4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A door can rotate about its hinges. A student pushes on the door handle with a force directed along the surface of the door, straight toward the hinges. Which statement about the torque that this force exerts on the door about the hinges is correct?
Answer and reasoning
AIt equals the force's magnitude times the handle's distance from the hinges. A student who thinks the whole force contributes, whatever its direction, multiplies the force by the distance. That product is the torque only for a force perpendicular to the door; a force directed along the door toward the hinges has no perpendicular component and exerts no torque.
BIt is zero, because the force has no component perpendicular to the door.Correct Torque comes only from the component of a force perpendicular to the position vector from the axis to the point of application. This force points along the door, parallel to that position vector, so its perpendicular component is zero and so is the torque: the push presses the door against its hinges without turning it.
CIt equals the force's magnitude, because a torque is a turning force. A student who treats torque as simply a turning force picks this. Torque and force are different quantities with different units (N·m and N), and a force directed along the line through the axis, as this one is, exerts no torque however large it is.
DIt is larger than it would be for the same push exerted nearer the hinges. A student who thinks a force exerted farther from the axis always exerts a greater torque picks this. Distance matters only through the perpendicular component of the force; this push has none, so it exerts zero torque wherever along the door it is exerted.
The diagram shows a rigid bar pivoted at O and a horizontal force of 50 N exerted on the bar at its end P. What is the magnitude of the torque that this force exerts on the bar about O?
Answer and reasoning
A30 N·mCorrect The lever arm is the perpendicular distance from O to the line of action of the force. The force is horizontal, so its line of action is the horizontal line through P, 0.60 m above O: τ = 50 N × 0.60 m = 30 N·m. The same result follows from τ = rF sin θ with r = 1.0 m and sin θ = 0.60.
B40 N·m A student who always uses the horizontal distance from the pivot, as in seesaw problems with vertical weights, gets 50 × 0.80 = 40 N·m. The lever arm must be perpendicular to the line of action; for a horizontal force that is the vertical distance, 0.60 m.
C50 N·m A student who takes the lever arm to be the distance from O to P, √(0.80² + 0.60²) = 1.0 m, gets 50 N·m. That would be right only if the force were perpendicular to the bar. The lever arm is the perpendicular distance from O to the force's line of action, 0.60 m.
D70 N·m A student who adds the horizontal and vertical distances, 0.80 m + 0.60 m = 1.4 m, gets 70 N·m. Perpendicular distances do not add as ordinary numbers, and only the distance perpendicular to the line of action, 0.60 m, is the lever arm.
Working The line of action of the horizontal force is the horizontal line through P. The lever arm is the perpendicular distance from O to that line, which is the vertical distance, 0.60 m. τ = (50 N)(0.60 m) = 30 N·m. Check with τ = rF sin θ: r = √(0.80² + 0.60²) = 1.0 m, and the angle θ between the bar and the horizontal force has sin θ = 0.60/1.0 = 0.60, so τ = (1.0 m)(50 N)(0.60) = 30 N·m.
A uniform rod is held horizontal by a hinge at its left end and a vertical string attached to its right end. A student is drawing a force diagram to analyze the torques exerted on the rod about the hinge. Which instruction for drawing the diagram is correct?
Answer and reasoning
ADraw every force on the rod from a single dot at the rod's center of mass. A student who carries over the free-body diagram convention draws every force from the center of mass. That hides what is needed for torques: the string's force and the hinge's force are exerted at the ends of the rod, and their distances from the hinge matter.
BInclude the forces that the rod exerts on the hinge and on the string. A student who lists every force present at a contact picks this. The rod's push on the hinge and its pull on the string are exerted on the hinge and on the string, not on the rod; only forces exerted on the rod belong in its diagram.
CLeave out the hinge's force, since it exerts no torque about the hinge. A student who drops forces that exert no torque about the chosen axis picks this. The hinge's force does exert zero torque about the hinge, but it is still a force exerted on the rod, and a force diagram shows every force exerted on the system.
DStart each force's arrow where that force is exerted on the rod.Correct A force diagram, like a free-body diagram, shows each force exerted on the rod by another object, with its relative magnitude and direction. It also shows where each force is exerted: the weight at the rod's center, the string's force at the right end and the hinge's force at the left end, because those locations determine the torques.
A student pushes horizontally on a door at its handle, with the force directed at 60° to the surface of the door. She then pushes at the same point with a force of the same magnitude directed at 30° to the surface of the door. The magnitude of the torque about the hinges in the second push is how many times its magnitude in the first push?
Answer and reasoning
A0.58Correct The angle between the force and the door is the angle θ between the force and the position vector from the hinges, so the torque is proportional to sin θ when r and F are fixed. The ratio is sin 30°/sin 60° = 0.50/0.87 ≈ 0.58: less of the force is perpendicular to the door in the second push.
B1.73 A student who uses cos θ with the given angles gets cos 30°/cos 60° = 0.87/0.50 ≈ 1.73. The angles are measured from the door, which lies along the position vector from the hinges, so sin θ is needed; cosine would apply only to angles measured from the perpendicular to the door.
C0.50 A student who takes the torque to be proportional to the angle itself gets 30°/60° = 0.50. The torque is proportional to sin θ, and sin 30° is more than half of sin 60°, so the ratio is about 0.58.
D1.00 A student who thinks a force's direction does not affect its torque picks 1.00, because the force's magnitude and the point where it is exerted are unchanged. Only the component of the force perpendicular to the door exerts a torque, and that component is smaller at 30° than at 60°.
Working The position vector from the hinges to the handle lies along the door, so each given angle is θ in τ = rF sin θ. r and F are unchanged, so τ₂/τ₁ = sin 30°/sin 60° = 0.500/0.866 = 0.58.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.3.A.1 Axis of rotation Fix
Axis of rotation
The line about which a rigid system rotates or would rotate. A torque is always stated about a particular axis: the same force can exert different torques about different axes.
Position vector from the axis (r⃗)
The vector from the axis of rotation to the point where a force is exerted. Its magnitude r is that distance (SI unit: m).
Perpendicular component of a force (F⊥)
The component of a force perpendicular to the position vector from the axis, F⊥ = F sin θ. Only this component produces a torque; the component along the position vector pushes or pulls along the line through the axis and produces none.
Students often think Torque is the same thing as force, so the larger force always exerts the larger torque, whatever its location or direction. In fact No. Torque depends on the force, on where it is exerted relative to the axis and on its direction: τ = rF sin θ. A small force far from the axis can exert a larger torque than a large force near it, and a force exerted at the axis exerts no torque at all.
Students often think The whole force contributes to the torque whatever its direction, so a force F exerted a distance r from the axis always exerts a torque of magnitude rF. In fact No. Only the component of the force perpendicular to the position vector from the axis contributes, so τ = rF sin θ. A force directed along the line through the axis exerts no torque, however large it is.
5.3.A.2 Line of action Fix
Line of action
The straight line through a force vector, extended in both directions. Moving a force's point of application along its line of action does not change its torque about a given axis.
Lever arm
The perpendicular distance from the axis of rotation to the line of action of a force (SI unit: m). The torque magnitude equals the force magnitude times the lever arm, which equals r sin θ.
Students often think The lever arm of a force is the distance from the axis to the point where the force is exerted, whatever the force's direction. In fact Only when the force is perpendicular to the position vector. In general the lever arm is the perpendicular distance from the axis to the force's line of action, r sin θ, which is shorter than r for a force at an angle.
Students often think The lever arm is always the horizontal distance from the axis to the point where the force is exerted. In fact No. The lever arm is measured perpendicular to the force's line of action. For a vertical force that is the horizontal distance; for a horizontal force it is the vertical distance; for a force in another direction it is neither.
5.3.B.1 Force diagram Fix
Force diagram
A diagram of a rigid system drawn as an extended object with its axis of rotation marked, showing each force exerted on the system as an arrow that starts where the force is exerted, with arrow lengths showing relative magnitudes.
Force diagram compared with a free-body diagram
Both show every force exerted on a system by its environment, with relative magnitudes and directions. A free-body diagram draws all the forces from one dot representing the center of mass, which is enough for translational motion; a force diagram also shows where each force is exerted relative to the axis, which is needed to analyze torques.
Point of application
The location on a system where a force is exerted. Contact forces are exerted where the contact is; the gravitational force on a rigid system is modeled as exerted at its center of mass.
Students often think In a force diagram used to analyze torques, as in a free-body diagram, every force can be drawn from a single dot at the center of mass. In fact No. A force diagram shows the extended object and starts each force arrow where that force is exerted, because the location of each force relative to the axis determines its torque.
Students often think A force diagram of an object includes the forces that the object exerts on other objects, such as its push on a support or its pull on a string. In fact No. A force diagram, like a free-body diagram, shows only the forces exerted ON the system by other objects. The object's push on a support or pull on a string is exerted on that support or string.
5.3.B.2 Torque (τ) Fix
Torque (τ)
The turning effect of a force exerted on a rigid system about an axis of rotation. Its magnitude is τ = rF⊥ = rF sin θ, where r is the distance from the axis to the point where the force is exerted and θ is the angle between the force and the position vector from the axis to that point. SI unit: N·m. AP Physics 1 works with torque magnitudes; the direction of torque as a vector is beyond the course.
Angle θ in τ = rF sin θ
The angle between the force vector and the position vector from the axis to the point of application. The torque is greatest when θ = 90° and zero when θ = 0° or 180°, that is, when the force points along the line through the axis.
Students often think Any angle shown between the force and a line in the diagram can be used as θ in τ = rF sin θ; which line the angle is measured from does not matter. In fact No. θ is the angle between the force and the position vector from the axis to the point of application. If the given angle is measured from another line, such as the perpendicular to the rod, sin and cos swap roles.
Students often think The r in τ = rF sin θ is the length of the object, such as the whole rod, rather than the distance from the axis to the point where the force is exerted. In fact No. r is the distance from the axis to the point where this particular force is exerted. The gravitational force on a uniform rod is modeled as exerted at its center of mass, halfway along it.
5 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 5
The diagram shows a top view of a wrench on a bolt at O. Forces F₁, F₂, F₃ and F₄, all of the same magnitude, are exerted on the wrench one at a time, in the plane of the diagram, at the points shown. Let τ₁, τ₂, τ₃ and τ₄ be the magnitudes of the torques that these forces exert on the wrench about O. Which ranking of the torque magnitudes is correct?
Answer and reasoning
Aτ₁ = τ₂ = τ₄ > τ₃ A student who thinks the whole force contributes whatever its direction gives every force at the end of the handle the torque LF, and F₃ the torque 0.5LF. Only the component perpendicular to the handle counts: F₂ gives 0.5LF and F₄, which points along the handle, gives nothing.
Bτ₁ = τ₂ = τ₃ = τ₄ A student who takes torque to be the same as force ranks the torques as equal because the forces are equal. The torques also depend on where each force is exerted and on its direction; they range from LF for F₁ down to zero for F₄.
Cτ₁ > τ₂ = τ₃ > τ₄Correct F₁ is perpendicular to the handle at distance L, so τ₁ = LF. F₂ is also exerted at distance L, but only its perpendicular component, F sin 30° = 0.5F, turns the wrench, so τ₂ = 0.5LF. F₃ is perpendicular but exerted at L/2, so τ₃ = 0.5LF too. F₄ points along the handle, away from O, and has no perpendicular component, so τ₄ = 0.
Dτ₁ > τ₃ > τ₂ = τ₄ A student who thinks only a force perpendicular to the handle exerts a torque gives F₂ zero torque, like F₄. F₂ is at 30° to the handle, and its perpendicular component, 0.5F, exerts a torque of 0.5LF about O, equal to τ₃.
Working Let each force have magnitude F and the handle length be L. τ₁ = LF sin 90° = LF. τ₂ = LF sin 30° = 0.5LF. τ₃ = (L/2)F sin 90° = 0.5LF. τ₄ = LF sin 0° = 0, because F₄ is parallel to the position vector from O. So τ₁ > τ₂ = τ₃ > τ₄.
Two students push, one at a time, on the same door, which can rotate about its hinges. Both forces are horizontal and have the same magnitude. Student 1 pushes perpendicular to the door at a point 0.20 m from the hinges. Student 2 pushes at the outer edge of the door, 0.80 m from the hinges, with a force directed at 30° to the surface of the door. Which claim about the magnitudes of the torques about the hinges, with its reasoning, is correct?
Answer and reasoning
AStudent 2's is four times as large, as student 2 pushes four times as far from the hinges. A student who takes the lever arm to be the distance to the point where the force is exerted compares 0.80 m with 0.20 m and gets a factor of four. Student 2's force is not perpendicular to the door, so its lever arm is only 0.80 m × sin 30° = 0.40 m.
BOnly student 1's push exerts a torque, as only it is perpendicular to the door. A student who thinks only a perpendicular force exerts a torque gives student 2 none. Student 2's force has a component perpendicular to the door, F sin 30° = 0.5F, exerted 0.80 m from the hinges, so it exerts a torque of 0.40F, twice student 1's 0.20F.
CThey are equal, as the two students push with forces of equal magnitude. A student who treats torque as the same thing as force compares only the force magnitudes. Torque also depends on the lever arm, which is 0.20 m for student 1 and 0.40 m for student 2, so the torques are not equal.
DStudent 2's is twice as large, as its line of action is twice as far from the hinges.Correct The lever arm is the perpendicular distance from the hinges to the line of action. For student 1 it is 0.20 m. For student 2 it is 0.80 m × sin 30° = 0.40 m, twice as long, so with equal force magnitudes student 2's torque is twice student 1's.
Working Lever arm for student 1: 0.20 m, because the force is perpendicular to the door. Lever arm for student 2: 0.80 m × sin 30° = 0.40 m. With equal force magnitudes F, τ₁ = 0.20F and τ₂ = 0.40F, so student 2's torque is twice student 1's.
The diagram shows four forces, A, B, C and D, exerted on a rod that can rotate about a fixed axis through point P, perpendicular to the page. The arrows are drawn to scale and start where each force is exerted. Which force exerts the torque of greatest magnitude about P?
Answer and reasoning
AForce A A student who equates torque with force picks the largest force. A is exerted at P itself, so its distance from the axis is zero and its torque about P is zero, however large the force.
BForce DCorrect D is perpendicular to the rod and exerted 2.5 m from P, so its torque is 2.5 m × 20 N = 50 N·m. That is more than B's (4.0 m × 10 N = 40 N·m), C's (3.0 m × 20 N × sin 30° = 30 N·m) or A's, which is zero because A is exerted at the axis.
CForce B A student who thinks the force exerted farthest from the axis always exerts the greatest torque picks B. B is small, 10 N, so its torque is 4.0 m × 10 N = 40 N·m, less than D's 50 N·m.
DForce C A student who uses the whole of C, ignoring its direction, gets 3.0 m × 20 N = 60 N·m and picks C. C is at 30° to the rod, so only its perpendicular component, 20 N × sin 30° = 10 N, exerts a torque: 30 N·m, less than D's 50 N·m.
Working τ = rF sin θ for each force. A: r = 0, so τ = 0. B: (4.0 m)(10 N)(sin 90°) = 40 N·m. C: (3.0 m)(20 N)(sin 30°) = 30 N·m. D: (2.5 m)(20 N)(sin 90°) = 50 N·m. D exerts the greatest torque.
A rod of length 0.60 m can rotate about an axle through one end. A string is attached to the rod 0.40 m from the axle and pulls on it with a force of 50 N directed at 37° to the rod. What is the magnitude of the torque exerted on the rod by the string's force about the axle? Use sin 37° = 0.60 and cos 37° = 0.80.
Answer and reasoning
A16 N·m A student who uses cos 37° gets 0.40 × 50 × 0.80 = 16 N·m. θ is the angle between the force and the position vector along the rod, which is the 37° given, so sin θ is needed; the cosine gives the component along the rod, which exerts no torque.
B18 N·m A student who takes r to be the length of the rod gets 0.60 × 50 × 0.60 = 18 N·m. r is the distance from the axle to the point where this force is exerted, 0.40 m, not the rod's length.
C12 N·mCorrect τ = rF sin θ with r = 0.40 m, the distance from the axle to where the string is attached, and θ = 37°, the angle between the force and the rod: 0.40 × 50 × 0.60 = 12 N·m.
D30 N·m A student who takes the torque to be the perpendicular component of the force, 50 N × 0.60 = 30 N, picks this. That quantity is a force, measured in newtons; it must be multiplied by the distance from the axle, 0.40 m, to give the torque, 12 N·m.
Working r is the distance from the axle to the point of application, 0.40 m, and θ is the angle between the force and the rod (the position vector from the axle), 37°. τ = rF sin θ = (0.40 m)(50 N)(0.60) = 12 N·m.
A uniform rod of mass M and length L hangs from a horizontal axle through one end. At one instant the rod makes an angle of 37° with the vertical. Which expression gives the magnitude of the torque exerted on the rod by the gravitational force about the axle at that instant? Use sin 37° = 0.60 and cos 37° = 0.80.
Answer and reasoning
A0.30MgLCorrect The gravitational force on a uniform rod is modeled as exerted at its center of mass, L/2 from the axle. The angle between the rod and the vertical force is 37°, so τ = (L/2)(Mg)(sin 37°) = (L/2)(Mg)(0.60) = 0.30MgL.
B0.60MgL A student who takes r to be the full length of the rod gets L × Mg × 0.60 = 0.60MgL, as if the rod's weight were exerted at its free end. For a uniform rod the gravitational force is modeled as exerted at the center of mass, L/2 from the axle.
C0.50MgL A student who uses the whole gravitational force, ignoring its direction relative to the rod, gets (L/2)(Mg) = 0.50MgL. That would be the torque only if the rod were horizontal; at 37° to the vertical only the component Mg sin 37° is perpendicular to the rod.
D0.40MgL A student who uses cos 37° gets (L/2)(Mg)(0.80) = 0.40MgL. The 37° is measured between the rod, which lies along the position vector from the axle, and the vertical gravitational force, so it is θ itself and sin θ is needed.
Working The gravitational force, of magnitude Mg, is modeled as exerted at the rod's center of mass, a distance L/2 from the axle along the rod. The position vector from the axle to the center of mass lies along the rod, at 37° to the vertical, and the gravitational force is vertical, so θ = 37°. τ = rF sin θ = (L/2)(Mg)(0.60) = 0.30MgL.
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
