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What is torque, how does it depend on force and lever arm, and why does it cause rotation?

Topic 5.3 Torque: calculate the torque produced by a force as tau = rF sin(theta), and identify the lever arm and the sense of rotation.

A focused answer to AP Physics 1 Topic 5.3, covering torque as the rotational effect of a force, the formula tau = rF sin(theta), the lever arm, the sense of rotation, and why where and how a force is applied matters, with full worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. What torque is
  3. The lever arm and the angle
  4. The sense of rotation
  5. Try this

What this topic is asking

The College Board (Topic 5.3) wants you to calculate the torque produced by a force about an axis, τ=rFsinθ\tau = rF\sin\theta, to identify the lever arm (the perpendicular distance from the axis to the line of the force), and to determine the sense of the rotation (clockwise or counterclockwise). Torque is the rotational analogue of force: it is what causes angular acceleration.

What torque is

Torque answers the question "how effectively does this force cause rotation?" It depends not just on how hard you push, but on where and in what direction. The same force can produce a large torque or none at all, depending on the distance from the axis and the angle.

The lever arm and the angle

This is why a door handle is placed far from the hinges and you push perpendicular to the door: both maximize the torque. Pushing near the hinges, or pushing the door edge-on toward the hinge, produces little or no turning effect. The lever-arm idea, force times perpendicular distance, is often the quickest route to a torque.

The sense of rotation

A torque also has a direction of turning: clockwise or counterclockwise about the axis. In AP Physics 1 you assign one sense as positive (commonly counterclockwise) and the other as negative, then add torques with signs to find the net torque, exactly as you add forces with signs along an axis. The net torque determines whether and how the object's rotation changes. The physical reason torque, not force alone, governs rotation is that rotation is about turning around an axis, and a force's ability to turn something depends on its line of action relative to that axis. This is the rotational counterpart of how force governs translation: where FnetF_{net} drives linear acceleration through Fnet=maF_{net} = ma, the net torque τnet\tau_{net} drives angular acceleration through τnet=Iα\tau_{net} = I\alpha (Topic 5.6). Two forces of equal magnitude can produce wildly different torques, or even cancel, depending on their lever arms and senses. This is what makes torque the right quantity for rotational dynamics and equilibrium: a balanced object has zero net torque (Topic 5.5), and an object with a net torque undergoes angular acceleration. Computing torques correctly, attending to distance, angle and sense, is the gateway skill for the rest of the unit.

Try this

Q1. A 5050 N force is applied perpendicular to a wrench 0.300.30 m from the bolt. Calculate the torque. [2 points]

  • Cue. τ=rFsin90=(0.30)(50)(1)=15\tau = rF\sin 90^\circ = (0.30)(50)(1) = 15 N\cdotm.

Q2. A force is applied along the line from the axis to its point of application. Calculate the torque it produces. [1 point]

  • Cue. θ=0\theta = 0, so τ=rFsin0=0\tau = rF\sin 0^\circ = 0. No torque.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)4 marksSection II (short FRQ, quantitative). A mechanic applies a 150150 N force to the end of a wrench 0.250.25 m long. (a) Calculate the torque if the force is applied perpendicular to the wrench. (b) Calculate the torque if the same force is applied at 4040 degrees to the wrench. (c) Explain why the perpendicular application produces more torque.
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A 4-point FRQ on torque and the angle of application.

(a) Perpendicular force (1 point): τ=rFsinθ=(0.25)(150)sin90=(0.25)(150)(1)=37.5\tau = rF\sin\theta = (0.25)(150)\sin 90^\circ = (0.25)(150)(1) = 37.5 N\cdotm.
(b) Force at 4040^\circ (2 points): τ=(0.25)(150)sin40=(0.25)(150)(0.643)=24.1\tau = (0.25)(150)\sin 40^\circ = (0.25)(150)(0.643) = 24.1 N\cdotm.
(c) Explain (1 point): torque is rFsinθrF\sin\theta, which is largest when θ=90\theta = 90^\circ (sinθ=1\sin\theta = 1). A perpendicular force has the full lever arm rr; at 4040^\circ only the component perpendicular to the wrench produces torque, so it is smaller.

Markers reward τ=rFsinθ\tau = rF\sin\theta for both cases and an explanation that the perpendicular component (or full lever arm) maximizes torque.

AP 2022 (style)1 marksSection I (multiple choice). To loosen a stubborn bolt with the least effort, you should push... (A) close to the bolt, perpendicular to the wrench (B) far from the bolt, perpendicular to the wrench (C) far from the bolt, along the wrench (D) close to the bolt, along the wrench. Justify your reasoning.
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A 1-point MCQ on maximizing torque. The answer is (B).

Torque is τ=rFsinθ\tau = rF\sin\theta, so it is greatest when rr is large (far from the bolt) and θ=90\theta = 90^\circ (perpendicular, sinθ=1\sin\theta = 1). Pushing far out and perpendicular maximizes the torque for a given force. Pushing along the wrench (θ=0\theta = 0) gives zero torque, and pushing close to the bolt gives a small lever arm. The trap is forgetting that both the distance and the perpendicular direction matter.

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