Unit 5 · Topic 5.3

5.3 Torque

Torque is the turning effect of a force about an axis. Its size depends on the force, where it's applied and its angle: $\tau = rF\sin\theta$ , or force times lever arm. Torque is what makes rigid objects start, stop or change their spin.

Key terms

torque
lever arm
line of action
axis of rotation
force diagram
cross product

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of one big Rotation unit (old Unit 5). That unit is now split in two: Unit 5 covers spinning motion, torque and rotational inertia, and Unit 6 covers rotational energy, angular momentum and rolling. Older videos and practice labeled "Unit 5: Rotation" still fit, but they may mix in Unit 6 ideas.

What makes something turn

Push on a door near the hinges and it barely moves. Push at the handle, straight into the door, and it swings easily. Push along the door toward the hinges and nothing turns at all. Torque captures all three effects.

Draw a position vector $\vec{r}$ from the axis to the point where the force acts. Only the part of the force perpendicular to $\vec{r}$ causes turning. The part along $\vec{r}$ just pushes or pulls on the axis. So the torque's size is $\tau = rF\sin\theta$ where θ is the angle between $\vec{r}$ and $\vec{F}$ . The unit is the newton-meter (N·m).

Lever arm and line of action

The line of action is the line that runs through the force arrow, extended in both directions. The lever arm $r_\perp$ is the perpendicular distance from the axis to that line. Then $\tau = r_\perp F$ . It's the same as $rF\sin\theta$ , because $r_\perp = r\sin\theta$ .

Use whichever is easier. If you know the angle between r and F, use rF sin θ. If you can see the perpendicular distance in a diagram, use the lever arm.

A force whose line of action passes through the axis has zero lever arm, so it makes zero torque. That's why the force from a hinge or axle exerts no torque about that hinge or axle.

Signs and force diagrams

On this exam, a torque's direction is clockwise or counterclockwise about the axis. Pick one as positive (usually counterclockwise), then give each torque a sign. The net torque is the signed sum.

A force diagram is like a free-body diagram, but it also shows where each force acts on the object. For a rod or beam, draw the object as a line and put each force arrow at its actual point of application. Gravity acts at the center of mass. You need those locations to find the lever arms.

Torque as a cross product

As a vector, torque is the cross product $\vec{\tau} = \vec{r} \times \vec{F}$ . The cross product of two vectors $\vec{A}$ and $\vec{B}$ has magnitude $AB\sin\theta$ and points perpendicular to both of them.

Find its direction with the right-hand rule. Point your fingers along $\vec{r}$ , curl them toward $\vec{F}$ , and your thumb points along $\vec{\tau}$ . A torque pointing out of the page (+z) turns things counterclockwise as you look at the page; one pointing into the page turns them clockwise.

You'll only need the right-hand rule qualitatively, and rotation will stay in one plane. Most of the time, clockwise or counterclockwise is all you'll write.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Torque on a wrench
You pull on the end of a 0.25 m wrench with an 80 N force. The force makes a 60° angle with the wrench handle. What torque do you exert about the bolt?
Show the solutionHide the solution
1. Step 1: θ is the angle between $\vec{r}$ (along the handle) and $\vec{F}$ , so θ = 60°.
2. Step 2: $\tau = rF\sin\theta = (0.25)(80)\sin 60^\circ \approx 17$ N·m.
3. Step 3: Pulling perpendicular to the handle would give the most torque: (0.25)(80) = 20 N·m.
Answer: About 17 N·m
Example 2Calculator allowed
Net torque on a rod
A light 2.0 m rod is pivoted at its left end and lies horizontally. Three forces act on it: 30 N straight up at the right end, 50 N straight down 0.80 m from the pivot, and 20 N at the right end pointing along the rod, away from the pivot. Find the net torque about the pivot. Take counterclockwise as positive.
Show the solutionHide the solution
1. Step 1: The 30 N upward force at 2.0 m turns the rod counterclockwise: +(2.0)(30) = +60 N·m.
2. Step 2: The 50 N downward force at 0.80 m turns it clockwise: −(0.80)(50) = −40 N·m.
3. Step 3: The 20 N force acts along the rod, so its line of action passes through the pivot. Its lever arm is zero, so its torque is 0.
4. Step 4: Net torque: 60 − 40 + 0 = +20 N·m.
Answer: 20 N·m counterclockwise
Example 3
Using the cross product
A force $\vec{F} = (3.0\,\hat{i} + 4.0\,\hat{j})$ N acts at position $\vec{r} = 0.50\,\hat{i}$ m from an axis along z. Find the torque and say which way it turns the object, viewed from +z.
Show the solutionHide the solution
1. Step 1: $\vec{\tau} = \vec{r} \times \vec{F} = (0.50\,\hat{i}) \times (3.0\,\hat{i} + 4.0\,\hat{j})$ .
2. Step 2: $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$ , so $\vec{\tau} = (0.50)(4.0)\,\hat{k} = 2.0\,\hat{k}$ N·m.
3. Step 3: Check with rF sin θ: only the 4.0 N y-component is perpendicular to r, giving (0.50)(4.0) = 2.0 N·m.
4. Step 4: +z points toward you, so the object turns counterclockwise as you look down from +z.
Answer: 2.0 N·m along +z, counterclockwise viewed from +z

Common mistakes

Measuring θ from the wrong line. It's the angle between r (from the axis to the force's point of application) and F, not the angle with the floor.
Multiplying a force by the distance along the object when the force isn't perpendicular. Use the lever arm, the perpendicular distance to the line of action.
Leaving out gravity on a beam, or placing it at the end. The object's weight acts at its center of mass.
Mixing up signs. Choose clockwise or counterclockwise as positive at the start and label each torque.

On the exam

Force diagrams on the free-response section need each force drawn where it acts on the object, with clear labels. Don't draw components as extra arrows.
When you compare torques in a multiple-choice question, check both the force and the lever arm. A smaller force farther out can win.

Connected topics

Videos

Topic 5.3 - Torque
Lessons With LondotWatch on YouTube (opens in a new tab)
AP Physics 1 - Unit 5 Lesson 4 - How to Calculate Torque in Rotational Motion
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Torque Introduction
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C - Torque
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
Finding torque for angled forces | Physics | Khan Academy
Khan Academy PhysicsWatch on YouTube (opens in a new tab)
30.5 Torque from Gravity
MIT OpenCourseWareWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 5.3 Torque. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A mechanic pulls on the end of a wrench 0.30 m from the bolt with a 40 N force. The force makes a 30° angle with the wrench handle. What is the magnitude of the torque on the bolt?

Question 2 of 4Calculator allowed

A door is hinged along one edge. Which force on the door produces zero torque about the hinge?

Question 3 of 4Calculator allowed

Four forces are applied, one at a time, to the same door, in the plane perpendicular to the hinge axis. Use sin 53° = 0.80 and sin 30° = 0.50. Which force produces the greatest torque about the hinge?

Question 4 of 4Calculator allowed

A light 1.0 m rod is pivoted at its center. Three vertical forces act on it: 6.0 N downward at the left end, 10 N downward 0.20 m to the right of the pivot, and 4.0 N upward at the right end. What is the net torque about the pivot?

0 of 4 answered