Corrected exercises: dynamics, Newton's laws and friction (203-SN1-RE)

Three forces act on the light: its weight $P=mg=(12)(9.8)=117.6$ N downward, the tension $T_{1}$ along cable 1 (upward and to the left, at 30° to the horizontal) and the tension $T_{2}$ along cable 2 (upward and to the right, at 45°).

Equilibrium: $\Sigma F_{x}=0$ gives $T_{2}\cos 45^{\circ}-T_{1}\cos 30^{\circ}=0$; $\Sigma F_{y}=0$ gives $T_{1}\sin 30^{\circ}+T_{2}\sin 45^{\circ}-117.6=0$.

From the first equation: $T_{2}=T_{1}\frac{\cos 30^{\circ}}{\cos 45^{\circ}}$. Substituting into the second: $T_{1}(\sin 30^{\circ}+\cos 30^{\circ}\tan 45^{\circ})=117.6$, that is $T_{1}(0.5+0.866)=117.6$, so $T_{1}\approx 86.1$ N and $T_{2}\approx 105.4$ N. The 45° cable is the more tense: the closer a cable is to the vertical, the larger the share of the weight it carries. Notice that $T_{1}+T_{2}\approx 191.5$ N far exceeds the weight: the horizontal components, which cancel each other, carry none of the weight but still put the cables under tension.

By symmetry, each cable carries half the weight vertically: $2T\sin 5^{\circ}=117.6$, so $T=\frac{117.6}{2\sin 5^{\circ}}\approx 675$ N, almost six times the weight of the light. As the angle tends to 0, $\sin\theta$ tends to 0 and the tension diverges: a horizontal cable has no vertical component and therefore cannot carry any weight, however tight it is. That is why load-bearing cables always sag.

Only two forces act on the person: the weight $\vec{F}_{g}=m\vec{g}$ downward and the normal force $\vec{N}$ upward. Positive axis upward: $N-mg=ma$, so $N=m(g+a)$. At rest ($a=0$): $N=65.0(9.80)=637$ N, reading $65.0$ kg. Starting upward ($a=+1.20$): $N=65.0(11.0)=715$ N, reading $73.0$ kg. Constant speed ($a=0$): $N=637$ N, reading $65.0$ kg. Braking at the end of the ascent ($a=-1.50$, since the elevator is moving up while slowing down): $N=65.0(8.30)=540$ N, reading $55.1$ kg.

No, mass is a property of the body and is $65.0$ kg in every situation. The true weight $mg=637$ N does not change either. What the scale measures is the normal force, that is, the contact force needed to give the person the elevator's acceleration. It is called the apparent weight: it is a contact force, not a property of the person.

Descent: when starting downward, the acceleration points DOWNWARD, so $N<mg$ and the reading is below $65.0$ kg. At constant speed, $a=0$ and the reading returns to $65.0$ kg. When braking at the end of the descent, the elevator moves down while slowing, so the acceleration points UPWARD, and the reading exceeds $65.0$ kg. The general rule: only the DIRECTION of the acceleration matters, never the direction of the motion. Acceleration upward, the scale over-reads; acceleration downward, it under-reads; zero acceleration, it shows the true mass, whether the elevator is stopped or moving at constant speed.

In free fall, $a=-g$, so $N=m(g-g)=0$: the scale shows zero. The person has lost nothing of their weight, which is still $637$ N and is precisely the force causing the fall. What has disappeared is the normal force: nothing opposes the fall any more, so nothing presses on the scale. This is exactly the situation of astronauts in orbit, who are constantly falling along with their station: their weight is very real, only the contact has disappeared.

We would need $N=80.0\times 9.80=784$ N, so $a=\frac{N}{m}-g=\frac{784}{65.0}-9.80\approx 2.26$ m/s$^{2}$ directed UPWARD. That is nearly double the acceleration of an ordinary elevator ($1$ to $1.5$ m/s$^{2}$), and the sensation would be very unpleasant: $23\,\%$ more apparent weight. Manufacturers deliberately limit the acceleration for comfort, not for safety.

Threshold of slipping: $f_{s,max}=\mu_{s}N=\mu_{s}mg=(0.50)(20)(9.8)=98$ N. Since 90 N $<$ 98 N, the crate does not move. The friction then equals exactly 90 N (not 98 N!): static friction adjusts itself to cancel the net force, and reaches its maximum only at the threshold of slipping.

120 N $>$ 98 N: the crate slides, and friction becomes kinetic: $f_{k}=\mu_{k}mg=(0.30)(196)=58.8$ N. Newton: $a=\frac{120-58.8}{20}=3.06$ m/s².

At constant speed, $\Sigma F=0$: it is enough to balance $f_{k}$, so 58.8 N. That is less than the 98 N needed to get started: this is why the hard part is setting an object in motion, not keeping it moving.

From 0 to 98 N, the graph follows the line $f=F$ (static friction equals the applied force: equilibrium). At 98 N, slipping begins and the friction drops abruptly to 58.8 N, then stays constant at that value whatever the applied force. The discontinuity reflects the change from the static regime ($\mu_{s}$) to the kinetic regime ($\mu_{k}<\mu_{s}$).

Pushing at 20° below the horizontal, the vertical component $F\sin 20^{\circ}$ adds to the weight: $N=mg+F\sin 20^{\circ}$. The threshold becomes $F\cos 20^{\circ}=\mu_{s}(mg+F\sin 20^{\circ})$, so $F=\frac{\mu_{s}mg}{\cos 20^{\circ}-\mu_{s}\sin 20^{\circ}}=\frac{98}{0.940-0.171}\approx 127.5$ N. About 30 % more is needed than for a horizontal push: pushing downward increases the normal force, and therefore the friction to overcome. Pushing down works against you.

On the way up, three forces act: the weight (components $-mg\sin\theta$ along the slope and $-mg\cos\theta$ perpendicular to it), the normal force $N=mg\cos\theta$, and the kinetic friction $f_{k}=\mu_{k}N$ directed DOWN the slope (it opposes the slipping, which is upward). Newton along the slope: $ma=-mg\sin\theta-\mu_{k}mg\cos\theta$, the mass cancels: $a=-9.8(0.5+0.25\times 0.866)\approx -7.02$ m/s².

With $v_{f}^{2}=v_{i}^{2}+2a\Delta x$: $0=64-2(7.02)\Delta x$, so $\Delta x=\frac{64}{14.04}\approx 4.56$ m.

It slides back. Criterion: a block at rest stays in place if $\tan\theta\leq\mu_{s}$. But $\tan 30^{\circ}\approx 0.577>0.30$: the maximum static friction ($\mu_{s}mg\cos\theta$) is not enough to hold back the component of the weight ($mg\sin\theta$), so the block slides down.

On the way down, the slipping is downward, so the friction now points UP the slope: $a=g(\sin\theta-\mu_{k}\cos\theta)=9.8(0.5-0.217)\approx 2.78$ m/s² (directed down the slope). Speed on return: $v=\sqrt{2(2.78)(4.56)}\approx 5.0$ m/s.

The weight does not change, but the friction changes direction: it always opposes the slipping. On the way up, weight and friction both pull down the slope (they add: strong deceleration); on the way down, they oppose each other (weak acceleration). Over the round trip, friction did negative work in both directions: the block returns with only $\frac{5.0^{2}}{8^{2}}\approx 40$ % of its initial kinetic energy, the rest having been dissipated as heat, hence a return slower than the departure.

6 kg block (on the table): weight 58.8 N downward, normal force $N$ upward, tension $T$ horizontal toward the pulley, kinetic friction $f_{k}$ opposite to the motion. 4 kg block (hanging): weight 39.2 N downward, tension $T$ upward.

The rope is inextensible: the two blocks travel the same distance in the same time, hence the same magnitude of acceleration $a$. The pulley is ideal (massless and frictionless): it changes the direction of the tension without changing its magnitude, hence the same $T$ at both ends. Block on the table (axis toward the pulley): $T-\mu_{k}m_{1}g=m_{1}a$. Hanging block (axis downward): $m_{2}g-T=m_{2}a$.

Adding the two equations, $T$ cancels: $m_{2}g-\mu_{k}m_{1}g=(m_{1}+m_{2})a$, so $a=\frac{39.2-(0.20)(58.8)}{10}=\frac{39.2-11.76}{10}\approx 2.74$ m/s². Tension: $T=m_{2}(g-a)=4(9.8-2.744)\approx 28.2$ N. Check with the other block: $28.2-11.76=16.5\approx(6)(2.744)$, consistent.

If $T$ were 39.2 N, the hanging block would feel a net force $m_{2}g-T=0$: it would not accelerate from rest, while the block on the table would feel $39.2-11.76>0$ and would accelerate. The two blocks would move in ways incompatible with an inextensible rope: contradiction. The error: $T=m_{2}g$ is true only at equilibrium ($a=0$). As soon as the system accelerates, the tension lies strictly between the friction force (11.8 N) and the weight of the hanging block (39.2 N).

The system stays at rest if static friction can balance the tension, which is then $T=m_{2}g=39.2$ N: we need $\mu_{s}m_{1}g\geq 39.2$, so $\mu_{s}\geq\frac{m_{2}}{m_{1}}=\frac{4}{6}\approx 0.67$.

The three blocks stay in contact, so they necessarily have the same acceleration: the second law can be applied to them as to a single body of mass $m_{1}+m_{2}+m_{3}=10.0$ kg. The contact forces between blocks are INTERNAL to the system defined this way and cancel in pairs in the overall balance. What remains is $F=(m_{1}+m_{2}+m_{3})a$, that is $a=\frac{30.0}{10.0}=3.00$ m/s$^{2}$ to the right.

Force between 1 and 2, isolating block 1: $F-F_{12}=m_{1}a$, so $F_{12}=30.0-2.00(3.00)=24.0$ N. Check by isolating the subsystem $\{2,3\}$: the only external horizontal force is $F_{12}$, so $F_{12}=(3.00+5.00)(3.00)=24.0$ N. Force between 2 and 3, isolating block 3: $F_{23}=m_{3}a=5.00(3.00)=15.0$ N. Check by isolating $\{1,2\}$: $F-F_{23}=(2.00+3.00)(3.00)=15.0$, so $F_{23}=30.0-15.0=15.0$ N. The two methods agree, which is the best check available on this type of problem.

Because they do not act on the same body. An action-reaction pair always acts on TWO different objects, and two forces can cancel only if they act on the same object. In the free-body diagram of block 1 appears only the $24.0$ N exerted BY block 2; in that of block 2 appears only the $24.0$ N exerted BY block 1. Neither diagram contains both. The confusion comes from mixing up the third law (two bodies, two forces) with equilibrium (one body, several forces).

The acceleration is unchanged, $a=\frac{30.0}{10.0}=3.00$ m/s$^{2}$, this time to the left: the overall balance sees only the total mass and the external force. The contact forces, however, change completely. Isolating block 3: $F-F_{32}=m_{3}a$, so $F_{32}=30.0-15.0=15.0$ N. Isolating block 1: $F_{21}=m_{1}a=6.00$ N. The values went from $24.0$ and $15.0$ N to $15.0$ and $6.00$ N. To remember: the overall acceleration does not depend on where the force is applied, but the internal forces depend on it entirely. That is why how the effort is applied matters: a train pulled from the front keeps its couplings in tension, whereas a push from the rear puts them in compression, which makes them buckle and risks derailing the train.

The horse's force on the cart and the cart's force on the horse do form an action-reaction pair, but they are INTERNAL to the complete horse-and-cart system: they therefore cannot change the motion of the whole. What moves the team is an external force: the horse's hooves push the ground backward, and by the third law the ground pushes the horse forward. This friction from the ground is the driving force. We can check by isolating the horse alone: it moves forward because the force of the ground on its hooves exceeds the pull that the cart exerts on it backward. On ice, this external force disappears and the team does not move, however much energy the horse puts in.

Axes: horizontal (toward the centre of the curve) and vertical. The normal force, perpendicular to the tilted road, has a horizontal component $N\sin\theta$ (toward the centre) and a vertical component $N\cos\theta$. Vertically there is no acceleration: $N\cos\theta=mg$. Horizontally, the net force provides the centripetal acceleration: $N\sin\theta=\frac{mv^{2}}{r}$. Dividing the two equations: $\tan\theta=\frac{v^{2}}{gr}$, so $v=\sqrt{gr\tan\theta}=\sqrt{(9.8)(50)(0.40)}=\sqrt{196}=14$ m/s (50.4 km/h). The mass has disappeared: banking lets ANY vehicle take the curve at this speed without needing any friction, hence its value on ice or in the rain.

Below the ideal speed, the horizontal component of the normal force is too large for the curvature required: the car tends to slide inward (down the slope), so friction acts UP the slope. Above the ideal speed, it is the opposite: the car tends to slide outward, and friction acts DOWN the slope, adding its centripetal contribution to that of the normal force.

At the threshold of skidding outward, $f_{s}=\mu_{s}N$ points down the slope. Its components: $\mu_{s}N\cos\theta$ horizontal (toward the centre) and $\mu_{s}N\sin\theta$ vertical (downward). The equations become: vertical $N\cos\theta-\mu_{s}N\sin\theta=mg$; horizontal $N\sin\theta+\mu_{s}N\cos\theta=\frac{mv_{max}^{2}}{r}$. Dividing the second by the first, $N$ and $m$ cancel: $\frac{v_{max}^{2}}{gr}=\frac{\sin\theta+\mu_{s}\cos\theta}{\cos\theta-\mu_{s}\sin\theta}=\frac{\tan\theta+\mu_{s}}{1-\mu_{s}\tan\theta}$. Numerically: $v_{max}=\sqrt{(9.8)(50)\times\frac{0.40+0.40}{1-(0.40)(0.40)}}=\sqrt{490\times\frac{0.80}{0.84}}\approx 21.6$ m/s (about 78 km/h).

On ice, the car can count only on the banking, which keeps the car on its path only at the ideal speed of 14 m/s. At 20 m/s, there is not enough centripetal force: the car skids toward the outside of the curve (it follows a path that is too straight). Conclusion: banking guarantees a safe curve at ONE design speed; friction provides the margin around that speed, here from 14 up to 21.6 m/s on a dry road, a margin that disappears with it.

At the top, two forces act on the water: its weight $mg$ downward and the force from the bottom of the bucket $N$, also downward since the bucket is upside down. Both therefore point toward the centre of the circle. The second law along the radial direction (positive toward the centre) gives $mg+N=\frac{mv^{2}}{R}$. The limiting condition is $N=0$, the bucket just ceasing to push: $mg=\frac{mv^{2}}{R}$, so $v_{min}=\sqrt{gR}=\sqrt{9.80(0.900)}\approx 2.97$ m/s. The mass disappears: the result is the same for a full or an empty bucket. The tension in the rope is then $T=0$ at the top: at exactly this speed the rope just goes slack, and the bucket and the water follow the circle under their weight alone.

Top: $T+mg=\frac{mv^{2}}{R}$, so $T=\frac{1.20(16.0)}{0.900}-1.20(9.80)=21.3-11.8\approx 9.57$ N. Lowest point: the weight now points away from the centre, so $T-mg=\frac{mv^{2}}{R}$ and $T=21.3+11.8\approx 33.1$ N. The ratio is about $3.5$: it is at the bottom that the rope may break, never at the top. In practice the gap is even larger, because the real speed at the bottom exceeds that at the top, gravity having done work during the descent. Assuming an identical speed is an idealization that the chapter on energy will let us remove.

Below $v_{min}$, the required centripetal acceleration $\frac{v^{2}}{R}$ becomes less than $g$. But at the top gravity alone provides an acceleration of $g$ toward the centre, and the bottom of the bucket can only PUSH the water, never hold it back: it cannot reduce the downward acceleration. The water can no longer follow the circle and leaves the path, becoming a simple projectile that falls. As for centrifugal force, it does not exist in an inertial frame: no object exerts it on the water, and it cannot appear in a free-body diagram. What we mistake for it is the inertia of the water, which tends to continue in a straight line while the bucket turns.

The ball moves in a horizontal circle at constant speed: its acceleration is purely centripetal, hence horizontal and directed toward the axis. The vertical is therefore in equilibrium: $T\cos\theta=mg$, so $T=\frac{0.250(9.80)}{\cos 25.0^{\circ}}\approx 2.70$ N. Radius: $R=L\sin\theta=1.20\sin 25.0^{\circ}\approx 0.507$ m. Horizontal: $T\sin\theta=\frac{mv^{2}}{R}$, so $v=\sqrt{\frac{T R\sin\theta}{m}}=\sqrt{\frac{2.70(0.507)(0.4226)}{0.250}}\approx 1.52$ m/s.

Derivation: dividing the horizontal equation by the vertical equation, $\tan\theta=\frac{v^{2}}{gR}$. With $R=L\sin\theta$ and $v=\frac{2\pi R}{T}$: $\tan\theta=\frac{4\pi^{2}R^{2}}{T^{2}gR}=\frac{4\pi^{2}L\sin\theta}{T^{2}g}$. Dividing by $\sin\theta$ (nonzero) leaves $\frac{1}{\cos\theta}=\frac{4\pi^{2}L}{T^{2}g}$, that is $T=2\pi\sqrt{\frac{L\cos\theta}{g}}$. Check: $2\pi\sqrt{\frac{1.20\cos 25.0^{\circ}}{9.80}}\approx 2.09$ s, which matches $\frac{2\pi R}{v}=\frac{2\pi(0.507)}{1.52}\approx 2.09$ s. Finally, a horizontal string would mean $\theta=90^{\circ}$ and $\cos\theta=0$: the vertical equation $T\cos\theta=mg$ would require an infinite tension. There must always be a vertical component of the tension to carry the weight, so the string always keeps some inclination, however small. That is exactly why no taut cable, no clothesline, is ever perfectly straight.

The only horizontal force on the top block is the friction exerted by the bottom block, directed FORWARD (in the direction of $F$). Without it, the bottom block would slide out from under the top block, which would stay in place: the slipping of the top relative to the bottom would be backward, and friction opposes this relative slipping, so it pushes the top block forward. Friction opposes the slipping BETWEEN the surfaces, not motion in general.

For the top block alone, Newton gives $f_{s}=m_{t}a$. Static friction is capped: $f_{s}\leq\mu_{s}m_{t}g$, so $a\leq\mu_{s}g=(0.60)(9.8)=5.88$ m/s², regardless of the masses. For the whole assembly (8 kg, frictionless floor): $F_{max}=(m_{t}+m_{b})\,a_{max}=(8)(5.88)\approx 47.0$ N.

$60>47.0$ N: the top block can no longer keep up, so it slides. Friction becomes kinetic: $f_{k}=\mu_{k}m_{t}g=(0.40)(2)(9.8)=7.84$ N. Top block: $a_{t}=\frac{7.84}{2}=3.92$ m/s². Bottom block: it feels $F=60$ N forward and the reaction of the friction, 7.84 N backward: $a_{b}=\frac{60-7.84}{6}\approx 8.69$ m/s². The bottom block pulls ahead ($a_{b}>a_{t}$): the top block falls back relative to it and will eventually fall off the rear.

The top block exerts on the bottom block: vertically a push of $m_{t}g=19.6$ N downward (the normal force, equal to the weight of the top block), and horizontally 7.84 N BACKWARD. This last force is the reaction (Newton's third law) to the 7.84 N forward friction that the bottom block exerts on the top block: same magnitude, opposite directions, acting on two different bodies. That is exactly why it slows the bottom block in the calculation of $a_{b}$.

Three forces: the weight $mg$ downward, the buoyant force $F_{B}$ upward and the drag $bv$ upward. Positive axis downward: $m\frac{dv}{dt}=mg-F_{B}-bv$. The novelty is crucial: the net force DEPENDS on the speed, so the acceleration is not constant and no constant-acceleration equation applies. We get a differential equation, and the motion is described by a transient regime followed by a steady regime.

At $t=0$, $v=0$ so the drag is zero: $a_{0}=\frac{mg-F_{B}}{m}=\frac{0.1176-0.0150}{0.0120}=\frac{0.1026}{0.0120}=8.55$ m/s$^{2}$. It is less than $g$ because of the buoyant force alone, which does not depend on the speed and acts from the very first instant. The drag does nothing yet.

The terminal speed is reached when the acceleration vanishes, that is, when the forces balance: $bv_{T}=mg-F_{B}$, so $v_{T}=\frac{0.1026}{0.0850}\approx 1.21$ m/s. The acceleration is then exactly zero: the ball afterwards descends at strictly constant speed. Motion at constant speed requires no net force, only equilibrium, in agreement with the first law.

$\tau=\frac{m}{b}=\frac{0.0120}{0.0850}\approx 0.141$ s. For $99\,\%$: $1-\mathrm{e}^{-t/\tau}=0.99$ gives $\mathrm{e}^{-t/\tau}=0.01$, so $t=\tau\ln 100\approx 0.141(4.61)\approx 0.650$ s. No, the ball never reaches exactly $v_{T}$: the exponential vanishes for no finite value of $t$, so the terminal speed is an asymptote. In practice, after a few $\tau$ the gap becomes smaller than any measurement precision, and we say the terminal speed has been reached.

Spread out: at terminal speed, $kv_{T}^{2}=mg$, so $k=\frac{80.0(9.80)}{55.0^{2}}=\frac{784}{3025}\approx 0.259$ kg/m. Parachute open: $k'=\frac{784}{6.00^{2}}=\frac{784}{36.0}\approx 21.8$ kg/m, about $84$ times larger, which is indeed the order of magnitude of the ratio of the areas. If this value applied all at once while the skydiver is still travelling at $55.0$ m/s: $F=k'v^{2}=21.8(3025)\approx 6.59\times 10^{4}$ N upward, net force $6.51\times 10^{4}$ N, so $a=\frac{6.51\times 10^{4}}{80.0}\approx 814$ m/s$^{2}$, or $83\,g$. No human body survives that. The conclusion is physical, not mathematical: a parachute CANNOT open instantly. Modern canopies deploy gradually, over about a second, thanks to a slider that holds back the opening, which brings the deceleration down to only a few $g$.