Torque - AP Physics 1

1. A rigid bar pivots about a point at its center. Two equal forces $F$ are applied at the right end: case 1 is upward (perpendicular to the bar), case 2 is downward (also perpendicular). The lever arm length is the same in both cases. Compared to case 1, the torque in case 2 is
   Answer: opposite direction and same magnitude This question tests understanding torque direction and magnitude in AP Physics 1. Torque is the product of force and perpendicular lever arm, with direction based on rotation sense. Both cases have equal F and lever arm, yielding equal magnitude torques. However, upward force in case 1 produces one direction, while downward in case 2 reverses it. Choice D misleads by claiming zero torque due to central pivot, but applied forces still generate torque. Use vector cross product thinking (τ = r × F) to determine both magnitude and direction consistently.
2. A sign is mounted on a horizontal beam that pivots about a hinge at the wall (pivot). A worker pulls straight down with $50\,\text{N}$ on a rope attached $0.60\,\text{m}$ from the hinge. The rope is vertical while the beam is horizontal, so the force is qualitatively perpendicular to the beam and the lever arm is $0.60\,\text{m}$. What is the torque magnitude about the hinge from this pull?
   Answer: $30\,\text{N}\cdot\text{m}$ This question tests direct calculation of torque magnitude. Torque is the product of force and perpendicular lever arm: τ = F × r⊥. With a vertical rope pulling on a horizontal beam, the force is perpendicular to the beam, so the lever arm equals the distance from hinge to rope attachment: r⊥ = 0.60 m. Therefore, τ = 50 N × 0.60 m = 30 N·m. Choice C incorrectly uses 50 N·m, perhaps confusing force magnitude with torque. The calculation strategy: when force is perpendicular to the lever arm, simply multiply force magnitude by distance from pivot to get torque magnitude.
3. A uniform beam pivots about a pin at its right end. A force $F$ is applied at the left end. The force's line of action is vertical upward, and the beam is horizontal so the force is perpendicular; the lever arm equals the beam's full length. Compared with applying the same $F$ at the beam's midpoint with the same direction, the torque magnitude is
   Answer: larger, because the lever arm is longer 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.
4. A seesaw pivots at its center. A student pushes straight down with force $F$ at a point close to the pivot, so the line of action is vertical and the lever arm is short. Another student pushes straight down with the same $F$ at the end, giving a longer lever arm. Which push produces the larger torque magnitude?
   Answer: The push at the end, because the lever arm is larger. 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.
5. A uniform horizontal door is hinged at its left edge (pivot). A student pushes on the door at its outer right edge. The force's line of action is vertical upward, and the push is applied perpendicular to the door's plane so the lever arm equals the door's full width. Which direction is the torque about the hinge?
   Answer: Counterclockwise 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.
6. A wrench pivots about the center of a bolt. A force is applied at the end of the handle. The line of action is perpendicular to the handle, and the lever arm is $0.20\,\text{m}$. If the same force is applied at a point closer to the bolt so the lever arm is $0.10\,\text{m}$, how does the torque magnitude change?
   Answer: It halves. 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.
7. A meterstick is pivoted at its left end on a frictionless pin. A student applies a force $F$ downward at the 1.0 m mark; the line of action is vertical through that point, so the lever arm equals 1.0 m. Which direction is the torque about the pivot?
   Answer: Clockwise. This question tests understanding of torque direction using the right-hand rule. Torque is calculated as force times perpendicular lever arm, with direction determined by the rotation the force would cause. When a downward force is applied at the 1.0 m mark with the pivot at the left end, the lever arm is 1.0 m and the force would rotate the stick clockwise (viewed from above). Using the right-hand rule: curl fingers in the rotation direction (clockwise), and the thumb points into the page, indicating clockwise torque. Choice C incorrectly assumes downward forces produce zero torque, confusing force direction with torque existence. To determine torque direction, visualize which way the object would rotate and apply the right-hand rule consistently.
8. A door pivots about a hinge. A student applies a force at a point $0.60\,\text{m}$ from the hinge. The force is horizontal and directed so its line of action is perpendicular to the door, producing a nonzero lever arm. If the student reverses the force direction (same magnitude, opposite direction) at the same point, what happens to the torque direction about the hinge?
   Answer: It reverses. This question assesses the concept of torque in rotational dynamics, investigating the effect of reversing force direction on a door. Torque is the product of force and perpendicular lever arm, with direction (clockwise or counterclockwise) depending on the force's orientation. Initially, the perpendicular force produces torque in one direction based on its push or pull. Reversing the force keeps the magnitude and lever arm the same but flips the direction, reversing the torque. Choice A is incorrect because it misses how force direction determines torque direction, not just the lever arm. A transferable strategy is to consider torque as a vector quantity and assess how changes in force vector affect both magnitude and direction.
9. A horizontal door rotates about a vertical hinge at its left edge (pivot). A student pushes with a $10\,\text{N}$ force at the doorknob, $0.80\,\text{m}$ from the hinge. The force's line of action is perpendicular to the door (qualitatively: "at right angles" to the door's surface), so the lever arm equals $0.80\,\text{m}$. Which direction is the torque about the hinge (viewed from above)?
   Answer: Counterclockwise This question tests understanding of torque direction. Torque is the product of force and the perpendicular lever arm (distance from pivot to force's line of action), calculated as τ = F × r⊥. Here, the force is 10 N applied perpendicular to the door at 0.80 m from the hinge, giving τ = 10 N × 0.80 m = 8 N·m. When viewed from above, pushing on a door handle away from the hinge causes counterclockwise rotation. Choice C incorrectly assumes torque depends only on force magnitude, ignoring the crucial role of lever arm distance. To determine torque direction, use the right-hand rule: point fingers along the lever arm toward the force, curl them in the force's direction, and your thumb points along the torque vector.
10. A rigid bar is pivoted at its left end. A force $F$ is applied at the right end, but the line of action is along the bar (qualitatively: parallel to the bar), passing through the pivot direction. The lever arm is therefore zero. What is the torque about the pivot due to this force?
    Answer: $\tau=0$. This question assesses the concept of torque in rotational dynamics, focusing on the role of the lever arm when a force is applied along a rigid bar. Torque is calculated as the product of the force and the perpendicular lever arm, which is the shortest distance from the pivot to the line of action of the force. Here, since the force is applied parallel to the bar and its line of action passes directly through the pivot, the perpendicular lever arm is zero. Therefore, the torque is zero, regardless of the force magnitude or the bar's length. Choice A is a common distractor because it mistakenly assumes that distance from the pivot alone produces torque without considering the perpendicular component. To solve similar problems, always visualize the line of action and check if it intersects the pivot to confirm a zero lever arm.
11. A bicycle pedal rotates about the crank axle (pivot). A rider applies a downward force at the pedal. When the crank arm is horizontal, the force's line of action is perpendicular to the crank (qualitatively "at $90^\circ$"), giving a maximum lever arm. When the crank arm is vertical, the force's line of action is along the crank (qualitatively "$0^\circ$"), giving zero lever arm. Compared to the horizontal position, the torque magnitude at the vertical position is
    Answer: zero This question tests understanding of how lever arm varies with rotation angle. Torque equals force times perpendicular lever arm: τ = F × r⊥. When the crank arm is horizontal and force is downward, they are perpendicular, giving maximum lever arm equal to the crank length. When the crank arm is vertical and force is still downward, they are parallel (aligned), making the perpendicular lever arm r⊥ = 0. Therefore, τ = F × 0 = 0, regardless of force magnitude. Choice B incorrectly assumes torque depends only on force, ignoring that parallel force and lever arm produce zero torque. Remember: torque varies cyclically during pedaling, reaching zero twice per revolution when force and crank arm align.
12. A uniform meterstick pivots about its left end. A $6\,\text{N}$ force is applied at the $0.50\,\text{m}$ mark. The force's line of action is along the stick (qualitatively: "parallel to the stick"), so the perpendicular lever arm about the pivot is $0$. What is the torque magnitude about the pivot due to this force?
    Answer: $0\,\text{N}\cdot\text{m}$ This question tests understanding of zero torque conditions. Torque is calculated as τ = F × r⊥, where r⊥ is the perpendicular distance from the pivot to the force's line of action. When a force is applied along (parallel to) the meterstick, its line of action passes through the pivot, making the perpendicular lever arm r⊥ = 0. Therefore, τ = 6 N × 0 m = 0 N·m, regardless of where along the stick the force is applied. Choice A incorrectly multiplies force by distance without considering that the force is parallel to the stick. The key insight: forces whose lines of action pass through the pivot produce zero torque, no matter their magnitude.
13. A rectangular hatch pivots about a hinge along its top edge. A person pushes on the hatch with a force whose line of action passes directly through the hinge line (qualitatively: "aimed at the hinge"). The point of application is far from the hinge, but the perpendicular lever arm about the hinge is $0$. The torque about the hinge due to this force is
    Answer: zero 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.
14. A seesaw pivots at its center. A child pushes straight down at a point far from the pivot, with the line of action vertical and the lever arm equal to the horizontal distance to the pivot. The child then moves closer to the pivot while pushing with the same force. How does the torque magnitude about the pivot change?
    Answer: It decreases. This question assesses the concept of torque in rotational dynamics, examining how proximity to the pivot affects torque on a seesaw. Torque is the product of the downward force and the perpendicular lever arm, which is the horizontal distance from the pivot to the force application point. Initially far from the pivot, the lever arm is large, producing greater torque. Moving closer reduces the lever arm, decreasing the torque magnitude for the same force. Choice B distracts by focusing only on the constant force, neglecting the lever arm's variation. When solving, isolate the changing variable in the torque equation to predict how the torque will adjust accordingly.
15. A meterstick is pivoted at its center. A downward force is applied at the right end. The line of action is vertical, so the lever arm is the horizontal distance from the pivot. What is the direction of the torque about the pivot (viewed from above)?
    Answer: Clockwise. This question assesses the concept of torque in rotational dynamics, determining the direction of torque on a pivoted meterstick. Torque is the product of the force and the perpendicular lever arm, with direction determined by the right-hand rule or conventional clockwise/counterclockwise notation. The downward vertical force at the right end creates a lever arm equal to half the meterstick's length, perpendicular to the force's line of action. Viewed from above, this setup produces a clockwise torque as the right end tends to rotate downward. Choice C is a distractor because it incorrectly assumes downward forces cannot produce torque, ignoring the lever arm. Always use the right-hand rule or visualize the rotation to determine torque direction in similar scenarios.
16. A rigid bar is pivoted at its left end. A force $F$ is applied at the right end at a right angle to the bar; the line of action is perpendicular, so the lever arm equals the full bar length $L$. If the bar is replaced by a longer bar of length $2L$ and the same $F$ is applied perpendicular at its end, how does the torque magnitude about the pivot change?
    Answer: It doubles because the lever arm doubles. This question tests understanding of torque calculation. Torque equals force times perpendicular lever arm (τ = F × r), making it directly proportional to lever arm when force remains constant. Initially, with bar length L and perpendicular force F at the end, torque is τ₁ = F × L. When the bar length doubles to 2L with the same perpendicular force F at the new end, the lever arm doubles, giving τ₂ = F × 2L = 2(F × L) = 2τ₁. The torque doubles because lever arm doubles while force stays constant—torque scales linearly with lever arm distance. Choice B incorrectly suggests longer bars reduce torque, contradicting the fundamental principle that increased lever arm amplifies rotational effect. To increase torque without changing force, maximize the perpendicular distance from axis to force—this explains why longer wrenches make loosening bolts easier.
17. A horizontal beam is hinged to a wall at its left end. A force $F$ is applied at the right end straight upward; the line of action is vertical. The lever arm is the beam's full length. What is the direction of the torque about the hinge (viewed from the front) due to this force?
    Answer: Counterclockwise. This question tests understanding of torque direction. Torque is a vector quantity with magnitude F × r and direction determined by the right-hand rule: curl fingers from position vector to force vector, and the thumb points in the torque direction. For a horizontal beam hinged at the left with upward force F at the right end, the position vector points rightward from hinge to force application point, and the force points upward. Using the right-hand rule (fingers curl from right to up), the thumb points out of the page, which corresponds to counterclockwise rotation when viewed from the front. Choice C incorrectly assumes vertical forces cannot create torque, but torque depends on the perpendicular distance between force and axis, not force direction alone. Always use the right-hand rule to determine torque direction: position vector to force vector gives rotation sense.
18. A wrench turns a bolt about its center. A force $F$ is applied at the end of the wrench handle, but the force is directed along the handle toward the bolt, so the line of action passes through the pivot and the lever arm is $0$. What is the torque about the bolt?
    Answer: Zero, because the lever arm is $0$. 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.
19. A horizontal rod is pivoted at its left end. A force $F$ is applied at the right end straight upward; the line of action is vertical through the end, so the lever arm equals the rod length $L$. If the force is doubled to $2F$ at the same point and direction, the torque magnitude about the pivot becomes
    Answer: doubles because torque is proportional to force. 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.
20. A sign is mounted on a horizontal rod that pivots at the wall. A student pulls straight downward on the rod at a point $0.50\,\text{m}$ from the pivot. The line of action is vertical and the rod is horizontal, so the force is qualitatively perpendicular to the rod. If the student pulls with the same force at $0.25\,\text{m}$ from the pivot, what happens to the torque magnitude about the pivot?
    Answer: It becomes half as large. 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.

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