04Dynamics: Newton's Laws of Motion

Look carefully at which quantity the first law is about, and check whether that quantity is the one being held constant here.

The first law says $\sum \vec F = 0$ exactly when the velocity is constant, and velocity is constant only if both its size and its direction hold still.

Here the size is fixed at 60 km/h but the direction is turning continuously as the road curves.

The passenger is wrong: the velocity is changing, so the acceleration is not zero and neither is the net force.

A third-law partner is found by swapping the two bodies in the phrase the force of A on B, not by looking for something that balances.

$\vec F_{A \text{ on } B} = -\,\vec F_{B \text{ on } A}$, and the two are always the same kind of force.

Here A is the earth and B is the book, and the force named is gravitational, so the partner must also be gravitational.

Swapping the two bodies gives the book's gravitational pull on the earth.

Ask what the first law actually promises when the forces cancel: a particular velocity, or an unchanging one?

$\sum \vec F = 0$ gives $\vec v = \text{constant}$, and a constant may be any value at all, including a large one.

Here nothing in the statement fixes what that constant is, so the probe may be drifting at 10 km/s just as legitimately as sitting still.

False: zero net force means the velocity does not change, not that it is zero.

Two different quantities are being asked about, and there is no rule forcing them to vanish together.

The only force acting is the weight, so $\sum F_y = -mg$ and $a_y = -g$ at every instant of the flight, whatever the velocity is doing.

Here the velocity passes through zero at the top, while the weight $mg$ is still there and still the only force.

The velocity is zero and the acceleration is 9.80 m/s² downward.

Two of the three ropes lie on the same line, so one column of the component table will be very short.

$\sum F_x = F_{1x}+F_{2x}+F_{3x}$ and likewise for $y$; then $\left|\sum \vec F\,\right| = \sqrt{(\sum F_x)^{2}+(\sum F_y)^{2}}$.

Here the x column is $30.0 + 0 - 20.0$ and the y column is $0 + 40.0 + 0$.

The net force is 41.2 N at 76.0° above the $+x$ axis.

The force is not given, so the motion has to be turned into an acceleration first; decide which of the constant-acceleration equations has only the given quantities in it.

$a = (v-v_0)/t$, then $\sum F = ma$.

Here $v_0 = 0$, $v = 27.0$ m/s, $t = 9.00$ s and $m = 1250$ kg.

The acceleration is 3.00 m/s² and the net force is 3.75 kN.

The planet's gravity is not given directly; it has to be extracted from the falling stone, which is a piece of last week's work.

$d = \tfrac12 g_p t^{2}$ gives $g_p = 2d/t^{2}$, and then weight is $mg$ on whichever planet you are standing on, while mass is $m$ everywhere.

Here $d = 1.85$ m in $t = 1.00$ s, and the astronaut has $m = 75.0$ kg.

The planet has $g_p = 3.70\ \mathrm{m/s^{2}}$; the astronaut's mass is 75.0 kg everywhere, weighing 278 N there and 735 N on earth.

The scale reading is one of the two forces in the vertical equation; decide which one, then the equation has a single unknown left.

$N - mg = ma_y$, so $a_y = N/m - g$.

Here $N = 830$ N and $m = 72.0$ kg, so $mg = 706$ N and $N$ exceeds it.

The acceleration is $1.73\ \mathrm{m/s^{2}}$ upward, and the direction of travel is not determined.

Tilt the axes before drawing anything, and note which component of the acceleration that makes zero.

Along the slope $a = g\sin\theta$; perpendicular to it $N = mg\cos\theta$; then $v^{2} = 2a d$ from rest.

Here $m = 8.00$ kg, $\theta = 35.0°$ and $d = 2.50$ m, with $mg = 78.4$ N.

The answers are $5.62\ \mathrm{m/s^{2}}$, $64.2\ \mathrm{N}$ and $5.30\ \mathrm{m/s}$.

Two different bodies are useful here at two different moments: one that makes the contact force disappear, and one that makes it the only unknown.

Take the pair as one body for $a = F/(m_1+m_2)$; then isolate the block that the applied force does not touch, for which $F_{\text{contact}} = m a$.

Here $F = 30.0$ N and $m_1+m_2 = 10.0$ kg, and the untouched block has mass 4.00 kg in the first case and 6.00 kg in the second.

The answers are $3.00\ \mathrm{m/s^{2}}$, $12.0\ \mathrm{N}$ and $18.0\ \mathrm{N}$.

Two bodies means two equations; the string supplies the two facts that let one symbol $a$ and one symbol $T$ serve both of them.

$m_1g - T = m_1a$ and $T - m_2g = m_2a$; adding them gives $a = (m_1-m_2)g/(m_1+m_2)$, and the pulley carries $2T$.

Here $m_1 - m_2 = 2.00$ kg and $m_1 + m_2 = 12.00$ kg, with $g = 9.80\ \mathrm{m/s^{2}}$.

The answers are $1.63\ \mathrm{m/s^{2}}$, $57.2\ \mathrm{N}$ and $114\ \mathrm{N}$.

One force is slanted and two are not, so exactly one splitting job is needed before either equation can be written.

$T\cos\theta = ma$ along the ground, and $N + T\sin\theta - mg = 0$ perpendicular to it.

Here $T = 120$ N, $\theta = 20.0°$ and $mg = (55.0)(9.80) = 539$ N.

The acceleration is $2.05\ \mathrm{m/s^{2}}$ and the normal force is $498\ \mathrm{N}$, smaller than the weight.

Two bodies are worth considering, and one of them turns the question into a single-unknown equation. Decide which before calculating anything.

Take both blocks as one body to get $a = F/(m_1+m_2)$, then isolate the front block, on which the contact force is the only horizontal force: $F_{\text{contact}} = m_2 a$.

Here $F = 18.0$ N, $m_1 + m_2 = 9.00$ kg and the front block has $m_2 = 6.00$ kg.

The acceleration is 2.00 m/s² and the contact force is 12.0 N.

Work out first whether the reading is above or below the passenger's weight, and note what that alone fixes.

$a_y = N/m - g$, and the sign of $a_y$ is the only thing this reading determines; the direction of travel is a separate matter.

Here $N = 480$ N against $mg = 588$ N, so the reading is 108 N short of the weight and $m = 60.0$ kg.

The acceleration is 1.80 m/s² downward, and the direction of travel cannot be deduced.

Draw the little right triangle at the block itself, with the weight as its hypotenuse, and read off which side lies along the slope.

With $\theta$ measured from the horizontal, the along-slope component of the weight is $mg\sin\theta$ and the perpendicular one is $mg\cos\theta$, so $a = g\sin\theta$ and $N = mg\cos\theta$.

Here $mg = 58.8$ N and $\theta = 40.0°$, with $\sin 40.0° = 0.643$ and $\cos 40.0° = 0.766$.

The correct pair is $a = 6.30\ \mathrm{m/s^{2}}$ and $N = 45.0\ \mathrm{N}$.

Before calculating, decide whether the tension can possibly equal the hanging block's weight while the block is accelerating downward.

$a = m_2g/(m_1+m_2)$ from the two equations added together, and then $T = m_1 a$ from the table block.

Here $m_2 g = 29.4$ N, $m_1+m_2 = 5.00$ kg and the table block has $m_1 = 2.00$ kg.

The acceleration is 5.88 m/s² and the tension is 11.8 N.

There are exactly two forces on the lift, and the answer depends on the difference between them rather than on either one alone.

$T - mg = ma_y$, so $a_y = T/m - g$.

Here $T = 1.40\times 10^{4}$ N, $mg = 1.176\times 10^{4}$ N and $m = 1200$ kg.

The acceleration is 1.87 m/s² upward.

Three sub-questions, but only one of them is about forces; the first two belong to earlier sections and have to be done first.

$1\ \mathrm{km/h} = 1/3.6\ \mathrm{m/s}$; then $v^{2} = v_0^{2}+2a d$ with $v = 0$ gives $a = -v_0^{2}/(2d)$; then $\sum F = ma$.

Here $v_0 = 90.0$ km/h, $d = 42.0$ m and $m = 1150$ kg, so $mg = 1.13\times 10^{4}$ N for the comparison.

The answers are 25.0 m/s, $-7.44\ \mathrm{m/s^{2}}$ and $8.56\times 10^{3}$ N, about 0.76 of the car's weight.

Ask which bodies are touching the ball once it has left the hand; that settles parts (a) and (b) before any arithmetic.

In flight the only force is the weight, so $\sum \vec F = m\vec g$ and $\vec a = \vec g$ everywhere. At the top the vertical velocity is zero while the horizontal one is unchanged at $v_0\cos\theta$.

Here $m = 0.145$ kg, $v_0 = 25.0$ m/s and $\theta = 40.0°$, with $\cos 40.0° = 0.766$.

The net force is 1.42 N downward, the acceleration is 9.80 m/s² downward, and the speed at the top is 19.2 m/s.

A correct equation must have the same units on both sides, so start by writing each side in base units and compare.

Replace every newton by $\mathrm{kg\,m/s^{2}}$ and simplify each side; an equation whose two sides differ in units is wrong whatever the numbers are.

Here the left side is an acceleration in $\mathrm{m/s^{2}}$, and the right side of the student's version is $\mathrm{kg\,m/s^{2}} \times \mathrm{kg}$.

The student's right side carries $\mathrm{kg^{2}}$, so the equation is impossible; the correct version gives $4.00\ \mathrm{m/s^{2}}$.

Component form removes the trigonometry from the first part entirely: the two vectors can be added column by column as they stand.

Add like components, then $\left|\sum \vec F\,\right| = \sqrt{(\sum F_x)^{2}+(\sum F_y)^{2}}$ and $\vec a = \sum \vec F/m$; equilibrium needs a third force equal to minus the present sum.

Here the x parts are $12.0$ and $-4.00$, the y parts are $5.00$ and $7.00$, and $m = 3.00$ kg.

The net force is $(8.00\,\hat{\imath} + 12.0\,\hat{\jmath})$ N of magnitude 14.4 N, the acceleration is 4.81 m/s² at 56.3°, and the balancing force is minus the net force.