06Using Newton’s Laws: Friction, Circular Motion, Drag Forces

The statement claims to give a value. Ask what the formula in the rule box is a statement about: a value, or a limit on a value.

The rule is written $F_{fr} \le \mu_s N$, an inequality, and the equality holds only at the instant the body is on the point of slipping.

Here the crate is not on the point of slipping; it is simply at rest under a push smaller than the ceiling, so the equilibrium condition $\sum F_x = 0$ is the equation that applies.

False: the friction equals the push, and $\mu_s N$ is only the largest it could have been.

Every arrow on a free-body diagram has to be traceable to a second body that is doing the pushing. Go through the candidates and ask, for each one, what is doing it.

Newton's first law says the passenger keeps a straight line unless a force acts, and the second law says the net force must point at the centre of the bend, which is to the left.

Here the bend is to the left, the passenger's acceleration is therefore to the left, and the only bodies touching the passenger are the seat and the door.

The door pushes the passenger inward, to the left, and that inward push is the whole horizontal net force.

The statement moves from a fact about the speed to a claim about the acceleration. Check whether that step is allowed by the definitions.

Acceleration is the rate of change of the velocity, which is a vector, and a vector changes if either its length or its direction changes.

Here the length is fixed at 25 km/h for the whole lap, but the direction is different at every point of the circle.

False: the acceleration is $v^{2}/r$ toward the centre and it is never zero.

Both are at terminal speed, so for each of them a balance holds. Write what is the same for the two of them and what is not.

At terminal speed $\tfrac12 \rho C_D A v_T^{2} = mg$, so $v_T = \sqrt{2mg/(\rho C_D A)}$ and everything except $m$ is shared.

Here the masses are 60.0 kg and 90.0 kg, a ratio of 1.50, and the mass sits inside a square root.

The heavier one falls faster, by a factor of $\sqrt{1.50} = 1.22$.

Three answers are wanted and they come in a fixed order, because each one is an input to the next. Decide which has to be first.

$\sum F_y = N - mg = 0$ gives $N$; then $F_{fr} = \mu_k N$; then $\sum F_x = F - F_{fr} = ma$.

Here $m = 35.0$ kg, $F = 140$ N and $\mu_k = 0.250$, so $mg = 343$ N.

$N = 343$ N, $F_{fr} = 85.8$ N and $a = 1.55\ \mathrm{m/s^{2}}$.

The mass of the car is not given, which is a hint about what will and will not survive into the answer.

With $N = mg$, friction gives $a = -\mu_k g$, and then $v^{2} = v_0^{2} + 2a\Delta x$ and $v = v_0 + at$ do the rest.

Here $\mu_k = 0.500$ and $v_0 = 30.0\ \mathrm{m/s}$, so $\mu_k g = 4.90\ \mathrm{m/s^{2}}$.

$4.90\ \mathrm{m/s^{2}}$, 91.8 m and 6.12 s.

Two very different questions in one situation: one asks for a limit, and the other asks for a value at a particular moment. Sort out which is which first.

The limit is $(F_{fr})_{\max} = \mu_s N$ with $N = mg$; the value while at rest comes from $\sum F_x = 0$.

Here $m = 40.0$ kg and $\mu_s = 0.550$, so $mg = 392$ N, and part (b) uses a push of 150 N.

About 216 N is needed to start it, and a 150 N push meets 150 N of friction.

Neither the mass nor any force is given, which tells you that both answers must come out of expressions in which the mass has cancelled.

At the slipping angle $\tan\theta = \mu_s$; once it is sliding, $a = g(\sin\theta - \mu_k\cos\theta)$.

Here $\theta = 22.0^{\circ}$, so $\tan 22.0^{\circ} = 0.404$, $\sin 22.0^{\circ} = 0.375$ and $\cos 22.0^{\circ} = 0.927$, with $\mu_k = 0.330$.

$\mu_s = 0.404$ and $a = 0.673\ \mathrm{m/s^{2}}$.

The ramp is tilted, so the first decision is where to put the axes. Choose them so that one component of the acceleration is known to be zero.

With $x$ down the slope and $y$ out of it: $N = mg\cos\theta$, $F_{fr} = \mu_k N$ up the slope, and $mg\sin\theta - F_{fr} = ma$.

Here $m = 4.00$ kg, $\theta = 35.0^{\circ}$ and $\mu_k = 0.200$, so $mg = 39.2$ N, $\cos 35.0^{\circ} = 0.819$, $\sin 35.0^{\circ} = 0.574$.

$N = 32.1$ N, $F_{fr} = 6.42$ N and $a = 4.02\ \mathrm{m/s^{2}}$.

A period is given, not a speed, so pick the route that starts from a period. There are two, and one of them skips part (a) entirely.

$v = 2\pi r/T$ and $a_R = v^{2}/r$, or in one step $a_R = 4\pi^{2} r/T^{2}$.

Here $r = 8.00$ m and $T = 6.00$ s, so one turn covers $2\pi(8.00) = 50.3$ m.

$v = 8.38\ \mathrm{m/s}$, $a_R = 8.77\ \mathrm{m/s^{2}}$, which is $0.895g$.

Part (c) is asking about the mass, so keep the mass as a symbol until the very end of part (a) and see whether it survives.

On a flat bend $\mu_s mg = mv_{\max}^{2}/r$, giving $v_{\max} = \sqrt{\mu_s g r}$, and the acceleration at the limit is $\mu_s g$.

Here $\mu_s = 0.400$ and $r = 120$ m, so $\mu_s g r = (0.400)(9.80)(120)$.

$21.7\ \mathrm{m/s}$, $3.92\ \mathrm{m/s^{2}}$, and the lorry has the same limit.

Part (a) is an equilibrium and part (b) is not. Decide for each part whether the right-hand side of the second law is zero before writing anything.

Terminal: $\tfrac12\rho C_D A v_T^{2} = mg$. At any other speed: $mg - \tfrac12\rho C_D A v^{2} = ma$.

Here $\tfrac12\rho C_D A = \tfrac12(1.21)(0.350)(4.20\times 10^{-3}) = 8.89\times 10^{-4}$ and $mg = 1.42$ N.

$v_T = 40.0\ \mathrm{m/s}$, and at 20.0 m/s the drag is 0.356 N and the acceleration is $7.35\ \mathrm{m/s^{2}}$.

An angled pull does two things at once, and they push the answer in opposite directions. Name both before computing either.

$\sum F_y = N + F\sin\theta - mg = 0$ gives $N$; then $F_{fr} = \mu_k N$ and $\sum F_x = F\cos\theta - F_{fr} = ma$.

Here $m = 25.0$ kg, $F = 120$ N, $\theta = 20.0^{\circ}$ and $\mu_k = 0.300$, so $mg = 245$ N, $\cos 20.0^{\circ} = 0.940$ and $\sin 20.0^{\circ} = 0.342$.

$N = 204$ N, $a = 2.06\ \mathrm{m/s^{2}}$ at 20.0°, against $1.86\ \mathrm{m/s^{2}}$ with a horizontal rope, so the angle wins here.

Two bodies means two diagrams and two equations, joined by the fact that the string forces one shared acceleration and carries one shared tension.

For the hanging block $m_2 g - T = m_2 a$; for the block on the table $T - \mu_k m_1 g = m_1 a$. Adding them removes $T$.

Here $m_1 = 3.00$ kg, $m_2 = 2.00$ kg and $\mu_k = 0.200$, so $m_2 g = 19.6$ N and $\mu_k m_1 g = 5.88$ N.

$a = 2.74\ \mathrm{m/s^{2}}$, $T = 14.1\ \mathrm{N}$, and it would have stayed still if $\mu_s$ had been at least 0.667.

At the design speed the inward part of the normal force is exactly what the turn needs. Ask what that same inward part is doing at a much lower speed.

The radial equation is $N\sin\theta \pm F_{fr,\text{horizontal}} = mv^{2}/r$; at a lower speed the right-hand side is smaller while $N\sin\theta$ is not.

Here $v = 8.00\ \mathrm{m/s}$ against a design speed of 25.0 m/s, so the required inward force is about a tenth of what the bank alone supplies.

Friction acts up the slope, holding the car back from sliding down the bank toward the inside of the bend.

The top and the bottom are two different problems, because the centre of the circle is on opposite sides of the rider in the two cases.

At the top, $N + mg = mv^{2}/r$ with $N \ge 0$; at the bottom, $N - mg = mv^{2}/r$.

Here $m = 60.0$ kg, $r = 6.00$ m, and part (b) uses $v = 12.0\ \mathrm{m/s}$, so $v^{2}/r = 144/6.00 = 24.0\ \mathrm{m/s^{2}}$.

$7.67\ \mathrm{m/s}$ at the top, $2.03\times 10^{3}\ \mathrm{N}$ at the bottom, which is 3.45 times the weight.

The feeling of lightness is a clue about the size of the answer relative to the weight. Decide before computing whether the answer should be larger or smaller than $mg$.

With down as the positive radial direction, $mg - N = mv^{2}/r$, so $N = m(g - v^{2}/r)$.

Here $m = 1500$ kg, $v = 15.0\ \mathrm{m/s}$ and $r = 50.0$ m, so $v^{2}/r = 225/50.0 = 4.50\ \mathrm{m/s^{2}}$ and $mg = 1.47\times 10^{4}\ \mathrm{N}$.

$N = 1500(9.80 - 4.50) = 7.95\times 10^{3}\ \mathrm{N}$, a little over half the weight.

Ask what changes at the edge of the table. Up to that point one force acts horizontally; after it, none does. Two separate calculations are needed, joined by one number.

On the table, $a = -\mu_k g$ and $v^{2} = v_0^{2} + 2a\Delta x$. Off the table, the horizontal velocity is constant and the fall time comes from $h = \tfrac12 g t^{2}$.

Here $v_0 = 4.00\ \mathrm{m/s}$, $\Delta x = 1.50\ \mathrm{m}$, $\mu_k = 0.200$ and the drop is $h = 1.20\ \mathrm{m}$.

The puck leaves at 3.18 m/s and lands 1.57 m from the base of the table.

The crate is accelerating, so the net force on it is not zero. Ask which body is touching the crate and could possibly supply that force.

For the crate, $F_{fr} = ma$, and while it does not slide $F_{fr} \le \mu_s N$ with $N = mg$, so $\mu_s \ge a/g$.

Here $a = 2.50\ \mathrm{m/s^{2}}$ in part (a), and $\mu_s = 0.200$ in part (c), with $g = 9.80\ \mathrm{m/s^{2}}$.

$\mu_s \ge 0.255$; the friction points forward; and with $\mu_s = 0.200$ the limit is $1.96\ \mathrm{m/s^{2}}$.

Part (a) is a first-law question and part (b) is a projectile question. Neither of them needs the radius, which is a hint that no circular formula is involved after the break.

With no force left, the velocity keeps its value and direction, which is tangential; then vertically $h = \tfrac12 g t^{2}$ while horizontally the speed stays at $6.00\ \mathrm{m/s}$.

Here $v = 6.00\ \mathrm{m/s}$ horizontally at the break and the fall is $h = 1.80\ \mathrm{m}$.

It goes off along the tangent, and lands 3.64 m away horizontally.

Two numbers are being turned into one, and then into a statement about how well it is known. Work out first what quantity each reading gives you directly.

Each reading gives $\mu_s = \tan\theta$; the best estimate is the mean of the two values and a fair uncertainty is half their spread.

Here $\tan 24^{\circ} = 0.445$ and $\tan 26^{\circ} = 0.488$, from readings of $24^{\circ}$ and $26^{\circ}$.

$\mu_s = 0.47 \pm 0.02$, about a 5 per cent uncertainty.