"Laws of Motion" — Practice

Question 1

A rod has one end on the floor and the other against a vertical wall. The floor end moves away from the wall at $2\ 	ext{m/s}$ when the rod makes $30°$ with the floor. Find the speed of the wall end.

Accepted Answer

Components along the rod must be equal:

$v_A\cos 30°=v_B\sin 30°$

$v_B=v_A\cot 30°=2(1.732)=\mathbf{3.46\ 	ext{m/s}}$, directed downward.

*Read the trend:* at $\alpha=45°$ the two speeds would be equal, and below $45°$ the wall end moves faster. As the rod flattens toward the floor, $\cot\alpha$ diverges and the top end accelerates without limit.

Question 2

A $10\ 	ext{kg}$ block sits on a rough floor with $\mu_s=0.5$. A horizontal force of $20\ 	ext{N}$ is applied and the block does not move. State the friction force, and explain why $\mu_sN$ is the wrong answer.

Accepted Answer

The block is in equilibrium, so the net horizontal force is zero and friction must exactly balance the applied force:

$f=\mathbf{20\ 	ext{N}}$

The **maximum available** friction is $\mu_sN=0.5(100)=50\ 	ext{N}$, which is merely the ceiling — the block is nowhere near it.

*The principle:* static friction is a constraint force. It takes whatever value prevents sliding, exactly like a normal force or a tension, and equals $\mu_sN$ **only** at the instant slipping is about to begin. Quoting $50$ N here would give a body with a $30$ N net force and no acceleration.

Question 3

On a banked road with friction, state the direction of the friction force on a car travelling at the **minimum** safe speed, and explain why.

Accepted Answer

Friction acts **up the slope**.

At the design speed $\sqrt{rg	an	heta}$ the horizontal component of the normal force alone provides exactly the centripetal requirement. Below that speed, less centripetal force is needed than the banking supplies, so the surplus inward push would drive the car **down the bank toward the centre**.

Friction opposes that impending slide, so it points **up** the slope, and its inward component subtracts from the excess.

*At the maximum speed the situation reverses:* the car tends to slide up and outward, so friction points down the slope. The two limits differ only by the sign of the friction term, which is exactly why $v_{min}$ and $v_{max}$ have $\mp\mu$ in the numerator.

Question 4

A block on the floor is pulled by a rope that passes over a pulley fixed above. When the rope makes $60°$ with the horizontal, the free end is drawn in at $5\ 	ext{m/s}$. Find the block's speed.

Accepted Answer

The rope is inextensible, so the rate at which it is drawn in equals the component of the block's velocity **along the rope**:

$v\cos 60°=5$

$v=\dfrac{5}{0.5}=\mathbf{10\ 	ext{m/s}}$

**The block moves faster than the rope is pulled**, which surprises most candidates. Only part of its motion shortens the rope; the rest swings it round.

*As the block approaches the pulley's base*, $	heta	o90°$ and $\cos	heta	o0$, so the required block speed diverges. In reality the rope goes slack or the block leaves the floor before that.

Question 5

A $1\ 	ext{kg}$ block rests on a $4\ 	ext{kg}$ block on a frictionless floor, with $\mu=0.5$ between them and $g=10$. Find the largest horizontal force that keeps them moving together, applied (a) to the lower block, (b) to the upper block.

Accepted Answer

**(a) Force on the lower block.** The upper block is accelerated by friction alone:

$f=ma\le\mu mg \Rightarrow a\le\mu g=5\ 	ext{m/s}^{2}$

$F_{max}=(M+m)a=5(5)=\mathbf{25\ 	ext{N}}$

**(b) Force on the upper block.** Now the **lower** block is dragged by friction:

$f=Ma\le\mu mg \Rightarrow a\le\dfrac{0.5(1)(10)}{4}=1.25\ 	ext{m/s}^{2}$

$F_{max}=5(1.25)=\mathbf{6.25\ 	ext{N}}$

*Four times the difference*, from the same two blocks and the same $\mu$. The ratio is $m/M$: friction can only ever muster $\mu mg$, and pushing from above forces that limited amount to move the heavy block instead of the light one.

Question 6

A coin lies $0.4\ 	ext{m}$ from the centre of a turntable with $\mu=0.25$ and $g=10$. Find the greatest angular speed before it slips, and solve it in both frames.

Accepted Answer

**Ground frame.** Friction is the only horizontal force, so it must supply the whole centripetal requirement:

$m\omega^{2}r\le\mu mg \Rightarrow \omega\le\sqrt{\dfrac{\mu g}{r}}=\sqrt{\dfrac{2.5}{0.4}}=\mathbf{2.5\ 	ext{rad/s}}$

**Rotating frame.** The coin is stationary, in equilibrium under friction inward and the centrifugal force $m\omega^{2}r$ outward — the identical inequality, reached without any dynamics.

That is $f=\omega/2\pi=0.40$ revolutions per second.

*Read the dependence:* $\omega_{max}\propto1/\sqrt{r}$, so coins near the rim slip first. This is exactly why a centrifuge separates material by radius, and why the outermost passengers feel a fairground ride hardest.

Question 7

A curve of radius $200\ 	ext{m}$ is banked at $15°$ with $\mu=0.5$ and $g=10$. Find the safe speed range, and comment on the result.

Accepted Answer

$	an 15°=0.268$, and $\mu=0.5$, so **$\mu>	an	heta$**.

$v_{min}^{2}=\dfrac{rg(	an	heta-\mu)}{1+\mu	an	heta}=\dfrac{2000(0.268-0.5)}{1.134}<0$

The expression is negative, so **there is no minimum speed at all**. Friction alone is more than enough to hold the car on this gentle bank, and it can be parked there without sliding.

$v_{max}=\sqrt{\dfrac{2000(0.268+0.5)}{1-0.134}}=\sqrt{\dfrac{1536}{0.866}}=\sqrt{1774}=\mathbf{42.1\ 	ext{m/s}}$

So the safe range is $\mathbf{0}$ to $\mathbf{42.1\ 	ext{m/s}}$.

*The design lesson:* a shallow bank with good grip is safe at every speed up to a high ceiling, whereas a steep bank with poor grip becomes dangerous when you go too **slowly**. That is why racing banks are steep and public roads are not.

Question 8

A uniform rope of mass $M$ and length $L$ lies coiled on the floor. It is lifted vertically by one end at constant speed $v$. Find the force applied when a length $x$ has been lifted.

Accepted Answer

Let $\lambda=M/L$. Two contributions, and both are needed.

**Weight of the part already lifted:** $\lambda xg$

**Force to set new rope moving.** Rope joins the moving section at rate $\dfrac{dm}{dt}=\lambda v$, and must be accelerated from rest to $v$:

$F_{thrust}=v\dfrac{dm}{dt}=\lambda v^{2}$

$$F=\lambda xg+\lambda v^{2}=\dfrac{M}{L}\left(gx+v^{2}ight)$$

*The mirror image of the falling chain*, where the pan must **destroy** momentum instead of supplying it. Note that the thrust term is constant while the weight term grows, so the lift starts out dominated by momentum and ends dominated by weight.

*Limit check:* an infinitely slow lift, $v	o0$, gives just $\lambda xg$, as it must.

Question 9

A block of mass $m$ slides on a frictionless wedge of mass $3m$ and angle $30°$, the wedge being free on a frictionless floor. Take $g=10$. Find the block's acceleration relative to the wedge and the wedge's acceleration.

Accepted Answer

With $M=3m$, $	heta=30°$, $\sin^{2}	heta=0.25$:

Denominator $=M+m\sin^{2}	heta=3m+0.25m=3.25m$

$a_{rel}=\dfrac{(4m)(10)(0.5)}{3.25m}=\mathbf{6.15\ 	ext{m/s}^{2}}$

$A=\dfrac{m(10)(0.5)(0.866)}{3.25m}=\mathbf{1.33\ 	ext{m/s}^{2}}$

*Compare the fixed wedge:* $g\sin 30°=5.0\ 	ext{m/s}^{2}$. The free wedge gives $23\%$ more, because the surface retreats beneath the block.

*Momentum check.* The block's absolute horizontal acceleration is $a_{rel}\cos	heta-A=5.33-1.33=4.00\ 	ext{m/s}^{2}$, carrying $4m$ one way; the wedge carries $3m(1.33)=4m$ the other. They cancel, as they must on a frictionless floor.

Question 10

A block rests on a wedge of angle $	heta$ with coefficient of friction $\mu$ between them. The wedge is accelerated horizontally. Find the range of accelerations for which the block does not slide.

Accepted Answer

In the wedge's frame the block is stationary under $N$, $mg$, the pseudo force $ma_0$ backward, and friction $f$ along the incline.

**Perpendicular to the incline:** $N=mg\cos	heta+ma_0\sin	heta$

**Along the incline:** friction must supply $m\left(g\sin	heta-a_0\cos	hetaight)$, and $|f|\le\mu N$.

**Lower limit** (block tends to slide down, friction acts up):

$g\sin	heta-a_0\cos	heta\le\mu\left(g\cos	heta+a_0\sin	hetaight) \Rightarrow a_0\ge\dfrac{g(	an	heta-\mu)}{1+\mu	an	heta}$

**Upper limit** (block tends to slide up, friction acts down):

$a_0\cos	heta-g\sin	heta\le\mu\left(g\cos	heta+a_0\sin	hetaight) \Rightarrow a_0\le\dfrac{g(	an	heta+\mu)}{1-\mu	an	heta}$

$$\frac{g(	an	heta-\mu)}{1+\mu	an	heta}\ \le\ a_0\ \le\ \frac{g(	an	heta+\mu)}{1-\mu	an	heta}$$

*Recognise the structure.* This is algebraically **identical** to the banked-road speed range, with $a_0$ in place of $v^{2}/r$ — because it is the same problem. Setting $\mu=0$ collapses the band to the single value $g	an	heta$.

Question 11

A uniform chain of length $L$ lies on a table with a length $l$ hanging over the edge. The coefficient of friction between chain and table is $\mu$. Find the greatest fraction $l/L$ that can hang without the chain slipping.

Accepted Answer

Let $\lambda$ be the linear density.

**Driving force** — the weight of the hanging part: $\lambda lg$

**Resisting force** — friction on the part still on the table, whose normal force is its own weight:

$f_{max}=\mu\lambda(L-l)g$

On the verge of slipping the two are equal:

$\lambda lg=\mu\lambda(L-l)g \Rightarrow l=\mu L-\mu l \Rightarrow l(1+\mu)=\mu L$

$$\frac{l}{L}=\frac{\mu}{1+\mu}$$

*Check the limits.* $\mu=0$ gives zero — any overhang at all starts it sliding, correct for a frictionless table. As $\mu	o\infty$ the fraction tends to $1$ but never reaches it, which is also right: however grippy the table, some chain must remain on it to grip.

*Worth noting:* the answer is **never more than a half**, since $\mu/(1+\mu)<1$ requires $\mu<1+\mu$, and reaching $1/2$ needs $\mu=1$.

Question 12

Two particles of equal mass $m$ lie on a smooth table joined by a slack inextensible string. One is projected with speed $u$ at $60°$ to the line joining them, and the string becomes taut. Find both velocities immediately after, and the fraction of kinetic energy lost.

Accepted Answer

Resolve the initial velocity along and perpendicular to the string:

$u_{\parallel}=u\cos 60°=\dfrac{u}{2}, \qquad u_{\perp}=u\sin 60°=0.866u$

**Perpendicular:** no impulsive force acts, so the moving particle keeps $0.866u$ and the other stays at rest in that direction.

**Along the string:** inextensibility forces a common component, and momentum along the string is conserved:

$m\left(\dfrac{u}{2}ight)=2mv_{\parallel} \Rightarrow v_{\parallel}=\dfrac{u}{4}$

**Projected particle:** $\sqrt{(u/4)^{2}+(0.866u)^{2}}=\mathbf{0.901u}$

**Other particle:** $\mathbf{0.25u}$, along the string.

**Energy.** Before, $	frac12mu^{2}$. After, $	frac12m(0.8125u^{2})+	frac12m(0.0625u^{2})=	frac12mu^{2}(0.875)$.

$$	ext{Fraction lost}=\mathbf{12.5\%}=	frac18$$

*Check with the reduced-mass formula:* the loss in any such jerk is $	frac12\left(\dfrac{m_1m_2}{m_1+m_2}ight)\left(\Delta v_{\parallel}ight)^{2}=	frac12\left(\dfrac{m}{2}ight)\left(\dfrac{u}{2}ight)^{2}=\dfrac{mu^{2}}{16}$, which is exactly one eighth of $	frac12mu^{2}$. A taut string is as dissipative as an inelastic collision, which is why real hoisting gear is deliberately given a little elasticity.

Practice — "Laws of Motion"

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