute into the equation that was not used for the final step: $m_2g - F_T = 19.6 - 11.8 = 7.8\ \mathrm{N}$, and $m_2a = 2.00(3.92) = 7.84\ \mathrm{N}$. They agree to the rounding. Two limits also check out: as $m_1$ tends to zero the acceleration tends to $g$, which is a block simply falling, and as $m_2$ tends to zero it tends to zero, which is nothing pulling.
Two diagrams, two equations, one addition. The vertical equation for the table block was written down and then never used, which is normal and worth noticing: a smooth horizontal surface makes that line decorative.
The tension came out at 11.8 N while the hanging block weighs 19.6 N. It must be smaller, because if the string pulled up with the full weight the hanging block would not accelerate at all. That inequality is a free check on every problem of this shape.
Choosing the hanging mass that gives an acceleration of 2.00 m/s squared
The same 3.00 kg block sits on the same smooth table with the same string and pulley, but the hanging block can be changed. Find the hanging mass that makes the system accelerate at 2.00 m/s squared, and the tension at that setting.
Given
Block on the table: $m_1 = 3.00\ \mathrm{kg}$, smooth surface
Same forces as before: tension on each block, weight on each block, normal force on the table block
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The hanging mass and the resulting tension.
Solution
The boxed general result is inverted rather than the two equations being solved again from scratch, because the unknown now sits in both the numerator and the denominator and the algebra is easier to do once, in symbols, than twice with numbers in the way.
Forward substitution into the original relation: $a = m_2g/(m_1+m_2) = 0.769(9.80)/3.769 = 7.54/3.769 = 2.00\ \mathrm{m/s^{2}}$, as required. The tension is also below the hanging weight of $0.769 \times 9.80 = 7.54\ \mathrm{N}$, as it must be for a block that is accelerating downward.
The formula $m_2 = am_1/(g-a)$ says something the numbers alone would not: as the required acceleration approaches $g$, the hanging mass runs away to infinity. A pulley system can never quite reach free fall, because there is always the table block's inertia to drag along.
Checkpoint
§05.4 — how the tension compares with the hanging weight●●○○○
Thirty seconds, no arithmetic. A 3.00 kg block on a smooth table is joined over a light pulley to a 2.00 kg block hanging beside the table, and the system is released from rest so that the hanging block descends.
Given
$m_1 = 3.00\ \mathrm{kg}$ on a smooth table, $m_2 = 2.00\ \mathrm{kg}$ hanging
Weight of the hanging block: $m_2g = 19.6\ \mathrm{N}$
Light inextensible string over a light free turning pulley
Find
(a) How does the tension in the string compare with 19.6 N?
Hint 1/4
Look only at the hanging block and ask what would have to be true of the two forces on it for it to speed up downward, rather than hover.
Hint 2/4
For the hanging block, with downward positive, $m_2g - F_T = m_2a$, and $a$ is positive because the block descends faster and faster.
Hint 3/4
Substitute the data from the question: $m_2g = 19.6\ \mathrm{N}$ and $m_2 = 2.00\ \mathrm{kg}$, so $F_T = 19.6 - 2.00a$ with $a > 0$.
Hint 4/4
So the tension is less than 19.6 N; with the numbers of this system it works out at 11.8 N.
Show solution
The argument is made on the hanging block alone, because that is the body on which both quantities being compared actually act; bringing in the table block would add algebra without adding information.
Write the descending block's equation
$$m_2g - F_T = m_2a,\qquad a>0$$
downward positive for this block, and the acceleration is positive because it descends ever faster from rest
Read the inequality off it
$$F_T = m_2(g-a) < m_2g$$
subtracting a positive quantity from $g$ can only reduce the product, so no arithmetic is needed for the comparison
Test the two extremes. With no block on the table, $a = g$ and the tension is zero, which is a block in free fall on a slack string. With an enormously heavy table block, $a \to 0$ and the tension rises to the full 19.6 N, which is a block hanging still. Every real case lies between those two, so the tension is always somewhere between zero and the weight.
This inequality is worth applying to every connected body answer you produce. A tension larger than the weight it is holding up, in a system that is accelerating downward, is arithmetically impossible and takes two seconds to spot.
⚠ Setting the tension equal to the hanging weight
the string visibly holds the block, and in every static problem so far it has held it with exactly its weight
5.5Four checks that catch a wrong answer before the marker does
Units, digits, size and limiting cases, applied in that order, each catching a different family of error.
The machine now runs in both directions. What it does not yet do is tell you when it has produced nonsense, and that is a separate skill with its own short list.
MethodMethod: the four checks, in order of how much they catch per second spent
Conditions
the checks are applied to the final answer and to any intermediate result that will be reused, not to every line
each check catches a different family of error, so passing one says nothing about the others
a limiting case is only a check if you already know what the limit should give, from a simpler problem you have solved
an order of magnitude comparison needs a landmark you actually believe, not a number invented on the spot
First ask whether both sides of your equation carry the same units, because that catches a wrong formula instantly. Then ask how many digits the least precise datum had, because that catches a wrong claim about precision. Then hold the number up against something you know the size of, because that catches a slipped factor or a slipped decimal point. Finally push one of the symbols to an extreme where you already know the answer, because that catches a structurally wrong formula that happens to give a plausible number.
Where answers live. The scale is logarithmic, so each step right multiplies by ten, and the landmarks are things you have felt: a train easing out, a lift starting, a car pulling away, a dropped stone. An answer landing where you expected is not proved right, but one that lands two decades from every landmark is certainly wrong.
Looks like this, but is not
This is a check doing its job: a student writes $a = 2x/t^{2}$ and tests the units, finding metres over seconds squared on both sides, so the formula survives and the arithmetic can be trusted to that extent.
This looks like a check and is not one: the same student writes $a = 2x/t$, gets metres per second, notices nothing because a speed and an acceleration both feel like rates, and moves on. A unit check works only if the units are written out and compared. It also has a blind spot no care removes: a factor of two, a wrong angle or a misplaced decimal point leaves the units untouched.
quantity
everyday landmark
value
acceleration
a lift starting or stopping
about $1\ \mathrm{m/s^{2}}$
acceleration
a car pulling away from lights
about $3\ \mathrm{m/s^{2}}$
acceleration
anything dropped
$9.80\ \mathrm{m/s^{2}}$
speed
a brisk walk
about $1.4\ \mathrm{m/s}$
speed
a car on a main road
$25\ \mathrm{m/s}$, which is $90\ \mathrm{km/h}$
force
the weight of a 1 kg bag
$9.8\ \mathrm{N}$
force
the weight of an adult
$600$ to $800\ \mathrm{N}$
force
the weight of a small car
about $1.2\times 10^{4}\ \mathrm{N}$
Eight numbers, and they are enough. Any acceleration you compute in this course should be recognisable against the first three: a tenth of free fall is gentle, a half is severe, ten times is a collision. Any force should be recognisable against the last three, and in particular a tension or a normal force on a body of a few kilograms belongs in the tens of newtons, not in the thousands and not in the tenths.
A tension of 19.6 N that cannot be right, and the check that says so
A solution to the two block problem reports an acceleration of 3.92 m/s squared and a tension of 19.6 N. Decide, without redoing the algebra, whether the pair can both be correct, and if not, repair it.
Given
System: $m_1 = 3.00\ \mathrm{kg}$ on a smooth table joined over a light pulley to $m_2 = 2.00\ \mathrm{kg}$ hanging
Reported: $a = 3.92\ \mathrm{m/s^{2}}$ and $F_T = 19.6\ \mathrm{N}$
Forces on the hanging block: tension up, weight $m_2g = 19.6\ \mathrm{N}$ down
$g = 9.80\ \mathrm{m/s^{2}}$
Find
Whether the reported pair is self consistent, and the correct tension if it is not.
Solution
The consistency test is done on the hanging block rather than on the table block, because both reported numbers appear in that one equation and a contradiction there settles the matter in a single line. Redoing the whole solution would also find the error, but it would take five times as long and could repeat the same slip.
Put both reported numbers into an equation that contains both
$$m_2g - F_T = m_2a$$
the hanging block's own equation, and the only one in the problem carrying both the tension and the acceleration
The bound check confirms the repair independently of the algebra: for a block accelerating downward the tension must lie strictly between zero and the block's weight, so it must be under 19.6 N, and 11.8 N is. The size check agrees too, since a few kilograms on a string should give a tension of tens of newtons.
One substitution and one comparison. No part of the original solution had to be re-derived, and the error was located as well as detected.
Notice which check did the work. Units would not have caught this, because 19.6 N is a perfectly good force; digits would not have caught it either. What caught it was substituting the answer back into an equation it had not been used to produce.
A stopping distance of 8.33 m, and the only check that catches it
A solution reports that a car travelling at 25.0 m/s, decelerating at 3.75 m/s squared, comes to rest in 8.33 m. Apply the four checks in order and say which of them, if any, rejects the answer.
Constant deceleration of magnitude $3.75\ \mathrm{m/s^{2}}$
Reported stopping distance $8.33\ \mathrm{m}$
Find
Which check rejects the reported answer, and what the distance should be.
Solution
The checks are applied in the stated order rather than the answer being recomputed first, because the point of the exercise is to see which of them has the power to catch this particular family of error. Recomputing would give the right number and teach nothing about the checks.
the decimal point had moved one place; the digits were right
$$a \to 2a \Rightarrow x \to x/2$$
limiting behaviour: doubling the braking force halves the distance, and the corrected formula does that while the reported number is not attached to any formula at all
Independent route to the same distance: the stopping time is $t = v_0/a = 25.0/3.75 = 6.67\ \mathrm{s}$ and the average speed over a constant deceleration is $12.5\ \mathrm{m/s}$, so the distance is $12.5 \times 6.67 = 83.3\ \mathrm{m}$. It also matches ordinary experience, since stopping from 90 km/h in the length of two buses would be a collision.
A slipped decimal point is invisible to two of the four checks and to any amount of rereading, because every digit on the page is correct. It is visible only to the check that leaves the page and compares the answer with the world.
Checkpoint
§05.5 — spotting the impossible answer without calculating●●○○○
Thirty seconds and no calculator. Four results are reported for the 2.00 kg block on the smooth 30.0 degree ramp of this section, released from rest. Exactly one of them can be rejected immediately, on a bound rather than on a calculation.
Given
$m = 2.00\ \mathrm{kg}$, so the weight is $mg = 19.6\ \mathrm{N}$
Smooth ramp at $\theta = 30.0^{\circ}$, released from rest
Forces: the weight and the normal force only
Find
(a) Which reported result is impossible?
Hint 1/4
Do not compute anything. Ask which of the four quantities has a ceiling or a floor that you can state in words before you know its value.
Hint 2/4
On a slope with nothing pressing the block down, the surface supports only the perpendicular part of the weight, so $F_N = mg\cos\theta \le mg$ for every angle.
Hint 3/4
Substitute the data from the question: $mg = 2.00 \times 9.80 = 19.6\ \mathrm{N}$, so any normal force above 19.6 N is out of bounds whatever the angle is.
Hint 4/4
So the reported $F_N = 21.0\ \mathrm{N}$ is the impossible one; the true value is $19.6\cos 30.0^{\circ} = 17.0\ \mathrm{N}$.
Show solution
The bound is derived from the cosine's range rather than from the numbers, so that it applies to every ramp problem and not only to this one.
Test the bound at both ends: at $\theta = 0$ it gives $F_N = 19.6\ \mathrm{N}$, a block on a level floor, and at $\theta = 90^{\circ}$ it gives zero, a block beside a vertical wall it is not touching. Both are known answers, so the bound is trustworthy.
The bound would change if something else pressed on the block, for instance a hand pushing it into the ramp, and then a normal force above the weight would be perfectly possible. Bounds are only as good as the list of forces they were derived from, so state the list.
⚠ Rounding at every intermediate line
each line looks finished when it is written, and carrying an extra digit feels like false precision rather than like protection
right$$x = 83.3\ \mathrm{m}\ \text{(units correct AND size plausible)}$$
5.6Reading an unseen problem and naming its type before writing anything
Three questions asked of the wording decide the route, and asking them takes less time than one wrong start.
Everything above assumed you already knew which end of the machine you were standing at. On an unseen paper you do not, and finding out is its own step.
MethodMethod: three questions that classify any problem from these four weeks
Conditions
the questions are asked of the wording, before any diagram and before any symbol is defined
the third question decides whether the constant acceleration formulas may be used at all, so it cannot be skipped
a problem with several parts may need the questions asked again for each part, since one part can run forwards and the next backwards
a problem that gives a quantity neither asked about nor needed is normal, and identifying the spare datum is part of the reading
$$\boxed{\;\text{(1) which end is given?}\;\longrightarrow\;\text{(2) how many bodies?}\;\longrightarrow\;\text{(3) is } a \text{ constant?}\;}$$
First, does the problem hand you the forces and ask about the motion, or hand you the motion and ask about a force? That fixes the direction of travel through the machine. Second, is there one body or are there several joined together, because several bodies means several diagrams and a constraint. Third, does anything change during the motion, because if a force switches on or off part way through, the interval has to be cut and each piece treated on its own.
The whole of the first four weeks as one decision. The two branches are the same journey walked in opposite directions, which is why the boxes at the foot of each say the same thing with the sources swapped, and both meet at the joint.
Looks like this, but is not
This is triage working: a crate is pulled by a rope at a known tension and the question asks how fast it is going after some time. Forces given, motion wanted, one body, nothing changing, so the route is diagram, components, second law, kinematics.
This looks like the same problem and is not: the rope is released once the crate reaches a certain speed, and the question asks where the crate is two seconds later. The third triage question now answers differently, because the tension switches off part way through: two intervals, two accelerations, and the end of the first is the start of the second. One application of the kit here is wrong with no warning.
signature in the question
what it is
first move
the usual trap
released from rest on a smooth slope
forces to motion, one body
resolve the weight along and across the slope
using $\cos$ where $\sin$ belongs
a scale reads, or a person feels heavier
forces to motion, one body
write $F_N - mg = ma$ with up positive
assuming the reading gives the direction of travel
joined by a string over a pulley
forces to motion, two bodies
one diagram per body, then add the equations
setting the tension equal to the hanging weight
comes to rest in a distance
motion to forces, one body
get $a$ from the time free formula
reaching for the second law before $a$ exists
pulled by a rope at an angle
forces to motion, one body
resolve the tension, then two axis equations
leaving the vertical component out of the normal force
speeds up, then coasts, then stops
two or three intervals
cut the motion where the forces change
using one equation across the whole time
a reading in km/h, or a mass in grams
a measurement question wearing a disguise
convert before anything else
carrying the wrong unit into the second law
The middle column decides everything and has only three values in this whole part of the course: forces to motion, motion to forces, or several intervals handled one at a time. Everything else is detail hanging off that decision, and placing a problem in the right row within thirty seconds is what keeps a long question from eating the clock.
A crate pulled by a rope at 25.0 degrees: forces given, motion wanted
A crate is dragged across a smooth level floor by a rope held at an angle above the horizontal. Find the acceleration and the normal force from the floor.
Given
Mass of the crate $m = 18.0\ \mathrm{kg}$, so its weight is $mg = 176\ \mathrm{N}$
Rope tension $F_T = 62.0\ \mathrm{N}$ at $25.0^{\circ}$ above the horizontal
The floor is smooth and level
Forces on the crate: the tension along the rope, the weight down, the normal force up
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The acceleration along the floor, and the normal force.
Solution
Horizontal and vertical axes are kept rather than turning them along the rope, because the acceleration is along the floor and not along the rope. Turning the axes to follow the rope would put a component of the acceleration on both axes and double the work.
Two independent checks. The bound: the normal force must lie between zero and the weight here, and 150 N sits below 176 N as the upward pull requires. The limit: if the rope were horizontal the vertical component would vanish and $F_N$ would rise to the full 176 N, while if the rope were vertical and strong enough the crate would lift off and $F_N$ would fall to zero. The answer moves the right way between those.
Two components, two equations, two divisions. The triage step at the top cost about ten seconds and fixed the axes, which is where the time is actually saved.
The angle costs twice. It reduces the useful horizontal pull to 56.2 N out of 62.0 N, and it also lifts, which on a smooth floor is simply wasted. On a rough floor the lift would help, which is why the best angle to pull a crate at is not zero, and that is next week's question.
The same crate timed instead: motion given, tension wanted
The same crate on the same smooth floor is pulled by the same rope at the same angle, but the tension is not known. Instead it is timed: from rest it reaches 4.00 m/s in 2.50 s. Find the tension and the normal force.
Given
Mass $m = 18.0\ \mathrm{kg}$, weight $mg = 176\ \mathrm{N}$
Rope at $25.0^{\circ}$ above the horizontal, tension unknown
From rest to $4.00\ \mathrm{m/s}$ in $2.50\ \mathrm{s}$, uniformly
Forces on the crate: tension along the rope, weight down, normal force up; the floor is smooth
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The tension in the rope and the normal force from the floor.
Solution
The acceleration is extracted from the motion data first, before any force equation is written, because with the tension unknown the horizontal equation has two unknowns in it and cannot be started. This is the reversal that triage is meant to detect.
Triage, and notice the reversal
$$\text{motion given}\ \to\ \text{force wanted}$$
the timing is the data and the tension is the unknown, so the machine runs backwards and the kinematics comes first
Run the answer forwards through the previous example's route: $31.8\cos 25.0^{\circ} = 28.8\ \mathrm{N}$ and $28.8/18.0 = 1.60\ \mathrm{m/s^{2}}$, which reproduces the acceleration the timing gave. Comparison also makes sense: the tension is about half the 62.0 N of the previous example and the acceleration is about half of 3.12, as a proportional relationship requires.
Same picture, same diagram, same two equations, and the only thing that changed was which symbol was unknown. Getting that decision right at the start is worth more marks than any algebra in the solution, because the wrong start leads to two unknowns in one equation and a blank three minutes.
Checkpoint
§05.6 — naming the first move on an unseen problem●●○○○
Thirty seconds, and no answer is wanted, only the first move. A car of mass 1500 kg travelling at 20.0 m/s is brought to rest in 45.0 m by a constant backward force from its brakes, on a level road.
Stopping distance $45.0\ \mathrm{m}$, level road, constant backward force
The question asks for the size of the braking force
Find
(a) What should be computed first?
Hint 1/4
Ask the first triage question of the wording: which end of the machine has been handed to you, the forces or the motion?
Hint 2/4
When the motion is given and a force is wanted, the acceleration is extracted from the motion first, and only then does $\sum F = ma$ have a single unknown in it.
Hint 3/4
Substitute the data from the question: with $v_0 = 20.0\ \mathrm{m/s}$, $v = 0$ and $x = 45.0\ \mathrm{m}$ and no time mentioned, the time free formula gives $a$ directly.
Hint 4/4
So compute the acceleration first: $a = -v_0^{2}/2x = -400/90.0 = -4.44\ \mathrm{m/s^{2}}$, and the force follows as $6.67\times 10^{3}\ \mathrm{N}$.
Show solution
The time free formula is chosen over finding the time first because no time is mentioned anywhere in the problem, which is the standard signal for that row of the selection table.
Motion first, since that is the end that was given
$$0 = v_0^{2} + 2a(x-x_0)$$
the final speed is zero and the time is absent from the problem, so this is the formula with nothing spare in it
Size check: $4.44/9.80 = 0.45$, so the car decelerates at about half of free fall, which is firm braking and well within what tyres can do. Force check: 6.67 kN against a car weight of about 14.7 kN is a ratio of 0.45 as well, which it must be, since both ratios are the same number written twice.
The whole difficulty was one decision made in ten seconds. Once the acceleration was in hand the rest was a single multiplication, and a student who started at the second law would have spent those ten seconds and several more staring at an equation with two unknowns.
⚠ Writing the second law before checking which end was given
the second law is the headline result of the four weeks, so it feels like the right place to start regardless of what the problem supplies
right$$a = -\frac{v_0^{2}}{2(x-x_0)}\ \text{first, then}\ F = ma$$
⚠ Assuming the mass always cancels
it cancelled on the ramp and in free fall, which are the two most memorable examples, and the cancellation is satisfying enough to be remembered as a rule
The rope at an angle, the archetype behind several questions here. Only the horizontal component accelerates the crate; the vertical component quietly cuts the normal force from 176 N to 150 N. Both come from one arrow, and forgetting the second is what turns a right acceleration into a wrong normal force.
From the words on the page to a number you can defend
Every problem in this part of the course, including the ones that look too short to need a procedure.
Read for the direction of travel
Decide in one sentence whether the forces are given and the motion is wanted, or the other way round. Write that sentence down; it is worth a mark of your own time and it prevents the commonest waste, which is starting an equation that has two unknowns in it.
Convert every unit before anything else
Kilometres per hour, grams, centimetres and minutes all get converted now, in one place, and the converted values are the only ones used afterwards. A conversion done halfway through a solution is a conversion that will be done twice or not at all.
Draw the situation, then draw the body
One quick sketch of what is happening, then a separate free body diagram for each body you will write an equation for. The free body diagram carries only forces, has no velocities and no accelerations in it, and includes nothing that acts on some other object.
Declare axes, origin, positive direction and time zero
In writing, before any equation. On a slope, turn the axes to follow it. If there are two bodies, each may have its own positive direction, and each declaration governs only its own equation.
Write one equation per axis per body
Resolve every arrow, then write the sum of the components along that axis equal to the mass times the acceleration component along that axis. Add the if two bodies are joined. Count unknowns against equations before solving; if they do not match, a force has been missed or a constraint has not been written.
Solve in symbols, substitute at the end
Rearranging with letters is faster, produces a formula you can reuse, and lets you see cancellations such as the mass disappearing from a ramp problem. Carry one guard digit through the substitution.
Run the four checks and box the answer with its unit
Units, significant figures, size against a landmark, and one limiting case. Then write the answer with its unit and, where it is a vector, its direction in words.
Where it goes wrong
Step three skipped because the situation looks simple. Every normal force error in this section comes from a diagram that was imagined rather than drawn.
Step four skipped because the direction seems obvious. It is obvious until the third line, where a minus sign has to be decided and there is nothing on the page to decide it against.
Step five started before the unknowns have been counted, which is how a solution reaches an equation with two unknowns and stalls.
Step seven skipped for lack of time, which is a false economy: the four checks take under a minute between them and they are the only part of the process that can find an error you did not know you had made.
Two bodies on one string: what to write, and in what order
Whenever two or more objects are joined so that they must move together, whether by a string, a rod or simple contact.
Decide which way the system will actually move
Guess if you have to; the algebra will correct you with a minus sign. What matters is that the guess is written down, because the positive direction for each body is then chosen to follow the motion, which keeps every acceleration positive.
Draw a separate diagram for each body
The tension appears in both diagrams, with the same magnitude and pointing away from each body along the string. Nothing else crosses between the diagrams.
Write the constraint before the equations
For an inextensible string over a fixed pulley the constraint is that the two accelerations have equal magnitude. Writing it first means the two equations can share one symbol $a$ from the start, instead of carrying $a_1$ and $a_2$ that must be reconciled later.
Add the equations to eliminate the tension
The tension enters one equation positively and the other negatively, so adding them removes it and leaves the acceleration alone. This is the shortcut that a system level equation would have given, and doing it this way keeps the tension available.
Substitute back for the tension, and check with the unused equation
Put the acceleration into whichever equation has fewest terms. Then substitute both results into the other equation as an independent check; a connected body problem always leaves you one equation to check with.
Where it goes wrong
Both tension arrows drawn on one diagram of the whole system, where they cancel and the tension becomes invisible.
The same compass direction taken as positive for both bodies, which makes one of the two accelerations negative and turns a sign error into a plausible looking wrong answer.
The constraint forgotten, leaving two equations in three unknowns.
The tension assumed equal to the hanging weight, which is only true when the system is not accelerating.
A ball at the top of its flight: zero velocity, full weight
A ball is thrown straight up and is examined at the single instant when it is momentarily at rest at the highest point. Find its acceleration and the net force on it at that instant, and how high it went.
Air resistance neglected, $g = 9.80\ \mathrm{m/s^{2}}$
At the instant considered, the ball is in free flight and nothing is touching it
Find
The acceleration and net force at the top, and the maximum height above the launch point.
Solution
The net force is written down from the free body diagram rather than deduced from the velocity, because the velocity at that instant is exactly the piece of information that cannot decide it.
List the forces at the top
$$\sum \vec{F} = m\vec{g}\ \text{only}$$
nothing is touching the ball, so the weight is the entire free body diagram at every instant of the flight, including this one
$$\sum F = 0.150(9.80) = 1.47\ \mathrm{N}\ \text{downward}$$
the full weight, unchanged from the moment of release
the time free formula with the final speed set to zero, which is what defines the highest point
Answer $$\boxed{a = 9.80\ \mathrm{m/s^{2}}\ \text{down},\quad \sum F = 1.47\ \mathrm{N}\ \text{down},\quad h = 3.27\ \mathrm{m}}$$
Check
If the acceleration really were zero at the top, the ball would stay there, which it does not. The height is also plausible: 3.27 m is about a storey, which is what an 8 m/s throw should manage.
The velocity is zero for one instant and the acceleration is 9.80 m/s squared throughout. They are different quantities and there is no instant at which one implies anything about the other.
A book resting on a table: zero velocity, zero net force
A book lies still on a horizontal table. Find its acceleration, the net force on it, and the force the table exerts on it.
Given
Mass $m = 1.20\ \mathrm{kg}$
The book is at rest on a level table and stays at rest
Forces on the book: its weight downward and the normal force from the table upward
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The acceleration, the net force, and the normal force from the table.
Solution
The net force is deduced from the motion here rather than from the diagram, because the motion is fully known and one of the two forces is not; this is the reverse of the ball, where the diagram was complete and the motion was in question.
Get the acceleration from the observed motion
$$v = 0\ \text{for all } t \;\Rightarrow\; a = 0$$
the velocity is not merely zero at an instant, it is zero over an interval, and that is what makes its rate of change zero too
Turn that into the forces
$$\sum F = ma = 0$$
the second law read in the direction from motion to force
Remove the table and the net force becomes the weight alone, 11.8 N downward, and the book falls at 9.80 m/s squared. That the answer changes when a force is removed is the sign that the force was doing something, which a zero net force can easily disguise.
Zero net force is not the absence of forces. Two forces of 11.8 N each are acting, and the table is being compressed by every one of them.
Both bodies have zero velocity at the moment described, and their net forces are 1.47 N and exactly zero: one is accelerating downward as hard as anything can near the ground, and the other is not accelerating at all.
How to tell them apart
Ask what is touching the body. If nothing is, the weight is the whole diagram and the acceleration is $g$ downward whatever the velocity does. If something is touching it and the velocity stays zero over an interval rather than at one instant, the acceleration is zero and the contact force is whatever makes the sum vanish. The dangerous phrase is at rest, which fits both cases and separates neither.
Scaffolding comes off
The common skeleton
Decide which end is given: the forces or the motion. Write the sentence down.
Draw a separate free body diagram for each body, carrying only the forces on that body.
Declare the axes and the positive direction for each body, turning the axes along the slope where there is one.
Write one equation per axis per body, and the constraint if two bodies are joined.
Solve in symbols, then substitute, keeping one guard digit.
Check the units, the digits, the size against a landmark, and one limiting case.
1 · fully worked
Fully worked: a 4.00 kg block on a table pulled by a 3.00 kg hanging block
A block on a smooth horizontal table is joined by a light inextensible string over a light free turning pulley to a block hanging beside the table. The system is released from rest. Find the acceleration and the tension, and how fast the hanging block is moving after it has descended 0.500 m.
Forces on $m_1$: tension toward the pulley, weight down, normal force up. Forces on $m_2$: tension up, weight down
Light inextensible string, light free turning pulley, released from rest
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The acceleration, the tension, and the speed after a 0.500 m descent.
Solution
The two body equations are used rather than the system shortcut, because the tension is asked for and an internal force cannot be extracted from a system equation. The last part is then a pure kinematics step, which is the joint being crossed.
Which end is given, and how many bodies
$$\text{forces given} \to \text{motion wanted},\quad \text{two bodies, one constraint}$$
the masses fix the forces and the questions are about acceleration, tension and speed, so the machine runs forwards and a constraint will be needed
One equation per body, with the constraint built in
$$m_1:\quad F_T = m_1a = 4.00a$$
the tension is the only horizontal force on the table block, and its positive direction is toward the pulley
Check with the equation not used for the final substitution: $m_2g - F_T = 29.4 - 16.8 = 12.6\ \mathrm{N}$ and $m_2a = 3.00(4.20) = 12.6\ \mathrm{N}$, which agree. The tension is below the hanging weight of 29.4 N as it must be, and $a = 4.20\ \mathrm{m/s^{2}}$ is below $g$, which it must also be, since the table block is being dragged along.
Two diagrams, three equations, one square root. The only place a sign could go wrong is in choosing a different positive direction for each body, and that was fixed in writing before any equation.
The three answers came from the three stations in order: forces, then acceleration, then motion. If a fourth part had asked for the time instead of the speed, nothing above the last subgoal would have changed.
2 · you write the reasoning
Easier than the one above, because there is one body, one axis and no constraint. A 55.0 kg person stands on a bathroom scale in a lift which accelerates upward at 1.20 m/s squared. Find the scale reading. The steps are given and the reasons are not; write the reason column yourself, then open each one and compare.
Two forces act on the person and only one axis is needed, because the person moves only vertically. Up is declared positive, which makes the normal force enter with a plus sign and the weight with a minus, and that declaration is what makes every sign in the following lines automatic rather than decided.
$$F_N = m(g + a_y)$$
reasoning
The unknown is isolated before any number is substituted. Solving in symbols means the same line will serve for an upward acceleration, a downward one and none at all, with only the sign of $a_y$ changing, and it makes the structure visible: the reading is the mass times an effective gravity.
$$F_N = 55.0(9.80 + 1.20) = 55.0(11.00)$$
reasoning
The acceleration is entered as $+1.20$ because it points upward and up was declared positive. This is the only place in the whole solution where the direction of the acceleration matters, and note that nothing was said about which way the lift is travelling, because that does not enter.
$$F_N = 605\ \mathrm{N}$$
reasoning
The weight is 539 N and the reading is 605 N, so the person feels about 12 per cent heavier, which matches $1.20/9.80 = 12$ per cent as it must. The answer carries three significant figures because both data do, and it is comfortably in the hundreds of newtons where an adult's weight belongs.
3 · find the buried error
Harder than the rung above: the axes have to be turned and a second force acts along the slope. A 5.00 kg block sits on a smooth ramp at 20.0 degrees, pulled up the slope by a rope parallel to it. The forces are the tension of 25.0 N up the slope, the weight of 49.0 N vertically down, and the normal force perpendicular to the surface. The work below was done badly: exactly two of the four steps are wrong. Find both, and say what each should have been.
Step 1. Take $x$ up the slope and $y$ perpendicular to the surface. The weight is $mg = 5.00(9.80) = 49.0\ \mathrm{N}$, directed vertically down.
Step 2. Along the slope, $\sum F_x = 25.0 - 49.0\cos 20.0^{\circ} = 25.0 - 46.0 = -21.0\ \mathrm{N}$, so $a = -21.0/5.00 = -4.20\ \mathrm{m/s^{2}}$, that is, down the slope.
Step 3. Perpendicular to the surface there is no acceleration, so $F_N = mg = 49.0\ \mathrm{N}$.
Step 4. From rest, after 2.00 s: $v = at = (-4.20)(2.00) = -8.40\ \mathrm{m/s}$, that is, 8.40 m/s down the slope.
the two buried errors (2)
⚠ step 2
The along slope component of the weight is $mg\sin\theta$, not $mg\cos\theta$. It should be $49.0\sin 20.0^{\circ} = 16.8\ \mathrm{N}$, so the net along slope force is $25.0 - 16.8 = 8.2\ \mathrm{N}$ and the acceleration is $+1.65\ \mathrm{m/s^{2}}$, up the slope. The sign of the answer, not only its size, was wrong.
On a ramp drawing the slope angle sits at the corner where the adjacent side is the base, so the picture suggests the cosine while the geometry of the weight requires the sine. It is also self concealing: the wrong answer stays a plausible acceleration in a plausible direction.
Perpendicular to the surface it is the perpendicular component of the weight that has to be balanced, not the whole weight. The correct line is $F_N = mg\cos\theta = 49.0(0.9397) = 46.0\ \mathrm{N}$. The rope is parallel to the slope, so it contributes nothing here, but the weight is not perpendicular to the surface and only part of it presses in.
On every level surface met before a ramp appears the two really are equal, so the equality gets stored as a property. The step also looks careful, because it states correctly that the perpendicular acceleration is zero before drawing the wrong conclusion from it.
right
$\sum F_y = F_N - mg\cos\theta = 0$, hence $F_N = 46.0\ \mathrm{N}$, which is below the weight as it must be on any slope.
4 · the bare problem
§05.6 — a crate dragged by a rope above the horizontal●●●○○
No scaffolding this time. A crate is dragged across a smooth level floor by a rope held above the horizontal, and both the acceleration and the force from the floor are wanted.
Given
Mass of the crate $m = 24.0\ \mathrm{kg}$, so its weight is $mg = 235\ \mathrm{N}$
Rope tension $F_T = 85.0\ \mathrm{N}$ at $32.0^{\circ}$ above the horizontal
Smooth level floor; the forces on the crate are the tension, the weight and the normal force
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the acceleration of the crate.
(b) Find the normal force the floor exerts on the crate.
(c) State one check that would reject an answer of 250 N for part (b).
Hint 1/4
The rope points one way and the crate moves another, so the tension has to be split before either equation can be written. Decide which component belongs to which axis before computing anything.
Hint 2/4
With the angle measured from the horizontal, $F_{Tx} = F_T\cos\theta$ and $F_{Ty} = F_T\sin\theta$; then $\sum F_x = ma_x$ along the floor and $\sum F_y = 0$ perpendicular to it, since the crate stays on the floor.
Hint 3/4
Substitute the data from the question, $F_T = 85.0\ \mathrm{N}$, $\theta = 32.0^{\circ}$, $m = 24.0\ \mathrm{kg}$: $\cos 32.0^{\circ} = 0.8480$ and $\sin 32.0^{\circ} = 0.5299$, so the horizontal pull is $72.1\ \mathrm{N}$ and the lift is $45.0\ \mathrm{N}$.
Hint 4/4
So $a = 72.1/24.0 = 3.00\ \mathrm{m/s^{2}}$ and $F_N = 235 - 45.0 = 190\ \mathrm{N}$.
Show solution
Horizontal and vertical axes are kept, because the acceleration is horizontal; aligning an axis with the rope would spread the acceleration over both axes and add an unknown to each equation.
Bound check: with an upward rope component and nothing pushing down, the normal force must lie between zero and the weight, and 190 N lies between 0 and 235 N. Limit check: at a rope angle of zero the lift vanishes and $F_N$ would rise to 235 N while the acceleration would rise to $85.0/24.0 = 3.54\ \mathrm{m/s^{2}}$, so tilting the rope costs acceleration and buys lift, which is the behaviour the formulas show.
Part (c) is the habit worth taking away. A bound stated before the arithmetic makes a whole class of wrong answers rejectable in a second, and on a ramp or under a rope the bound on the normal force is the most useful one in this part of the course.
Full exam-style question
Full exam style: a car braking from a radar reading, in five partsexam format
A car is recorded by a roadside radar and then brakes to a stop on a level road. The question chains a measurement step, a kinematics step and a dynamics step, which is what a long question on a paper is made of. Work through the parts in order; each one uses the result of the last.
Given
Mass of the car $m = 1.20\times 10^{3}\ \mathrm{kg}$
Speed recorded by the radar: $90.0\ \mathrm{km/h}$, constant until the driver reacts
Driver's reaction time $0.750\ \mathrm{s}$, during which the car does not slow at all
Constant backward force from the brakes once they are applied: $4.50\times 10^{3}\ \mathrm{N}$
Forces on the car while braking: the weight down, the road's normal force up, the braking force backward; air resistance neglected, road level
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) the speed in SI units, (b) the distance covered during the reaction time, (c) the acceleration while braking, (d) the braking distance, (e) the total stopping distance, and (f) the normal force from the road.
Solution
The motion is cut into two intervals at the instant the brakes come on, because the forces are different in the two intervals and the constant acceleration formulas cannot span them. Within each interval the formula is chosen by the variable nobody mentions: no time is asked for in the braking phase, so the time free formula is used there.
Size: $3.75/9.80 = 0.38$, firm braking rather than an emergency stop. Landmark: about 100 m from 90 km/h is what road safety tables quote, so the answer agrees with something measured off this page. Alternative route to (d): the braking time is $6.67\ \mathrm{s}$ and the average speed $12.5\ \mathrm{m/s}$, giving the same $83.3\ \mathrm{m}$.
Six parts, one conversion, one interval split, and one crossing of the joint in each direction. Part (f) could have been skipped without changing anything else, and it is there because it is the one line where the normal force really does equal the weight.
The reaction distance is 18 per cent of the total and scales with the speed, while the braking distance scales with its square. At twice the speed one doubles and the other quadruples, so the total at 180 km/h is not twice 102 m but nearer four times it.
Practice
A · concept 3 questions
1§05.1 — zero net force and the state of motion●●○○○
A statement about the first law, of the kind that appears as a two mark opener. Decide whether it is true, and be ready to justify it in one sentence, because the justification is where the marks are.
Given
The claim: if the net force on an object is zero, then the object must be at rest.
Find
(a) True or false, and why?
Hint 1/4
The first law is a statement about a change, not about a value. Ask what quantity it says stays the same, and then ask whether that quantity has to be zero.
Hint 2/4
With $\sum \vec{F} = 0$ the second law gives $\vec{a} = 0$, which means the velocity is constant; constant includes non zero constant.
Hint 3/4
Substitute a counterexample: a puck sliding on smooth ice at $3.00\ \mathrm{m/s}$ has zero net force and is not at rest, which is enough to settle the claim.
Hint 4/4
So the statement is false; zero net force guarantees only that the velocity does not change.
Show solution
A counterexample is used rather than a general argument, because one clear case that contradicts the claim settles it and a general argument would only restate the first law.
Test the converse too, since it is the version that is true in one direction only: a body at rest on a table does have zero net force, so rest implies balance for a body that stays at rest, while balance does not imply rest. Only one of the two implications holds.
The phrase at rest belongs to a chosen frame of reference as well. A passenger sitting still in a moving train is at rest for one observer and moving for another, and neither observer sees a net force on her.
2§05.3 — whether the normal force always equals the weight●●○○○
A claim about the normal force, phrased the way it usually gets remembered rather than the way it is true. Decide whether it holds in general.
Given
The claim: the normal force on a body always has the same magnitude as the body's weight.
Find
(a) True or false, and give one situation that settles it.
Hint 1/4
The normal force is never quoted from a formula; it is solved for. So ask what equation it comes out of, and what else lives in that equation.
Hint 2/4
Perpendicular to the surface, $\sum F_\perp = ma_\perp$, and the weight is only one of the terms on the left; a slope, a rope, or an accelerating floor changes the rest.
Hint 3/4
Substitute a case from this section: a $2.00\ \mathrm{kg}$ block on a $30.0^{\circ}$ ramp has weight $19.6\ \mathrm{N}$ and normal force $19.6\cos 30.0^{\circ} = 17.0\ \mathrm{N}$.
Hint 4/4
So the claim is false, and it fails on a ramp, in an accelerating lift, and under a rope pulled at an angle.
Show solution
Three counterexamples are given rather than one, because they fail the claim for three different reasons: a tilted surface, a vertical acceleration, and an extra vertical force.
a rope pulling upward takes over part of the support, so the surface provides less
Answer $$\boxed{\text{False}:\ F_N = mg\ \text{only on a level surface with } a_y=0\ \text{and no other vertical force}}$$
Check
The three counterexamples move the normal force in both directions, one below the weight, one above it and one below it again for a different reason, which shows that the equality is not even an upper or a lower bound in general.
Whenever a problem asks for a normal force, expect it not to be the weight. The cases where it is equal are the ones that need justifying with a line of working, not the other way round.
3§05.1 — identifying the partner of an action and reaction pair●●●○○
A book lies at rest on a level table. The Earth pulls the book downward with a gravitational force. The third law says this force has a partner, equal in size and opposite in direction, and the whole difficulty is naming which force that is.
Given
A book of mass $1.20\ \mathrm{kg}$ resting on a level table
The force in question: the Earth's gravitational pull on the book, $11.8\ \mathrm{N}$ downward
The book is not accelerating
Find
(a) Which force is the third law partner of the Earth's pull on the book?
Hint 1/4
Do not look for the force that balances this one. Look for the force that involves the same two bodies with the roles swapped, which is a different question and has a different answer.
Hint 2/4
If A exerts $\vec{F}$ on B, then B exerts $-\vec{F}$ on A. The two bodies in the named force are the Earth and the book, so the partner must also involve exactly the Earth and the book.
Hint 3/4
Substitute the bodies from the question: the named force is the Earth pulling the book with $11.8\ \mathrm{N}$ downward, so the partner is the book pulling the Earth with $11.8\ \mathrm{N}$ upward.
Hint 4/4
So the partner is the book's gravitational pull on the Earth, a force that acts on the Earth and therefore never appears in the book's free body diagram.
Show solution
The two bodies are identified first and the direction second, because the commonest error here is to pick a force with the right direction and the wrong bodies.
Name the two bodies in the given force
$$\vec{F}_{\text{Earth on book}} = 11.8\ \mathrm{N}\ \text{down}$$
the subscript makes both bodies explicit, which is the only reliable way to keep third law pairs straight
Swap the roles
$$\vec{F}_{\text{book on Earth}} = -\vec{F}_{\text{Earth on book}} = 11.8\ \mathrm{N}\ \text{up}$$
same two bodies, same interaction, opposite direction, acting on the Earth
Distinguish it from the balancing force
$$\vec{F}_{\text{table on book}} = 11.8\ \mathrm{N}\ \text{up},\ \text{acting on the book}$$
equal in size here only because the book is not accelerating; it is a different pair, and its own partner is the book pressing down on the table
Answer $$\boxed{\vec{F}_{\text{book on Earth}} = 11.8\ \mathrm{N}\ \text{upward, acting on the Earth}}$$
Check
Test the two candidates by imagining the table removed. The book falls, the Earth's pull and the book's pull on the Earth are unchanged and still equal, while the normal force has vanished. Only the genuine partner survives, which is what makes the test decisive.
The habit worth keeping is to write every force with two subscripts. Once forces are named as A on B, third law pairs identify themselves and never end up in the same free body diagram.
B · computation 5 questions
1§05.1 — a block released on a smooth 18.0 degree ramp●●○○○
A block is released from rest at the top of a smooth ramp and slides down. Everything asked for follows from one acceleration, which has to be produced from the forces first.
Given
Mass $m = 4.50\ \mathrm{kg}$
Ramp angle $\theta = 18.0^{\circ}$ above the horizontal, surface smooth
Released from rest at $t = 0$
Forces on the block: the weight $mg = 44.1\ \mathrm{N}$ down, and the normal force perpendicular to the surface
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the acceleration along the slope.
(b) Find the normal force from the ramp.
(c) Find how far the block has slid, and how fast it is moving, 2.20 s after release.
Hint 1/4
Parts (a) and (b) are a force problem and part (c) is a motion problem. Do not start part (c) until part (a) has produced a number.
Hint 2/4
With the axes turned along the slope, $mg\sin\theta = ma$ and $F_N = mg\cos\theta$, and then $x = \tfrac{1}{2}at^{2}$ and $v = at$ from rest.
Hint 3/4
Substitute the data from the question, $m = 4.50\ \mathrm{kg}$, $\theta = 18.0^{\circ}$, $t = 2.20\ \mathrm{s}$: $\sin 18.0^{\circ} = 0.3090$ and $\cos 18.0^{\circ} = 0.9511$.
Hint 4/4
So $a = 3.03\ \mathrm{m/s^{2}}$, $F_N = 41.9\ \mathrm{N}$, and after 2.20 s the block has slid $7.33\ \mathrm{m}$ at $6.66\ \mathrm{m/s}$.
Show solution
The axes are turned along the slope so that the acceleration has one component instead of two; with horizontal axes both equations would contain the unknown normal force and would have to be solved together.
Force half
$$mg\sin\theta = ma \;\Rightarrow\; a = 9.80\sin 18.0^{\circ} = 9.80(0.3090) = 3.03\ \mathrm{m/s^{2}}$$
the normal force has no component along the slope, so the weight component is the whole net force there
from rest, so the initial velocity term is absent; a guard digit is kept in the acceleration
$$v = at = (3.0284)(2.20) = 6.66\ \mathrm{m/s}$$
the same acceleration and the same clock
Answer $$\boxed{a = 3.03\ \mathrm{m/s^{2}},\ F_N = 41.9\ \mathrm{N},\ x = 7.33\ \mathrm{m},\ v = 6.66\ \mathrm{m/s}}$$
Check
Independent route to the speed: $v^{2} = 2ax = 2(3.0284)(7.33) = 44.4$, giving $v = 6.66\ \mathrm{m/s}$, from a formula that was not used to produce it. The normal force also passes its bound, since $41.9 < 44.1\ \mathrm{N}$.
Compare with the 30 degree ramp earlier in the section: 18 degrees gives 3.03 m/s squared against 4.90, and the ratio is the ratio of the sines. The mass never appeared in either acceleration.
2§05.2 — a stone thrown downward from a bridge●●●○○
A stone is thrown straight down from a bridge into the water below, so its initial velocity is not zero and it is in the same direction as the acceleration. The quadratic that results has one physical root.
Height of the bridge above the water: $28.0\ \mathrm{m}$
Air resistance neglected, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) How long does the stone take to reach the water?
(b) How fast is it moving when it arrives?
Hint 1/4
Declare the origin and the positive direction first. Taking down as positive here makes the initial velocity, the acceleration and the displacement all positive, and removes every minus sign from the algebra.
Hint 2/4
With down positive, $y = v_0t + \tfrac{1}{2}gt^{2}$, which gives a quadratic in $t$; the speed then follows from $v = v_0 + gt$ or from the time free formula.
Hint 3/4
Substitute the data from the question, $v_0 = 5.00\ \mathrm{m/s}$ and $y = 28.0\ \mathrm{m}$: $4.90t^{2} + 5.00t - 28.0 = 0$.
Hint 4/4
So $t = 1.93\ \mathrm{s}$, taking the positive root, and the impact speed is $24.0\ \mathrm{m/s}$.
Show solution
Down is taken as positive rather than up, because every quantity in this problem points down and the choice removes three minus signs; nothing physical depends on it, as the alternative choice gives the same time.
Set up and reduce to a quadratic
$$\text{down positive, origin at the bridge}:\quad 28.0 = 5.00t + 4.90t^{2}$$
the position formula, chosen because the impact speed is not needed for part (a) and is therefore the absent variable
$$4.90t^{2} + 5.00t - 28.0 = 0$$
arranged with a positive leading coefficient before the formula is applied
using the time free formula makes this an independent calculation rather than a continuation of the last one
$$v = \sqrt{573.8} = 24.0\ \mathrm{m/s}$$
positive, meaning downward under the declared convention
Answer $$\boxed{t = 1.93\ \mathrm{s},\qquad v = 24.0\ \mathrm{m/s}}$$
Check
Cross check with the other route: $v = v_0 + gt = 5.00 + 9.80(1.9341) = 24.0\ \mathrm{m/s}$, agreeing with the time free result. Size: a stone falling roughly nine storeys should arrive at some tens of metres per second, and 24 m/s is about 86 km/h.
Throwing the stone down at 5.00 m/s instead of dropping it saves only about 0.46 s out of 2.39 s, because most of the speed at impact was going to come from the fall anyway. That is the kind of comparison a symbolic answer makes easy and a numerical one hides.
3§05.3 — finding a lift's acceleration from a scale reading●●○○○
A person weighs themselves in a lift and watches the reading change as the lift starts. The reading is steady for a few seconds, which is long enough to take it.
Given
Mass of the person $m = 78.0\ \mathrm{kg}$
Steady scale reading during the interval: $890\ \mathrm{N}$
Forces on the person: the weight downward and the normal force from the scale upward, nothing else
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the person's weight.
(b) Find the acceleration of the lift, in magnitude and direction.
(c) State whether the lift is going up or down, or explain why that cannot be decided.
Hint 1/4
The reading is bigger than the weight. Before computing anything, decide which way that forces the net force to point, and therefore which way the acceleration points.
Hint 2/4
With up positive, $F_N - mg = ma_y$, and this single equation contains everything the problem gives and everything it asks for except part (c).
Hint 3/4
Substitute the data from the question, $m = 78.0\ \mathrm{kg}$ and $F_N = 890\ \mathrm{N}$: $890 - 764.4 = 78.0a_y$.
Hint 4/4
So the weight is $764\ \mathrm{N}$, the acceleration is $1.61\ \mathrm{m/s^{2}}$ upward, and the direction of travel cannot be determined.
Show solution
The sign of the answer is predicted before the arithmetic, from the fact that the reading exceeds the weight, so that a slip in the subtraction would be caught immediately.
Weight, and the prediction
$$mg = 78.0(9.80) = 764\ \mathrm{N}$$
the weight is a property of the person and the planet and does not change inside the lift
$$F_N = 890 > 764 \Rightarrow a_y > 0$$
the upward force wins, so the net force and therefore the acceleration point upward, before any division is done
positive, matching the prediction, and directed upward
What the sign does not say
$$a_y > 0 \nRightarrow v_y > 0$$
a lift starting upward and a lift arriving at a lower floor both have upward accelerations, and no measurement of the normal force can separate them
Answer $$\boxed{mg = 764\ \mathrm{N},\quad a = 1.61\ \mathrm{m/s^{2}}\ \text{up},\quad \text{direction of travel undetermined}}$$
Check
Substitute back: $F_N = m(g + a_y) = 78.0(9.80 + 1.6103) = 78.0(11.41) = 890\ \mathrm{N}$, the given reading. Size: the reading is 16 per cent above the weight and $1.61/9.80 = 16$ per cent, which is the same statement twice.
Part (c) is worth as many marks as part (b) on most papers, and it is the part students leave out. A quantity that the physics cannot determine is a legitimate answer, and saying so is not hedging.
4§05.4 — a lighter hanging block on a smooth table system●●●○○
The connected block arrangement again, with a hanging block much lighter than the one on the table, which makes the acceleration small and the tension close to the hanging weight.
Given
Table block $m_1 = 5.00\ \mathrm{kg}$ on a smooth horizontal table
Hanging block $m_2 = 1.50\ \mathrm{kg}$, so its weight is $14.7\ \mathrm{N}$
Light inextensible string over a light free turning pulley; released from rest
Forces on $m_1$: tension toward the pulley, weight down, normal force up. Forces on $m_2$: tension up, weight down
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the acceleration of the system.
(b) Find the tension in the string.
(c) Find the speed of the hanging block after it has descended 0.800 m.
Hint 1/4
Two bodies means two diagrams and two equations. Before writing them, decide which way the system moves and take that direction as positive for each block separately.
Hint 2/4
$F_T = m_1a$ for the table block and $m_2g - F_T = m_2a$ for the hanging block; adding them removes the tension. Then $v^{2} = 2a(x - x_0)$ from rest.
Hint 3/4
Substitute the data from the question, $m_1 = 5.00\ \mathrm{kg}$, $m_2 = 1.50\ \mathrm{kg}$ and a descent of $0.800\ \mathrm{m}$: $14.7 = 6.50a$.
Hint 4/4
So $a = 2.26\ \mathrm{m/s^{2}}$, $F_T = 11.3\ \mathrm{N}$, and after 0.800 m the speed is $1.90\ \mathrm{m/s}$.
Show solution
Two separate body equations are written rather than a system equation, because the tension is asked for and is invisible to the system approach.
Two equations, one shared acceleration
$$m_1:\ F_T = 5.00a$$
the tension is the only horizontal force on the table block
$$m_2:\ 14.7 - F_T = 1.50a$$
downward positive for the hanging block, with the same $a$ because the string does not stretch
Solve
$$14.7 = 6.50a \;\Rightarrow\; a = 2.26\ \mathrm{m/s^{2}}$$
the driving weight divided by the total mass; adding the equations eliminated the tension
$$F_T = 5.00(2.2615) = 11.3\ \mathrm{N}$$
back into the shorter equation, with a guard digit carried
Cross into kinematics
$$v^{2} = 2a(x-x_0) = 2(2.2615)(0.800) = 3.618$$
from rest, distance given, no time mentioned, so the time free formula is the one with nothing spare
$$v = 1.90\ \mathrm{m/s}$$
the positive root, the block descending in its own positive direction
Check with the equation not used at the end: $14.7 - 11.3 = 3.4\ \mathrm{N}$ and $m_2a = 1.50(2.2615) = 3.39\ \mathrm{N}$, agreeing to the rounding. Bound check: the tension 11.3 N is below the hanging weight 14.7 N, as it must be for a descending block.
Compare with the earlier system, where 3.00 kg pulled by 2.00 kg gave 3.92 m/s squared. Making the hanging block lighter lowers the acceleration and pushes the tension towards the hanging weight, and in the limit of a very light hanging block nothing moves and the tension becomes exactly its weight.
5§05.5 — a vertical throw, with one answer to reject●●○○○
A ball is thrown straight up from ground level with a speed that makes the arithmetic clean. Two quantities are asked for, and then one reported answer has to be rejected using a check rather than a recalculation.
Air resistance neglected, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the maximum height reached above the launch point.
(b) Find the net force on the ball at the highest point.
(c) A classmate reports a maximum height of 100 m. Name the check that rejects it, and say what went wrong.
Hint 1/4
The highest point is defined by one condition on the velocity, not by the time or the height. Write that condition down first and the formula picks itself.
Hint 2/4
At the top $v = 0$, so $0 = v_0^{2} - 2gh$; and the net force there is the weight alone, since nothing is touching the ball.
Hint 3/4
Substitute the data from the question, $v_0 = 14.0\ \mathrm{m/s}$ and $m = 0.450\ \mathrm{kg}$: $h = 196/19.6$ and the weight is $0.450 \times 9.80$.
Hint 4/4
So $h = 10.0\ \mathrm{m}$ and the net force at the top is $4.41\ \mathrm{N}$ downward; the reported 100 m fails the size check by a factor of ten.
Show solution
The time free formula is used rather than finding the time to the top first, because the time is not asked for anywhere and computing it would add a step and a rounding.
work the reported answer backwards to the launch speed it implies and compare it with the one that was given
$$44.3 \gg 14.0$$
the implied speed is more than three times the actual one, so the size check rejects the answer without any need to find the error
Answer $$\boxed{h = 10.0\ \mathrm{m},\quad \sum F = 4.41\ \mathrm{N}\ \text{down},\quad 100\ \mathrm{m}\ \text{fails the size check}}$$
Check
Independent route to the height: the time to the top is $14.0/9.80 = 1.43\ \mathrm{s}$, and the average speed on the way up is $7.00\ \mathrm{m/s}$, giving $7.00 \times 1.43 = 10.0\ \mathrm{m}$. Landmark: 10 m is about three storeys, which is a good hard throw and matches experience.
Part (b) catches more people than part (a). At the top the ball is momentarily at rest and the net force is at its usual full value, which is the pair of facts the contrast earlier in this section was built around.
C · exam level 3 questions
1§05.3 — cable tension on an accelerating lift cage●●●○○
A lift cage is hauled upward by a single cable. The cage is speeding up, and the question is what the cable has to carry. Take the cage alone as the body; the cable and the motor are not part of it.
Given
Mass of the loaded cage $m = 1.40\times 10^{3}\ \mathrm{kg}$, so its weight is $1.37\times 10^{4}\ \mathrm{N}$
Upward acceleration $a = 1.50\ \mathrm{m/s^{2}}$
Forces on the cage: the cable tension upward and the weight downward, nothing else
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What is the tension in the cable?
Hint 1/4
Decide first whether the tension should come out larger or smaller than the weight, using only the direction of the acceleration. That single decision eliminates half the options.
Hint 2/4
With up positive, $F_T - mg = ma$, so $F_T = m(g+a)$, and the bracket is an effective gravity larger than $g$ whenever the acceleration points up.
Hint 3/4
Substitute the data from the question, $m = 1.40\times 10^{3}\ \mathrm{kg}$ and $a = 1.50\ \mathrm{m/s^{2}}$: $F_T = 1400(9.80+1.50) = 1400(11.30)$.
Hint 4/4
So the tension is $1.58\times 10^{4}\ \mathrm{N}$, about 15 per cent above the weight.
Show solution
The cage alone is chosen as the body rather than the cage and the cable together, because the tension is an internal force of that larger system and would not appear in its equation.
Ratio check: $F_T/mg = 11.30/9.80 = 1.153$, so the tension exceeds the weight by 15.3 per cent, which is the same as $a/g = 1.50/9.80 = 15.3$ per cent, as the formula requires. Limit check: setting $a = 0$ returns $1.37\times 10^{4}\ \mathrm{N}$, the static value.
Lift cables are rated for the accelerating case, not the hanging case, and the margin above is why. The same equation with a minus sign gives the tension while the cage slows on its way up, which is when the cable is least loaded.
2§05.4 — locating the first wrong step in a connected body solution●●●●○
Below is a student's solution to a connected body problem. A 6.00 kg block on a smooth table is joined by a light string over a light pulley to a 4.00 kg block hanging beside the table, and the system is released from rest. Exactly one step is where the solution first goes wrong; the later steps are faithful to it.
Given
Table block $m_1 = 6.00\ \mathrm{kg}$ on a smooth table, hanging block $m_2 = 4.00\ \mathrm{kg}$
Step 1: both blocks share one acceleration $a$, because the string does not stretch.
Step 2: the string is holding the hanging block, so $F_T = m_2g = 39.2\ \mathrm{N}$.
Step 3: for the table block, $F_T = m_1a$, so $a = 39.2/6.00 = 6.53\ \mathrm{m/s^{2}}$.
Step 4: check, $6.53\ \mathrm{m/s^{2}}$ is below $g$, so the answer is reasonable.
Find
(a) At which step does the solution first go wrong?
Hint 1/4
Read each step as a claim and ask what would have to be true for it to hold. One of them is true only in a situation the problem explicitly rules out.
Hint 2/4
For the hanging block, $m_2g - F_T = m_2a$, so $F_T = m_2(g-a)$, and the tension equals the weight only when $a = 0$.
Hint 3/4
Substitute the data from the question, $m_1 = 6.00\ \mathrm{kg}$ and $m_2 = 4.00\ \mathrm{kg}$: the correct pair is $a = 39.2/10.00 = 3.92\ \mathrm{m/s^{2}}$ and $F_T = 6.00(3.92) = 23.5\ \mathrm{N}$.
Hint 4/4
So the solution first goes wrong at step 2, and the correct tension is $23.5\ \mathrm{N}$, not $39.2\ \mathrm{N}$.
Show solution
The steps are tested in order rather than the problem being resolved from scratch, because the question asks where the solution first fails and a fresh solution would not answer that.
Test each claim in turn
$$\text{Step 1}: |a_1| = |a_2| \ \checkmark$$
an inextensible string moves both ends by the same amount in the same time, so this is the constraint stated correctly
$$\text{Step 2}: F_T = m_2g \iff a = 0 \ \times$$
the hanging block's own equation makes this equivalent to zero acceleration, which contradicts the system being released from rest and moving
Redo from the repaired step
$$m_2g - F_T = m_2a \ \text{and}\ F_T = m_1a$$
one equation per body, with the tension left as an unknown rather than assumed
$$39.2 = (6.00+4.00)a \;\Rightarrow\; a = 3.92\ \mathrm{m/s^{2}}$$
adding the two removes the tension
$$F_T = 6.00(3.92) = 23.5\ \mathrm{N}$$
back into the table block's equation, which is the one with a single force in it
Check the repaired pair in the equation not used to produce the tension: $m_2g - F_T = 39.2 - 23.5 = 15.7\ \mathrm{N}$ and $m_2a = 4.00(3.92) = 15.7\ \mathrm{N}$. The bound also holds now, since $23.5 < 39.2\ \mathrm{N}$, whereas the student's tension sat exactly on the bound, which is the signature of the equilibrium assumption.
A check that only tests whether a number is below $g$ will pass almost anything. A check worth writing has to be one the wrong answer would fail, and here the tension bound is exactly that check.
3§05.3 — a block hauled up a smooth incline by a rope●●●●○
A block on a smooth incline is pulled up the slope by a rope lying along the slope. The question runs the machine forwards for the first three parts and backwards for the last one, which is what a full exam question usually does.
Given
Mass $m = 7.50\ \mathrm{kg}$, so the weight is $mg = 73.5\ \mathrm{N}$
Incline angle $\theta = 24.0^{\circ}$ above the horizontal, surface smooth
Rope tension $F_T = 48.0\ \mathrm{N}$, directed up the slope and parallel to it
Forces on the block: the tension up the slope, the weight vertically down, the normal force perpendicular to the surface
Released from rest at $t = 0$; $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the acceleration, stating its direction along the slope.
(b) Find the normal force from the incline.
(c) Find how far the block moves in the first 1.80 s.
(d) What tension would hold the block at rest on the slope?
Hint 1/4
Turn the axes along the slope before writing anything, so that the acceleration lies on one axis and the normal force on the other. Then parts (a) and (b) are two separate one line equations.
Hint 2/4
Along the slope $F_T - mg\sin\theta = ma$; perpendicular to it $F_N - mg\cos\theta = 0$. Part (d) is the same first equation with $a$ set to zero.
Hint 3/4
Substitute the data from the question, $m = 7.50\ \mathrm{kg}$, $\theta = 24.0^{\circ}$, $F_T = 48.0\ \mathrm{N}$, $t = 1.80\ \mathrm{s}$: $\sin 24.0^{\circ} = 0.4067$ and $\cos 24.0^{\circ} = 0.9135$.
Hint 4/4
So $a = 2.41\ \mathrm{m/s^{2}}$ up the slope, $F_N = 67.1\ \mathrm{N}$, the block moves $3.91\ \mathrm{m}$, and a tension of $29.9\ \mathrm{N}$ would hold it still.
Show solution
The axes are turned along the slope, which puts the whole acceleration on one axis and the whole normal force on the other; with horizontal axes both unknowns would appear in both equations and a two by two system would have to be solved for no gain.
Along the slope
$$F_T - mg\sin\theta = ma$$
the normal force is perpendicular to this axis and contributes nothing, leaving two forces and one unknown
Bound check on the normal force: it must be below the weight of 73.5 N on any slope, and 67.1 N is. Consistency check between (a) and (d): the excess tension above the holding value is $48.0 - 29.9 = 18.1\ \mathrm{N}$, and dividing it by the mass reproduces the acceleration, which means parts (a) and (d) cannot both be wrong in different ways.
Part (d) is the same equation as part (a) with one symbol set to zero, and examiners like it for exactly that reason. Recognising that the last part of a question is an earlier part with a condition imposed saves a whole re derivation.
D · interleaved 3 questions
1§05.5 — a caught ball, from a radar reading to a force●●●○○
Mixed practice, and the type is not announced. A ball arrives at a catcher's glove and is brought to rest over a short distance as the glove gives way. The speed was recorded by a radar gun in kilometres per hour.
Given
Mass of the ball $m = 0.145\ \mathrm{kg}$
Recorded speed on arrival: $144\ \mathrm{km/h}$
The glove brings the ball to rest over $0.250\ \mathrm{m}$, with a constant force
$g = 9.80\ \mathrm{m/s^{2}}$; the ball is moving horizontally, so gravity may be ignored over this short stop
Find
(a) Find the arrival speed in metres per second.
(b) Find the ball's acceleration while it is being stopped.
(c) Find the average horizontal force the glove exerts on the ball.
Hint 1/4
Two of the three parts are last month's material and one is this month's. Decide which end of the machine the data sit at before writing anything.
Hint 2/4
Convert first, then $v^{2} = v_0^{2} + 2a(x-x_0)$ with $v = 0$ gives the acceleration, and $F = ma$ turns it into a force.
Hint 3/4
Substitute the data from the question, $144\ \mathrm{km/h}$, $x = 0.250\ \mathrm{m}$ and $m = 0.145\ \mathrm{kg}$: the speed is $40.0\ \mathrm{m/s}$ and $0 = 1600 + 2a(0.250)$.
Hint 4/4
So $v_0 = 40.0\ \mathrm{m/s}$, $a = -3.20\times 10^{3}\ \mathrm{m/s^{2}}$, and the force is $464\ \mathrm{N}$ backwards.
Show solution
The conversion is done first and once, because every later line uses SI units, and the time free formula is chosen because no time is given or asked for anywhere in the question.
the mass is in kilograms and the distance in metres, so the speed has to join them before any force can be computed
Motion to acceleration
$$0 = v_0^{2} + 2a x \;\Rightarrow\; a = -\frac{1600}{2(0.250)} = -3.20\times 10^{3}\ \mathrm{m/s^{2}}$$
the ball stops, so the final speed is zero, and no time appears anywhere in the data
Cross the joint
$$F = ma = 0.145(3.20\times 10^{3}) = 464\ \mathrm{N}$$
the magnitude is reported and the direction stated in words, opposite to the ball's motion
Answer $$\boxed{v_0 = 40.0\ \mathrm{m/s},\quad a = 3.20\times 10^{3}\ \mathrm{m/s^{2}},\quad F = 464\ \mathrm{N}}$$
Check
Size check on the acceleration: $3.20\times 10^{3}/9.80 = 327$, so the ball decelerates at 327 times free fall, which is enormous but correct for a 25 cm stop from 40 m/s, and it is why the answer to (c) is hundreds of newtons rather than a few. Alternative route: the stopping time is $2x/v_0 = 0.500/40.0 = 0.0125\ \mathrm{s}$, and $F = m\Delta v/\Delta t = 0.145(40.0)/0.0125 = 464\ \mathrm{N}$.
Notice what makes the force large: not the mass and not even the speed on its own, but the shortness of the stopping distance. Moving the glove backwards while catching, which doubles the distance, halves the force, and that is the whole physics of catching well.
2§05.1 — two perpendicular pushes on one box●●●○○
Mixed practice, type not announced. Two people push a box across a smooth level floor at the same time, from directions at right angles to each other, and the box moves off in neither of their directions.
Given
Mass of the box $m = 12.0\ \mathrm{kg}$
First push: $40.0\ \mathrm{N}$ due east, horizontal
Second push: $30.0\ \mathrm{N}$ due north, horizontal
Smooth level floor; the vertical forces on the box are its weight and the normal force, and they cancel
Find
(a) Find the magnitude of the net horizontal force.
(b) Find the magnitude and direction of the acceleration.
(c) Find the speed of the box 3.00 s after the pushes begin, starting from rest.
Hint 1/4
Two forces at right angles cannot be added as numbers. Decide what has to be done to them before the second law can be used at all.
Hint 2/4
Add by components: $R = \sqrt{R_x^{2}+R_y^{2}}$ and $\tan\phi = R_y/R_x$; then $\vec{a} = \sum \vec{F}/m$, which points along the resultant.
Hint 3/4
Substitute the data from the question, $40.0\ \mathrm{N}$ east and $30.0\ \mathrm{N}$ north with $m = 12.0\ \mathrm{kg}$: $R = \sqrt{1600+900}$ and $\tan\phi = 30.0/40.0$.
Hint 4/4
So the net force is $50.0\ \mathrm{N}$ at $36.9^{\circ}$ north of east, the acceleration is $4.17\ \mathrm{m/s^{2}}$ in that same direction, and after 3.00 s the speed is $12.5\ \mathrm{m/s}$.
Show solution
East and north are kept as the axes rather than rotating to align with the resultant, because the data are already given along those directions and rotating would mean resolving both forces for no benefit.
the acceleration points along the net force, so it inherits the 36.9 degree direction without further work
$$v = at = (4.1667)(3.00) = 12.5\ \mathrm{m/s}$$
from rest, in the direction of the acceleration, which does not change during the push
Answer $$\boxed{R = 50.0\ \mathrm{N},\quad a = 4.17\ \mathrm{m/s^{2}}\ \text{at}\ 36.9^{\circ}\ \text{N of E},\quad v = 12.5\ \mathrm{m/s}}$$
Check
Bound check on the resultant: it must lie between $\vert 40.0-30.0\vert = 10.0\ \mathrm{N}$ and $40.0+30.0 = 70.0\ \mathrm{N}$, and 50.0 N does. Direction check: the eastward push is the larger, so the resultant should lie closer to east than to north, and 36.9 degrees is indeed under 45.
The three, four, five triangle is worth recognising on sight, because examiners use it constantly: forces of 30 and 40 at right angles give 50, and so do 60 and 80 giving 100.
3§05.4 — a dropped ball and the force from the ground●●●●○
Mixed practice, type not announced. A ball is dropped from a measured height onto a hard floor, and the floor brings it to rest over a very short distance as it deforms. Two intervals with different forces, so the motion has to be cut where they change.
Given
Mass of the ball $m = 0.500\ \mathrm{kg}$
Dropped from rest from a height of $20.0\ \mathrm{m}$
On landing it is brought to rest over a distance of $0.0400\ \mathrm{m}$, with a constant force from the ground
Air resistance neglected; $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of the ball just before it touches the floor.
(b) Find its acceleration while it is being stopped.
(c) Find the force the floor exerts on the ball during the stop.
Hint 1/4
There are two motions here with different forces acting, so the constant acceleration formulas are used twice and the speed at the floor is what connects them.
Hint 2/4
Falling: $v^{2} = 2gh$. Stopping: $0 = v^{2} + 2a d$. Then the free body diagram during the stop has two forces, so $F_{\text{floor}} - mg = ma$ with up positive.
Hint 3/4
Substitute the data from the question, $h = 20.0\ \mathrm{m}$, $d = 0.0400\ \mathrm{m}$ and $m = 0.500\ \mathrm{kg}$: $v^{2} = 392$ and then $a = 392/0.0800$.
Hint 4/4
So $v = 19.8\ \mathrm{m/s}$, $a = 4.90\times 10^{3}\ \mathrm{m/s^{2}}$ upward, and the floor pushes with $2.45\times 10^{3}\ \mathrm{N}$.
Show solution
The fall and the stop are treated as two separate intervals, because the forces are different in each; the impact speed is the state that both intervals share and it is the only quantity carried across the join.
Ratio check: the stopping distance is 500 times shorter than the fall, so the deceleration should be about 500 times $g$, and $4900/9.80 = 500$ exactly. Force check: $2.45\times 10^{3}\ \mathrm{N}$ is about 500 times the ball's weight, which is the same statement, and it is the reason a dropped ball can crack a tile.
The whole answer is controlled by one number, the ratio of the fall height to the stopping distance. That is why a ball dropped onto a carpet is harmless and the same ball dropped onto stone is not, and neither the mass nor the drop height alone decides it.
Mistake ledger (17 entries)
⚠ Using the whole weight as the along slope force
the weight is the only force with a memorable formula, so on a picture with one obvious arrow the whole of it gets used
⚠ Starting from a kinematic equation before the acceleration exists
the question mentions a distance and a time, so the formula containing a distance and a time looks like the place to start, and the acceleration slot gets filled with the nearest available number
on every level surface met before a ramp appears the two really are equal, so the equality gets stored as a property of the normal force rather than as one line of arithmetic
⚠ Putting the acceleration into the diagram as a force arrow
it has a direction and a magnitude and it is what the problem is about, so it looks like it belongs on the diagram with the forces
wrong$$\sum F = F_N - mg - ma = 0$$
right$$\sum F = F_N - mg = ma$$
⚠ Reading the direction of travel from the sign of the acceleration
both are directions attached to the same moving body and in the simplest examples they agree, so the distinction never gets tested until an exam tests it
wrong$$a_y < 0 \Rightarrow \text{the lift is descending}$$
right$$a_y < 0 \Rightarrow \vec{a}\ \text{is downward};\ \vec{v}\ \text{may point either way}$$
⚠ Setting the tension equal to the hanging weight
the string is visibly holding the block, and in every static problem so far it has held it with exactly its weight
⚠ Putting both ends of the string into one diagram so the tension cancels
on a picture of the whole system the two tension arrows really are equal and opposite, and cancelling equal and opposite arrows is a good habit everywhere else
⚠ Treating a passed unit check as a finished answer
the unit check is taught first and feels most like physics, so passing it carries more confidence than it has earned, and it is blind to every factor of ten
⚠ Resolving the weight on a slope with the cosine instead of the sine
the slope angle sits at the corner where the base is the adjacent side, so the picture suggests the cosine while the geometry of the weight requires the sine; the wrong answer also stays plausible in size and direction
the acceleration must be constant over the whole interval; each formula omits one of $x$, $v$, $t$, $a$, and you pick the one omitting the quantity nobody mentions
Free fall near the ground
$$a_y = -g = -9.80\ \mathrm{m/s^{2}}\ \text{with up positive}$$
nothing touching the body and air resistance neglected; the constant $g$ itself is always positive
Block on a smooth incline
$$a = g\sin\theta,\qquad F_N = mg\cos\theta$$
smooth surface, nothing acting along the slope except the weight, $\theta$ measured from the horizontal; the mass cancels from the acceleration but not from the normal force
smooth table, light inextensible string, light free turning pulley, and the string taut throughout; the formulas fail the instant it goes slack
The tension bound for a descending block
$$0 \le F_T < m_2g$$
the hanging block accelerates downward; equality with the weight would mean no acceleration at all
The normal force bound on a slope
$$0 \le F_N \le mg$$
only the weight and the surface act, with nothing pressing the body into the surface or lifting it off
The four checks
$$[\text{left}]=[\text{right}];\quad n_{\mathrm{sf}} = \min_i n_i;\quad \text{compare with a landmark};\quad \text{push a symbol to a limit}$$
applied to the final answer and to any result another part will be built on; each catches a different family of error
Triage, the three questions
$$\text{(1) which end is given?}\quad \text{(2) how many bodies?}\quad \text{(3) is } a \text{ constant?}$$
asked of the wording, before any diagram; a multi part question may need them asked again for each part
Check yourself
Close the page and write down, from memory: the quantity that carries information between forces and motion, and the equations on either side of it; the four constant acceleration formulas with the variable each leaves out; what a free body diagram may and may not contain; three situations where the normal force is not the weight, with a number for each; the two equations and the constraint for two blocks on a string, plus the bound on the tension; the four checks; and the three triage questions. Then reopen and compare. The gaps are your reread list, and none of it is scored.
State which quantity joins a force problem to a motion problem, and say for a given question whether the calculation runs from forces to motion or the other way?
c-acceleration-hinge
Declare an origin, a positive direction and a time zero in writing, and then choose the right constant acceleration formula by naming the variable the problem never mentions?
c-setup-protocol
Draw a free body diagram for a body on a slope, in a lift, or under a rope at an angle, and produce a normal force that is not the weight?
c-fbd-to-equations
Write one equation per block for two bodies on a string, state the constraint that joins them, and check the tension against its bound?
c-two-body-constraints
Apply the four checks to a finished answer and say, for a given wrong answer, which check would have caught it and which would not?
c-answer-checking
Read an unseen problem, name its type in one sentence, and say what the first quantity to compute is before writing any equation?
c-triage
Glossary (14 terms)
net forcenet kuvvet
The vector sum of all the forces acting on one chosen body. It is what the second law relates to the acceleration, and it can be zero while several large forces are acting.
free body diagramserbest cisim diyagramı
A drawing of one body alone with an arrow for every force acting on it, and nothing else on it: no velocities, no accelerations, and no forces that the body exerts on other things.
inertiaeylemsizlik
The tendency of a body to keep whatever velocity it already has, measured by its mass. A body with more inertia needs more net force for the same change of velocity.
weightağırlık
The gravitational force on a body, equal to the mass times the free fall acceleration, measured in newtons. It is a force and not a mass, and a body carried elsewhere keeps its mass while its weight changes.
normal forcenormal kuvvet
The push a surface exerts on a body in contact with it, always perpendicular to the surface and always outward. It has no formula of its own and is solved for from the equation perpendicular to the surface.
tensiongerilme
The pull a taut string exerts along its own length on whatever is attached to it. In a light string over a light free turning pulley it has the same magnitude everywhere along the string.
apparent weightgörünen ağırlık
The force a support such as a scale or a floor pushes a body with, which is what the body feels. It equals the true weight only when the vertical acceleration is zero.
equilibriumdenge
The state in which the net force on a body is zero, so its velocity does not change. It includes bodies at rest and bodies moving in a straight line at a steady speed.
etki tepki çifti
Two forces of equal magnitude and opposite direction that two bodies exert on each other. They act on different bodies, so they never appear in the same free body diagram and never cancel each other there.
constraint equationkısıt denklemi
A relation between the motions of two connected bodies that comes from the geometry rather than from any force, such as two blocks joined by an inextensible string having accelerations of equal magnitude.
ideal ip
A string treated as having no mass and no stretch. The first assumption makes the tension the same at both ends, and the second makes the two connected bodies share one acceleration magnitude.
ideal makara
A pulley treated as having no mass and turning without resistance, so that it changes the direction of a string's pull without changing its magnitude.
limiting casesınır durumu
An extreme value of one symbol in a formula, such as an angle going to zero or a mass going to zero, at which the answer is already known from a simpler problem and can therefore be used to test the formula.
smooth surfacesürtünmesiz yüzey
A surface idealised as exerting no force along itself, so that its only action on a body is the normal push perpendicular to it. It is an assumption stated in the problem and not a property of any real material.
What comes next
§06 · Using Newton's Laws: Friction, Circular Motion, Drag Forces
Every surface in this review was declared smooth, and each time that word appeared it was removing an arrow from the diagram. Next week the word comes off. A surface that resists sliding adds an arrow with no fixed size: it adjusts itself up to a limit set by the normal force, which is why this section spent so long on getting the normal force right. The same week adds motion on a curved path and forces that grow with speed. Nothing here is put away; the diagram, the two component equations and the four checks are all used on the first page, with one more arrow in the picture.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition, the chapters covering measurement, kinematics in one and two dimensions, and Newton's laws of motion The required book for the course. This section reviews material already covered in the first four weeks and introduces no new result, so it quotes no section numbers: the week line in the syllabus reads only Catch up and Review and names no chapter, and inventing numbers would be worse than leaving them out.
PHYS 101 syllabus: the week five line and the published assessment weights The week line reads Catch up and Review. The coverage table maps that line onto the four weeks already taught, and the sixty second card quotes only the published weights, with nothing claimed beyond them about how any particular topic is examined.
The International System of Units, for the metre, the kilogram, the second and the newton Used for the unit symbols and for the conversion factors between kilometres per hour and metres per second, which are exact by definition and therefore never limit the number of significant figures in an answer.