AP Physics 1 Quiz: Torque

What this quiz covers

This quiz focuses on Torque, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Zero, because the lever arm is $0$. (correct answer)
2. $F$ times the wrench length, because the force is applied at the end.
3. Nonzero, because any force at a distance causes torque.
4. It depends only on the bolt’s mass.

Explanation: This question tests understanding of torque when force is directed through the pivot. Torque equals force times the perpendicular lever arm (τ = F × r⊥), where the lever arm is the shortest distance from the pivot to the force's line of action. When force is applied along the wrench handle toward the bolt, the line of action passes directly through the pivot point, making the perpendicular distance (lever arm) equal to zero. Since torque equals F × 0 = 0, no rotation occurs regardless of force magnitude. Choice B incorrectly assumes any force at the handle's end creates torque, ignoring that direction matters. Remember that torque depends on both where and in what direction force is applied—forces directed through the pivot produce zero torque.

Question 2

1. greater, because the force is applied at the same point.
2. the same, because only $F$ matters.
3. smaller, because the perpendicular lever arm is smaller. (correct answer)
4. dependent only on the rod’s mass distribution.

Explanation: This question tests understanding of how force angle affects torque through the lever arm. Torque equals force times the perpendicular lever arm (τ = F × r⊥), where r⊥ is the shortest distance from pivot to the force's line of action. When force is applied at an oblique angle closer to along the rod than perpendicular, the perpendicular lever arm becomes much smaller than the distance to the application point. Compared to applying the same force perpendicular (which maximizes lever arm), the oblique force produces smaller torque magnitude. Choice B incorrectly ignores the angle's effect on lever arm, assuming only force magnitude matters. Always find the perpendicular distance from pivot to line of action—angled forces reduce the effective lever arm and thus the torque.

Question 3

1. decreases because the beam is longer.
2. stays the same because the force is unchanged.
3. increases because the lever arm increases. (correct answer)
4. depends only on the sign’s weight.

Explanation: This question tests understanding of how changing lever arm affects torque when force remains constant. Torque equals force times perpendicular lever arm (τ = F × r⊥), making torque directly proportional to lever arm when force is fixed. Moving the cable attachment point farther from the hinge increases the horizontal distance (lever arm) while the upward pull force stays constant, thus increasing the torque about the hinge. This principle explains why longer wrenches make loosening bolts easier—increased lever arm means more torque for the same applied force. Choice B incorrectly assumes torque depends only on force magnitude, ignoring the lever arm's contribution. When analyzing torque changes, consider both force magnitude and lever arm distance—doubling either doubles the torque.

Question 4

1. Counterclockwise.
2. Clockwise. (correct answer)
3. Zero, because the force is horizontal.
4. Cannot be determined without $R$.

Explanation: This question tests understanding of torque direction for tangential forces on rotating objects. Torque is force times perpendicular lever arm, with direction indicating rotation sense. A rightward tangential force at the disk's top, perpendicular to the radius, has lever arm equal to radius R and would rotate the disk clockwise when viewed from above. Applying the right-hand rule: curl fingers in the clockwise rotation direction, and the thumb points into the page, confirming clockwise torque. Choice C incorrectly claims zero torque for horizontal forces, confusing force direction with the existence of a lever arm. For circular motion problems, tangential forces always produce maximum torque (lever arm = radius), while radial forces produce zero torque.

Question 5

1. It increases. (correct answer)
2. It decreases.
3. It stays the same because the force magnitude is unchanged.
4. It becomes zero because the door’s mass does not change.

Explanation: This question assesses the concept of torque in rotational dynamics, specifically how the direction and point of application of a force affect the torque on a door. Torque is defined as the product of the applied force and the perpendicular distance from the pivot point to the line of action of the force, known as the lever arm. In the initial setup, the oblique angle results in a lever arm shorter than 0.80 m because only the component of the force perpendicular to the door contributes to the torque. When the force is applied perpendicularly, the lever arm becomes the full 0.80 m, maximizing the torque for the given force magnitude and distance. Choice C is incorrect because it ignores the change in the effective lever arm due to the angle of application. A useful strategy is to always calculate the perpendicular lever arm by considering the geometry of the force application to determine the torque accurately.

Question 6

1. Clockwise
2. Zero, because the door’s mass is unchanged
3. Counterclockwise (correct answer)
4. Cannot be determined without the door’s weight

Explanation: This question assesses the skill of determining the direction of torque in AP Physics 1. Torque is defined as the rotational equivalent of force, calculated as the product of the applied force and the perpendicular distance from the pivot point to the line of action of the force, known as the lever arm. In this scenario, the vertical upward force at the right edge creates a lever arm equal to the door's width, resulting in a torque that tends to rotate the door around the hinge. The direction is determined by the right-hand rule: pointing fingers in the direction of the force and curling toward the pivot indicates counterclockwise rotation. Choice A is incorrect because it suggests clockwise torque, which would occur if the force were downward instead. To analyze similar problems, always visualize the rotation direction by imagining the object's response to the force and confirm with the right-hand rule.

Question 7

1. half as large
2. unchanged
3. twice as large (correct answer)
4. four times as large

Explanation: This question evaluates scaling of torque with lever arm in AP Physics 1. Torque is the product of force and perpendicular lever arm, directly proportional to the lever arm when force is constant and perpendicular. Initially, τ = F r; moving to 2r doubles the lever arm. Thus, new torque is 2 F r, twice as large. Choice A distracts by halving instead, which would occur if r halved. Apply the formula τ = F d_perp systematically, isolating variables to see multiplicative effects.

Question 8

1. Counterclockwise
2. Clockwise (correct answer)
3. Zero, because the force is vertical
4. Cannot be determined without the rod’s length

Explanation: This question examines the skill of identifying torque direction in AP Physics 1. Torque is computed as the force multiplied by the perpendicular lever arm, with direction determined by the potential rotation. The downward force at the midpoint creates a lever arm half the rod's length, producing a torque that rotates the rod. Using the right-hand rule, the downward force to the right of the pivot results in clockwise torque. Choice A is incorrect as it would apply to an upward force, reversing the direction. Always define a consistent convention, like positive for counterclockwise, to determine torque signs in problems.

Question 9

1. It becomes four times as large.
2. It becomes half as large. (correct answer)
3. It is unchanged because the force is unchanged.
4. It depends only on the sign’s weight.

Explanation: This question assesses the concept of torque in rotational dynamics, exploring the effect of changing the point of force application on a sign's rod. Torque equals the force multiplied by the perpendicular lever arm, which here is the horizontal distance from the pivot to the force point since the force is vertical and perpendicular. Initially at 0.50 m, the torque is F times 0.50 m. Moving to 0.25 m halves the lever arm, thus halving the torque magnitude. Choice C is misleading as it focuses only on the unchanged force, disregarding the lever arm's role. For such problems, systematically apply the torque formula and note how each factor changes to predict outcomes.

Question 10

1. Student A’s, because $1.5\,\text{m}$ is larger
2. Student B’s, because $300\,\text{N}$ is larger
3. They are equal (correct answer)
4. Cannot be compared without the board’s mass

Explanation: This question tests calculating and comparing torque magnitudes. Torque equals force times perpendicular lever arm: τ = F × r⊥. For Student A: τ_A = 200 N × 1.5 m = 300 N·m (clockwise). For Student B: τ_B = 300 N × 1.0 m = 300 N·m (counterclockwise). Both torques have the same magnitude of 300 N·m, though they act in opposite rotational directions. Choice A incorrectly focuses only on distance, while choice B incorrectly focuses only on force magnitude. When comparing torques, always calculate the full product F × r⊥; neither factor alone determines which torque is larger.

Question 11

1. nonzero because the force is applied far from the hinge
2. nonzero because the hatch has mass
3. zero (correct answer)
4. maximum because the hinge is at an edge

Explanation: This question tests recognizing zero torque from forces through the pivot. Torque equals force times perpendicular lever arm: τ = F × r⊥, where r⊥ is the perpendicular distance from pivot to the force's line of action. When a force's line of action passes through the pivot point (hinge), the perpendicular lever arm r⊥ = 0, regardless of where the force is actually applied. Therefore, τ = F × 0 = 0, no matter how large the force or how far the application point is from the hinge. Choice A incorrectly focuses on application point distance rather than lever arm. Key principle: any force aimed directly at or away from the pivot produces zero torque about that pivot.

Question 12

1. $2Fr$
2. $Fr$
3. $0$ (correct answer)
4. Depends only on the rod’s mass

Explanation: This question tests understanding of net torque from multiple forces. Torque is a vector quantity with both magnitude and direction, calculated as τ = F × r⊥. Each force produces torque magnitude Fr since they're perpendicular to the rod at distance r. However, forces on opposite sides create opposite rotational effects: one clockwise, one counterclockwise. When calculating net torque, we assign opposite signs to opposite rotations, giving τ_net = Fr - Fr = 0. Choice A incorrectly adds magnitudes without considering direction. The principle for multiple forces: calculate each torque with proper sign for direction, then sum algebraically to find net torque.

Question 13

1. The push near the pivot, because it is closer to the support.
2. The push at the end, because the lever arm is larger. (correct answer)
3. They produce equal torque because the forces are equal.
4. Torque is larger for the heavier student regardless of where they push.

Explanation: This question tests understanding of how lever arm distance affects torque magnitude. Torque equals force times perpendicular lever arm (τ = F × r⊥), making it directly proportional to the distance from pivot to where force is applied (when force is perpendicular). The student pushing at the seesaw's end has a longer lever arm than the one pushing near the pivot, so despite equal forces, the end push produces larger torque magnitude. This explains why door handles are placed far from hinges—maximizing lever arm makes rotation easier. Choice C incorrectly assumes equal forces always produce equal torques, ignoring the lever arm's role. To maximize torque with given force, apply it as far from the pivot as possible and perpendicular to the line from pivot to application point.

Question 14

1. twice as large because the midpoint is closer to the center of mass.
2. the same because $F$ is the same.
3. half as large because the lever arm is half as long. (correct answer)
4. determined only by the door’s mass.

Explanation: This question tests understanding of how application point affects torque through lever arm changes. Torque equals force times perpendicular lever arm (τ = F × r⊥), so when the same perpendicular force F is applied at the door's midpoint instead of the handle, the lever arm becomes half the door width. Since torque is directly proportional to lever arm, halving the lever arm halves the torque magnitude, making the door harder to open. This explains why doorknobs are placed at the edge—maximizing distance from hinge maximizes mechanical advantage. Choice A incorrectly relates lever arm to center of mass position rather than distance from pivot. Remember that torque depends on where force is applied relative to the pivot, not relative to the object's center of mass.

Question 15

1. The torque is zero because the force is applied horizontally.
2. The torque magnitude is $F$ divided by the door width.
3. The torque magnitude is $F$ times the door width, producing an outward rotation. (correct answer)
4. The torque depends only on the door’s mass, not on where the force is applied.

Explanation: This question tests understanding of torque calculation. Torque is the product of force and the perpendicular distance from the pivot point to the line of action of the force (lever arm), expressed as τ = F × r⊥. When the student pushes perpendicular to the door at the handle (farthest from the hinge), the lever arm equals the full door width, giving torque magnitude F times the door width. The perpendicular push tends to rotate the door outward (opening it). Choice A incorrectly claims zero torque despite the non-zero lever arm, while choice B incorrectly divides instead of multiplying. To solve torque problems, always identify the pivot point, find the perpendicular distance to the force's line of action, and multiply by the force magnitude.

Question 16

1. unchanged because the lever arm is unchanged.
2. doubles because torque is proportional to force. (correct answer)
3. halves because the rod resists rotation.
4. dependent only on the rod’s mass.

Explanation: This question tests understanding of torque's linear dependence on force magnitude. Torque equals force times perpendicular lever arm (τ = F × r⊥), making it directly proportional to force when lever arm is constant. When the upward force at the rod's end doubles from F to 2F while maintaining the same vertical direction and application point, the lever arm remains L and the torque doubles from FL to 2FL. This linear relationship means doubling any factor in the torque equation doubles the result. Choice C incorrectly suggests the rod's resistance affects the mathematical relationship between force and torque. To find how torque changes, identify which factors change—if only force doubles while lever arm stays constant, torque doubles proportionally.

Question 17

1. larger for the steep-angle force because the perpendicular lever arm is larger. (correct answer)
2. larger for the shallow-angle force because it points more toward the pivot.
3. the same because the force magnitude and point of application are the same.
4. set only by the lever’s weight.

Explanation: This question tests understanding of how force angle affects torque through perpendicular lever arm. Torque equals force times perpendicular lever arm (τ = F × r⊥), where r⊥ depends on the angle between force direction and the line from pivot to application point. A steep-angle force nearly perpendicular to the lever creates a large perpendicular lever arm (close to full lever length), while a shallow-angle force nearly along the lever creates a small perpendicular lever arm. Despite equal force magnitudes at the same point, the steep-angle force produces larger torque due to its larger lever arm. Choice B incorrectly associates pointing toward the pivot with larger torque, reversing the relationship. Maximum torque occurs when force is perpendicular to the lever—remember that lever arm is the perpendicular distance, not just any distance.

Question 18

1. It doubles.
2. It is unchanged because the force is unchanged.
3. It halves. (correct answer)
4. It becomes zero because the wrench’s mass is unchanged.

Explanation: This question assesses the concept of torque in rotational dynamics, examining how changing the lever arm affects torque on a wrench. Torque is the product of the force and the perpendicular lever arm, where the lever arm is the distance from the pivot to the point of force application when the force is perpendicular. Initially, with a 0.20 m lever arm, the torque is F times 0.20 m. When the application point moves to 0.10 m, the lever arm halves, so the torque becomes half its original value for the same force. Choice B is incorrect as it overlooks the dependency of torque on the lever arm length, not just the force. A transferable strategy is to identify variables in the torque formula and assess how changes in one, like the lever arm, impact the result while others remain constant.

Question 19

1. Case 2, because the force is applied closer to the bolt’s direction
2. Case 1, because the lever arm is larger when the force is perpendicular (correct answer)
3. They are equal because the force magnitude is the same
4. Case 2, because torque depends only on the wrench’s mass

Explanation: This question tests the skill of understanding how lever arm affects torque in AP Physics 1. Torque equals the force times the perpendicular lever arm, which is maximized when the force is perpendicular to the line from the pivot to the application point. In case 1, the perpendicular application gives the full handle length as the lever arm, while case 2's shallow angle reduces it significantly. Thus, case 1 produces larger torque due to the greater lever arm. Choice C is misleading because it ignores the lever arm's role, assuming equal torques from equal forces alone. Remember to calculate the effective lever arm using sinθ to compare torques in angled force applications.

Question 20

1. the same, because $F$ is the same
2. smaller, because the beam’s mass is unchanged
3. larger, because the lever arm is longer (correct answer)
4. zero, because the force is upward

Explanation: This question assesses comparing torque magnitudes based on lever arm in AP Physics 1. Torque is the product of force and perpendicular lever arm, independent of the object's mass for applied forces. Applying F at the left end uses the full beam length as the lever arm, while the midpoint halves it. Therefore, the full-length application yields larger torque. Choice A distracts by suggesting torque remains the same due to constant F, ignoring the lever arm change. To solve efficiently, sketch the setup and measure lever arms before calculating torques.