Corrected exercises: Newton's laws and free-body diagrams, PHYS 101

a) The inventory is made by walking around the crate and asking what touches it, then asking what field reaches it. Two objects touch the crate: the floor, which pushes perpendicular to the surface, and the hand, which pushes horizontally. One field reaches it: gravity, which pulls it toward the centre of the Earth. So the diagram carries exactly three arrows, the normal force $\vec{N}$ from the floor, the applied force $\vec{F}$ from the hand, and the weight $\vec{W}$ from the Earth. Naming the source of each arrow is not decoration, it is the check: an arrow whose source cannot be named is an arrow that does not belong.

b) A forward force would have to come from something. Nothing is in front of the crate, the hand pushes from behind, and the floor is frictionless, so no object supplies it. The first law says exactly this: an object keeps its velocity by itself, and a force is what CHANGES a velocity, not what maintains one. The crate would keep sliding at constant speed with no push at all on this frictionless floor. The wrong arrow, often written $\vec{F}_{\text{motion}}$, is the single most common error on a first-year diagram and it costs the whole question, because every equation written afterwards inherits it.

c) The weight is $W = mg = 18 \times 9.80 = 176.4$ N, pointing straight down. The crate does not accelerate vertically, it slides along the floor, so the vertical equation is $\Sigma F_{y} = 0$, that is $N - W = 0$ and $N = 176.4$ N. The normal force equals the weight HERE, and only because the push is horizontal and the floor is level. That coincidence is the reason so many students memorise $N = mg$ as a law; exercises 4, 5 and 6 each break it.

d) Horizontally, only the push acts, so $\Sigma F_{x} = F = ma$, giving $a = \dfrac{65}{18} = 3.61$ m/s$^2$. Sanity check on the order of magnitude: 65 N on 18 kg is about 3.6 N per kilogram, roughly a third of $g$, which is a brisk push, not an unreasonable one. Notice that $N$ and $W$ never entered the horizontal equation. Projecting on two axes is what keeps them out: a force perpendicular to an axis contributes zero to that axis, and writing one scalar equation per axis is what turns a picture into arithmetic.

a) The mass is a property of the book: $m = 6.0$ kg, and the number is read straight from the statement. The weight is a force: $W = mg = 6.0 \times 9.80 = 58.8$ N. Writing the weight as 6.0 kg is the error the marker is looking for, and it costs the mark every time, because a newton and a kilogram measure different physical quantities. A useful mental anchor: on Earth one kilogram weighs about ten newtons.

b) The mass does not change, because nothing was added to or removed from the book: $m = 6.0$ kg on the Moon as on Earth. The weight does change, because the field changed: $W = mg_{M} = 6.0 \times 1.62 = 9.72$ N. The ratio is $\dfrac{9.72}{58.8} = 0.165$, about one sixth, the familiar figure for the Moon.

c) A spring scale measures a FORCE, the pull of the spring, and then divides by the Earth value of $g$ printed on its dial. On the Moon it feels $9.72$ N and divides by $9.80$, so it displays $\dfrac{9.72}{9.80} = 0.99$ kg, which is wrong by a factor of six as a mass. A two-pan balance compares the book with reference masses, and both pans sit in the same field, so the field cancels: it reads $6.0$ kg on the Moon, on Earth, and in orbit. The instrument decides what is measured, which is why a laboratory report must say which one was used.

d) The second law is $\Sigma F = ma$, and $g$ appears nowhere in it. On the Moon, $F = 6.0 \times 2.5 = 15$ N; on Earth, the same $F = 15$ N. Pushing the book sideways is exactly as hard on the Moon as on Earth, because the resistance to a change of motion is the mass, not the weight. Only lifting it is easier, since lifting has to beat the weight. This single sentence, mass resists acceleration and weight is a force, settles most of the confusion students carry into the midterm.

a) The partner of the force exerted BY A ON B is the force exerted BY B ON A. It has the same magnitude, 96 N, the opposite direction, backwards, and it acts on A. The reliable way to name a partner is to read the force as a sentence and swap the two nouns: A pushes B becomes B pushes A. Any answer that keeps the same body, such as the weight of B or the normal force on B, is not a partner, it is just another force that happens to act there.

b) Each skater gets their own diagram and their own equation. For B, $\Sigma F = 96$ N and $a_{B} = \dfrac{96}{48} = 2.00$ m/s$^2$ forward. For A, $\Sigma F = 96$ N backwards and $a_{A} = \dfrac{96}{72} = 1.33$ m/s$^2$ backwards. The forces are equal, the accelerations are not, and their ratio is exactly the inverse mass ratio: $\dfrac{a_{B}}{a_{A}} = \dfrac{72}{48} = 1.5$. Equal forces, unequal effects, because the second law divides by the mass.

c) Two forces cancel only when they are added in the SAME equation, and an equation belongs to one body. The 96 N on B appears in the equation of B; the 96 N on A appears in the equation of A. They are never on the same line, so they never cancel. This is why the rule of one body at a time is not a style preference: drawing both skaters on one picture is what produces the false cancellation, and drawing two separate diagrams makes the error impossible.

d) The weight of B is exerted by the Earth on B, so its partner is exerted by B on the Earth: a 470.4 N pull upward on the planet, which the planet answers with an acceleration too small to measure. The force that BALANCES the weight while B stands still is the normal force from the ice, 470.4 N upward. Same number, completely different status: the partner acts on another body and is never in the equation of B, while the normal force acts on B and is exactly what makes $\Sigma F_{y} = 0$. If B stepped onto a scale in an accelerating elevator, the normal force would change and the weight would not, which settles the question once and for all.

a) With the angle measured from the horizontal, the adjacent side is horizontal and the opposite side is vertical: $T_{x} = T\cos 30^{\circ} = 85 \times 0.8660 = 73.6$ N and $T_{y} = T\sin 30^{\circ} = 85 \times 0.5000 = 42.5$ N. The check that costs nothing: $\sqrt{73.6^{2} + 42.5^{2}} = 85$ N, the components rebuild the original. The sine and cosine are swapped in about one copy in four, and the symptom is visible without any calculation, a $30^{\circ}$ rope should have a horizontal component LARGER than its vertical one.

b) The sled slides along the ground, so it has no vertical acceleration and $\Sigma F_{y} = 0$. Three forces have a vertical component: the normal force up, the weight down, the tension partly up. So $N + T_{y} - mg = 0$, which gives $N = mg - T_{y} = 22 \times 9.80 - 42.5 = 215.6 - 42.5 = 173.1$ N. The rope carries part of the load, so the ground carries less. Writing $N = mg = 215.6$ N here is an error of 42.5 N, and it would matter in every friction problem later in the course.

c) Horizontally, only $T_{x}$ acts: $\Sigma F_{x} = T_{x} = ma$, so $a = \dfrac{73.6}{22} = 3.35$ m/s$^2$. Note that $N$ never appears, because it is perpendicular to the direction of motion. A useful consistency check: pulling flat along the ground with the same 85 N would give $a = 3.86$ m/s$^2$, so tilting the rope costs acceleration but relieves the ground, which is exactly the trade a mover makes.

d) Pushing downward at $30^{\circ}$ reverses the sign of the vertical component: $N = mg + T_{y} = 215.6 + 42.5 = 258.1$ N. Same rope, same angle, same magnitude, and a normal force that differs by 85 N between the two cases, which is why the normal force can never be read off a formula and must come from the vertical equation every time. To lift the sled the vertical equation would need $T\sin 30^{\circ} \ge mg$, that is $T \ge \dfrac{215.6}{0.5} = 431$ N. With 85 N the sled stays firmly on the ground.

a) At rest the acceleration is zero, so $\Sigma F_{y} = 0$ gives $N = mg = 62 \times 9.80 = 607.6$ N. At a constant $3.0$ m/s the acceleration is STILL zero, so the reading is the same $607.6$ N. This is the first law in its most useful form: constant velocity and rest are the same mechanical state, and the speed itself never enters a force equation. A reading that depended on the speed would be a reading that contradicted the first law.

b) Taking up as positive, the two forces on the student are $N$ up and $mg$ down, and the acceleration is $+1.30$ m/s$^2$: $N - mg = ma$, so $N = m(g + a) = 62 \times (9.80 + 1.30) = 62 \times 11.10 = 688.2$ N. The student feels heavier by 80.6 N, about 8 kg on a dial calibrated in kilograms, because the scale now has to do two jobs, hold the student up AND accelerate them upward. The weight is still 607.6 N: nothing about the student changed.

c) Moving up while slowing down means the acceleration points DOWN, so $a = -1.60$ m/s$^2$ with up positive: $N = m(g + a) = 62 \times (9.80 - 1.60) = 62 \times 8.20 = 508.4$ N. This is the step that separates the students who draw an arrow for the acceleration from those who guess: going up and speeding up, or going down and slowing down, both give a reading above the weight; the other two combinations give a reading below it. The direction of the velocity is irrelevant, only the direction of the acceleration counts.

d) In free fall $a = -g$, so $N = m(g - g) = 0$ N. The scale reads zero and the student floats inside the cabin. The weight has not vanished, it is still 607.6 N and it is precisely what makes the student fall with the cabin. Apparent weightlessness in a spacecraft is this same situation, an object and its support falling together, and not the absence of gravity. General formula worth memorising: $N = m(g + a)$ with $a$ counted positive upward, which returns all four answers of this exercise.

a) The weight is $W = mg = 12 \times 9.80 = 117.6$ N, straight down. With the $x$ axis pointing down the slope and the $y$ axis perpendicular to it, the weight is the only force that is oblique to these axes. The angle between the weight and the perpendicular equals the angle of the incline, so $W_{\parallel} = mg\sin 28^{\circ} = 117.6 \times 0.4695 = 55.2$ N and $W_{\perp} = mg\cos 28^{\circ} = 117.6 \times 0.8829 = 103.8$ N. Sine or cosine is decided once, by a limiting case: a flat surface, $\theta = 0$, must give zero pull along the slope, and $\sin 0 = 0$, so the along-slope component carries the sine.

b) Perpendicular to the slope the block neither sinks into the surface nor lifts off it, so $\Sigma F_{y} = 0$ and $N = W_{\perp} = 103.8$ N. Writing $N = mg = 117.6$ N is wrong by 13.8 N, and the reason is geometric, not numerical: on an incline the surface no longer has to hold the whole weight, because part of the weight is aimed along the surface and the surface does nothing about it. The steeper the slope, the smaller the normal force, and at $90^{\circ}$ it is zero.

c) Along the slope only $W_{\parallel}$ acts, since the surface is frictionless and the normal force is perpendicular: $\Sigma F_{x} = mg\sin 28^{\circ} = ma$, so $a = \dfrac{55.2}{12} = 4.60$ m/s$^2$ down the slope. Order of magnitude check: $4.60$ is less than $9.80$, as it must be, since a block on a slope falls more slowly than a block dropped in the air, and the ratio $\dfrac{4.60}{9.80} = 0.47$ is exactly $\sin 28^{\circ}$.

d) Dividing the along-slope equation by $m$ gives $a = g\sin\theta$: the mass cancels because both the driving force and the inertia are proportional to it. A 2 kg block and a 200 kg block released together on the same slope stay side by side. For the speed, the motion is straight with constant acceleration, so $v^{2} = v_{0}^{2} + 2a\Delta x = 0 + 2 \times 4.60 \times 3.0 = 27.6$, and $v = 5.25$ m/s. The equations of motion of the first chapters are reused here without change: the dynamics supplies $a$, the kinematics supplies the rest.

a) The block on the table carries three forces, its weight 39.2 N down, the normal force 39.2 N up, the tension $T$ horizontal toward the pulley. It moves horizontally, so $\Sigma F_{x} = T = m_{1}a$, and vertically $N - m_{1}g = 0$. The hanging block carries two forces, its weight $m_{2}g = 29.4$ N down and the tension $T$ up. Taking down as positive for that block, $m_{2}g - T = m_{2}a$. Two bodies, two diagrams, two equations, and the two unknowns $a$ and $T$ appear in both.

b) Adding the two equations eliminates the tension, which is why the equations are written before anything is substituted: $m_{2}g = (m_{1} + m_{2})a$, so $a = \dfrac{m_{2}g}{m_{1} + m_{2}} = \dfrac{3.0 \times 9.80}{7.0} = 4.20$ m/s$^2$. The result reads well: the hanging weight drives the motion, the total mass resists it. With a very heavy table block the acceleration would tend to zero, with a very light one it would tend to $g$, and both limits are correct.

c) Substituting into the first equation, $T = m_{1}a = 4.0 \times 4.20 = 16.8$ N. Verify with the other equation, which was not used to find $T$: $m_{2}g - T = 29.4 - 16.8 = 12.6$ N and $m_{2}a = 3.0 \times 4.20 = 12.6$ N, they agree. Using one equation to solve and the other to check is free and it catches sign errors that no amount of rereading will.

d) If $T$ were $29.4$ N, the hanging block would have $m_{2}g - T = 0$, so zero acceleration, so it would hang motionless. But then the table block would also be motionless while a 29.4 N tension pulled it along a frictionless table, which contradicts the second law. The tension equals the hanging weight only when the system is in equilibrium, for instance if someone holds the blocks. As soon as the system accelerates, $T < m_{2}g$ strictly, here 16.8 N against 29.4 N, and that inequality is the fastest check on the whole problem.

1) TRUE. Constant velocity means zero acceleration, and the second law then gives $\Sigma \vec{F} = m\vec{a} = \vec{0}$. Careful with the converse in everyday language: zero net force does not mean no force at all, it means the forces balance. A crate sliding at constant speed on a frictionless floor has a weight of several hundred newtons and a normal force of the same size; their sum is zero.

2) FALSE. The normal force is whatever the perpendicular equation requires. Corrected: the normal force equals the weight only when the surface is horizontal, the object has no vertical acceleration, and no other force has a vertical component. Three counterexamples from this set: on the $28^{\circ}$ incline of exercise 6, $N = mg\cos 28^{\circ} = 103.8$ N instead of 117.6 N; with a rope pulling at $30^{\circ}$ in exercise 4, $N = 173.1$ N instead of 215.6 N; in the accelerating elevator of exercise 5, $N = 688.2$ N instead of 607.6 N.

3) FALSE. Corrected: the two forces of a pair act on two different bodies, so they never appear in the same equation and cannot cancel. Only forces on the SAME body can cancel. In exercise 3 the 96 N on B accelerates B at $2.00$ m/s$^2$ and the 96 N on A accelerates A at $1.33$ m/s$^2$; both skaters move, and the pair is the reason.

4) FALSE. Corrected: $m\vec{a}$ is not a force and has no source, it is the RESULT of the sum of the real forces. The diagram carries only forces with an identifiable source, contact or field. Drawing $m\vec{a}$ leads to writing something like $F - mg - ma = 0$, which counts the acceleration twice and turns a correct list of forces into a wrong equation. The acceleration belongs beside the diagram, as an arrow clearly marked as an acceleration, never among the forces.

5) TRUE, and it is worth knowing what ideal means: the pulley has no mass, so no net torque is needed to spin it, and its axle is frictionless. It changes the direction of the tension and nothing else. In exercise 7 that is what lets a single symbol $T$ appear in the equation of the table block and in the equation of the hanging block. A real pulley with mass would give a larger tension on the side that pulls harder, which is a topic of the rotation chapter.

a) The hanging mass is in equilibrium, so the cord holding it pulls up with exactly its weight: $T = mg = 3.2 \times 9.80 = 31.4$ N. The pulleys are ideal, so this same 31.4 N is the tension everywhere in the cord, including in the two sections that reach the sling. That is the whole point of the arrangement: a hanging mass is a tension you can read off a label, and changing the pull means changing the mass.

b) Put the $x$ axis along the leg, pointing toward the pulleys, and the $y$ axis perpendicular to it. Each section contributes $T\cos 25^{\circ} = 31.4 \times 0.9063 = 28.4$ N along $x$, and $\pm T\sin 25^{\circ} = \pm 13.3$ N along $y$. The two perpendicular components are opposite and cancel, which is exactly what the symmetry is for, so the resultant is $2T\cos 25^{\circ} = 56.8$ N along the axis of the leg. Since the leg is in equilibrium, the hip and the body supply 56.8 N in the opposite direction: that is the stretch the treatment is after.

c) Adding 31.4 and 31.4 treats the two tensions as if they pulled in the same direction. They do not, they make an angle of $50^{\circ}$ with each other, so part of each one is spent pulling sideways and is cancelled by its neighbour. The vector sum is 56.8 N, not 62.7 N, an overestimate of 9.4 percent, and the gap grows fast with the angle. Forces add head to tail, never by adding magnitudes, unless they are parallel.

d) Set $2mg\cos 25^{\circ} = 65$, so $m = \dfrac{65}{2 \times 9.80 \times 0.9063} = 3.66$ kg. Opening the angle to $40^{\circ}$ with the original 3.2 kg gives $2 \times 31.4 \times \cos 40^{\circ} = 48.0$ N, a loss of 8.8 N, because a wider V spends more of each tension sideways. The geometry of the frame is therefore part of the prescription, not a detail of installation: the same mass with a different angle is a different treatment.

a) The weight is $mg = 78 \times 9.80 = 764.4$ N. Along the ramp it contributes $mg\sin 15^{\circ} = 764.4 \times 0.2588 = 197.8$ N, directed down the slope. Constant speed means zero acceleration, so the along-ramp equation is $T - mg\sin 15^{\circ} = 0$ and $T = 198$ N. That is about a quarter of the weight, which is precisely why a ramp exists: rolling the stretcher up costs 198 N instead of the 764 N needed to lift it straight into the ambulance.

b) Nothing changes in the diagram, only the right-hand side of the equation: $T - mg\sin 15^{\circ} = ma$, so $T = 197.8 + 78 \times 0.45 = 197.8 + 35.1 = 233$ N. The extra 35 N is the price of the acceleration, about 18 percent more than the constant-speed pull. The structure of the calculation is worth keeping: the equilibrium term first, the $ma$ term added on top, so that a sign error on $a$ is visible at once.

c) Perpendicular to the ramp the stretcher neither rises nor sinks, so $N = mg\cos 15^{\circ} = 764.4 \times 0.9659 = 738$ N. Writing $N = mg = 764$ N would be wrong by 26 N here; on a $15^{\circ}$ ramp the error is small, which is exactly what makes the habit dangerous, since the same reflex gives 117.6 N instead of 103.8 N on the $28^{\circ}$ incline of exercise 6, and much worse on a steeper slope.

d) A horizontal cable is oblique to the tilted axes, so it splits too. Along the ramp, $T\cos 15^{\circ} = mg\sin 15^{\circ}$, which gives $T = mg\tan 15^{\circ} = 764.4 \times 0.2679 = 205$ N, slightly more than before, since only part of the horizontal pull works along the slope. Perpendicular, the cable now also presses the stretcher onto the ramp: $N = mg\cos 15^{\circ} + T\sin 15^{\circ} = 738 + 53 = 791$ N. So the horizontal arrangement costs more tension AND loads the ramp more, and on a rough ramp it would be worse still, since a larger normal force means more friction. Pulling parallel to the surface is the efficient choice, and the two equations say so without any hand waving.