AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/5/5-3)
Unit 5 · Topic 5.3
5.3 Torque
Torque is the turning effect of a force. It depends on how big the force is, how far from the axis it acts and its direction, and you calculate its size with τ = rF sin θ or as force times lever arm. In this course you give a torque's size and whether it turns things clockwise or counterclockwise.
Key terms
- torque
- lever arm
- line of action
- axis of rotation
- force diagram
What torque measures
Try opening a door by pushing near the hinges. It's hard. Push at the handle, far from the hinges, and it swings easily. Push straight toward the hinges and nothing turns at all. Torque captures all three effects.
The size of a torque is τ = rF sin θ. Here r is the distance from the axis to the point where the force acts, F is the size of the force, and θ is the angle between the force and the line from the axis to that point. Torque is measured in newton-meters (N·m).
Only the part of the force perpendicular to that line causes rotation. The part pointing along the line just pushes or pulls on the axis.
Three ways to get the same torque
You can read rF sin θ in two helpful ways. Group F sin θ together and it's the perpendicular component of the force, F⊥. Group r sin θ together and it's the lever arm r⊥: the shortest (perpendicular) distance from the axis to the force's line of action. The line of action is the line through the force arrow, extended in both directions. All three methods give the same answer, so use whichever fits the picture.
The lever-arm method is quickest when a force is drawn at an angle and you can see the perpendicular distance to its line. The component method is quickest when the force is easy to split.
| Method | Formula | Best when |
|---|---|---|
| Angle | τ = rF sin θ | you know r, F and the angle between them |
| Perpendicular force | τ = rF⊥ | the force splits easily into components |
| Lever arm | τ = r⊥F | the perpendicular distance to the line of action is easy to see |
When a force makes no torque
The biggest torque from a given force happens at θ = 90°, with the force perpendicular to the line from the axis. A force gives no torque at all in any of these cases:
- The force acts at the axis (r = 0), like the hinge force on a door.
- The force points straight toward or away from the axis (θ = 0° or 180°), so sin θ = 0.
- The force's line of action passes through the axis, which means the lever arm is zero.
Clockwise, counterclockwise and net torque
In AP Physics 1 a torque has a size and a sense: clockwise or counterclockwise. Torque is actually a vector with a direction along the axis, but that direction is beyond the course.
To find the net torque, pick one sense as positive, give each torque a sign and add. To decide a torque's sense, imagine the object pinned at the axis and ask which way that force alone would turn it.
Force diagrams for extended objects
A free-body diagram draws an object as a dot. For torque you need a force diagram instead. It shows the object's shape and the axis, and draws each force at the spot where it acts. The arrows should still show each force's relative size and direction.
Gravity on a rigid object acts as if all of it pulled at the center of mass. For a uniform rod or plank that's the middle. Drawing the weight at the wrong spot is one of the most common torque errors.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A wrench at two angles
You pull with 50 N on the end of a wrench, 0.30 m from the center of the bolt. Find the torque (a) when you pull perpendicular to the wrench and (b) when your pull makes a 30° angle with the wrench handle.
Show the solutionHide the solution
- Step 1: (a) θ = 90°, so τ = rF sin 90° = (0.30)(50)(1) = 15 N·m.
- Step 2: (b) θ = 30° between the force and the handle (which runs along the line from the bolt). τ = (0.30)(50) sin 30° = (0.30)(50)(0.5) = 7.5 N·m.
- Step 3: Same force, half the torque, because only the perpendicular component, 50 sin 30° = 25 N, turns the bolt.
Answer: (a) 15 N·m (b) 7.5 N·m
- Example 2Calculator allowed
Lever arm on a door
You push on a door with 40 N at a point 0.80 m from the hinges. Your push makes a 60° angle with the door's surface. Find the lever arm and the torque about the hinges.
Show the solutionHide the solution
- Step 1: The line from the hinge to your hand runs along the door, so θ = 60°.
- Step 2: Lever arm: r⊥ = r sin θ = (0.80) sin 60° ≈ 0.69 m.
- Step 3: Torque: τ = r⊥F ≈ (0.693)(40) ≈ 28 N·m. Same as rF sin θ = (0.80)(40)(0.866) ≈ 27.7 N·m.
Answer: Lever arm ≈ 0.69 m; torque ≈ 28 N·m
- Example 3Calculator allowed
Net torque with signs (classic trap)
A light rod is pivoted at its center. Counterclockwise is positive. Forces: 12 N straight down at 0.40 m left of the pivot; 8.0 N straight down at 0.30 m right of the pivot; 10 N straight up at 0.20 m right of the pivot; and 15 N pulling straight along the rod toward the pivot at its left end. Find the net torque.
Show the solutionHide the solution
- Step 1: 12 N down on the left side turns the rod counterclockwise: +(0.40)(12) = +4.8 N·m.
- Step 2: 8.0 N down on the right side turns it clockwise: −(0.30)(8.0) = −2.4 N·m.
- Step 3: 10 N up on the right side turns it counterclockwise: +(0.20)(10) = +2.0 N·m. The trap is giving every force on the right side the same sign; the sense depends on the force's direction too.
- Step 4: The 15 N pull runs along the rod through the pivot, so its torque is zero.
- Step 5: Net torque = 4.8 − 2.4 + 2.0 + 0 = +4.4 N·m.
Answer: 4.4 N·m counterclockwise
Common mistakes
- Using the wrong angle. θ is between the force and the line from the axis to the point of application, not between the force and the floor.
- Using the distance to the point of application when the force is at an angle. Use r sin θ (the lever arm) or the perpendicular force component.
- Drawing an object's weight at the end or at the pivot. For a uniform object it acts at the center.
- Mixing up N·m of torque with joules. They share base units, but torque is not energy, so write N·m.
On the exam
- You may be asked to draw a force diagram: show each force at its point of application, with arrow lengths that match their relative sizes.
- Expect ranking or comparison questions (which force gives the largest torque about the hinge?). Compare lever arms rather than distances.
Connected topics
Videos
Check yourself
4 questions on 5.3 Torque. Pick an answer to see if you got it, and why.
A student pushes on the end of a 0.30 m wrench with a 40 N force. The force makes a 30° angle with the wrench handle. What is the magnitude of the torque about the bolt?
A door is hinged along one vertical edge. Which of the following forces, applied at the doorknob, exerts zero torque about the hinges?
A rod of length L can pivot about its left end. Four separate forces are considered, each in the plane of rotation: Force 1 has magnitude F, acts at the right end, perpendicular to the rod. Force 2 has magnitude 2F, acts at the midpoint, perpendicular to the rod. Force 3 has magnitude 2F, acts at the right end, at 30° to the rod. Force 4 has magnitude 3F, acts L/4 from the pivot, perpendicular to the rod. Which ranking of the torques about the pivot is correct?
A uniform 2.0 kg rod, 1.2 m long, pivots about one end. It is held at rest at 60° above the horizontal. What is the magnitude of the torque that gravity exerts on the rod about the pivot? Use g = 9.8 m/s².
0 of 4 answered