Unit 5.2 | Torque
- 15 minutes
- 1 formulas
- 23 questions
- 1 videos
Same force, different result: where and how you push decides how much it turns. This lesson covers torque and the four torque problem types on the AP exam.
Unit 5 Breakdown
In this lesson:
- What torque is, and why where and how you push matters as much as how hard
- Torque from one force: [katex] \tau = rF\sin\theta [/katex], and the lever arm
- Signs: counterclockwise positive, clockwise negative, and how to add torques
- The 4 torque problem types: one force at an angle, comparing pushes, net torque, and torque from an object’s own weight
Introduction
This lesson covers torque: what it is, how to calculate it, and the four kinds of torque problems you will see on the AP exam.
In Unit 5.1 you described rotation. Now: what changes it? In Unit 2, a net force changes linear motion. For rotation, the job belongs to torque.
Try opening a door by pushing right next to the hinge. It barely moves. Push just as hard at the handle and it swings open. Same force, different result. That difference is torque.
What Is Torque?
Torque ([katex] \tau [/katex]) measures how well a force turns an object about a pivot (also called the axis). It depends on three things: how hard you push, how far from the pivot you push, and the angle you push at.
[katex] \tau = rF\sin\theta [/katex]
- [katex] r [/katex] is the distance from the pivot to the point where the force acts.
- [katex] \theta [/katex] is the angle between [katex] r [/katex] (the rod) and the force.
- Torque is measured in newton-meters, [katex] \text{N·m} [/katex].
Two ways to read the same formula
Group it as [katex] r(F\sin\theta) [/katex]: the distance times the perpendicular part of the force. That is the picture above.
Or group it as [katex] (r\sin\theta)F [/katex]: the force times the lever arm, the perpendicular distance from the pivot to the line the force acts along. This view is faster when a force is slanted, like a ladder or a cable: extend the force into a line and measure the shortest distance to the pivot.
Torque has a direction
A torque turns an object counterclockwise or clockwise. Use counterclockwise as positive and clockwise as negative. To find the direction, imagine that force acting alone: put a pencil on the pivot, push it with your finger in the direction of the force, and watch which way it turns.
A force makes zero torque if it acts at the pivot ([katex] r = 0 [/katex]) or points straight toward or away from the pivot ([katex] \theta = 0^\circ [/katex] or [katex] 180^\circ [/katex]).
Every torque question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. One force at an angle | a wrench, a door, a cable “at an angle” | [katex] \tau = rF\sin\theta [/katex], with [katex] \theta [/katex] between [katex] r [/katex] and [katex] F [/katex] |
| 2. Comparing pushes | “greatest torque”, “rank”, “ratio” | Compare [katex] rF\sin\theta [/katex] factor by factor |
| 3. Net torque from several forces | several forces on one rod or disk | Sign each torque (+ counterclockwise, − clockwise), then add |
| 4. Torque from an object’s own weight | a beam, rod or meter stick “of mass M” | The weight acts at the center of mass |
Problem Type 1: One force at an angle
How to spot it: One force pulls or pushes on a wrench, door, rod or beam at an angle, and the question asks for its torque.
Setup: Find [katex] r [/katex] from the pivot to where the force acts. Find the angle between the rod and the force. Then [katex] \tau = rF\sin\theta [/katex].
A mechanic pulls on the end of a [katex] 0.25\,\text{m} [/katex] wrench with an [katex] 80\,\text{N} [/katex] force, at [katex] 60^\circ [/katex] to the handle. What torque does she apply to the bolt?
[katex] \tau = rF\sin\theta = (0.25)(80)\sin 60^\circ \approx 17 \text{ N·m} [/katex]. Pulling perpendicular to the handle would give [katex] (0.25)(80) = 20 \text{ N·m} [/katex], the most this force can do.
AP trap: Using the wrong angle. [katex] \theta [/katex] is between the force and the rod, not between the force and the horizontal or the vertical. If a problem gives the angle from the other line, the angle you need is [katex] 90^\circ [/katex] minus it.
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Problem Type 2: Comparing pushes
How to spot it: Several doors, wrenches or forces are compared: “which exerts the greatest torque”, “rank the torques”, “what is the ratio”.
Setup: Write [katex] \tau = rF\sin\theta [/katex] for each case and compare factor by factor. For a ratio, write [katex] \frac{\tau_2}{\tau_1} [/katex] and cancel what the two cases share.
Push 1 is a force [katex] F [/katex] at the doorknob, perpendicular to the door. Push 2 is [katex] 2F [/katex] at the middle of the door, also perpendicular. Push 3 is [katex] 2F [/katex] at the doorknob, at [katex] 45^\circ [/katex] to the door. Rank the torques.
Call the door width [katex] w [/katex]. [katex] \tau_1 = wF [/katex]. [katex] \tau_2 = \tfrac{w}{2}(2F) = wF [/katex]. [katex] \tau_3 = w(2F)\sin 45^\circ \approx 1.4\,wF [/katex]. So [katex] \tau_3 > \tau_1 = \tau_2 [/katex].
AP trap: Thinking the bigger force always wins. Push 2 is twice as strong as push 1 but makes the same torque, because it acts half as far from the hinge.
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Problem Type 3: Net torque from several forces
How to spot it: Several forces act on one rod, beam or disk. The question asks for the net torque, the way it will turn, or the torque needed to stop it from turning.
Setup: Pick the pivot. For each force find [katex] rF\sin\theta [/katex] and give it a sign: + counterclockwise, − clockwise. Then add: [katex] \sum\tau [/katex].
A [katex] 1.0\,\text{m} [/katex] rod is pivoted at its center. A [katex] 10\,\text{N} [/katex] force pushes down on the left end, an [katex] 8\,\text{N} [/katex] force pushes down [katex] 0.30\,\text{m} [/katex] to the right of the pivot, and a [katex] 6\,\text{N} [/katex] force pushes up on the right end. Find the net torque.
- Left end pushed down: the rod turns counterclockwise, [katex] +(0.50)(10) = +5.0 \text{ N·m} [/katex].
- Pushed down right of the pivot: clockwise, [katex] -(0.30)(8) = -2.4 \text{ N·m} [/katex].
- Right end pushed up: counterclockwise, [katex] +(0.50)(6) = +3.0 \text{ N·m} [/katex].
[katex] \sum\tau = 5.0 – 2.4 + 3.0 = +5.6 \text{ N·m} [/katex], so the rod starts to turn counterclockwise.
AP trap: Adding the sizes and forgetting the signs. Decide each force’s direction on its own, as if it were the only force on the rod.
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Problem Type 4: Torque from an object’s own weight
How to spot it: The rod, beam, board or meter stick has mass: “a uniform beam of mass M”, “a 200 g meter stick”.
Setup: Treat the whole weight [katex] Mg [/katex] as one force at the center of mass. For a uniform object that is the middle, so its lever arm is the distance from the pivot to the middle.
A uniform [katex] 4.0\,\text{m} [/katex] beam of mass [katex] 30\,\text{kg} [/katex] sticks straight out from a wall, bolted at one end. A [katex] 20\,\text{kg} [/katex] sign hangs from its far end. Find the torque about the bolt.
Beam: [katex] (30)(9.8)(2.0) = 588 \text{ N·m} [/katex]. Sign: [katex] (20)(9.8)(4.0) = 784 \text{ N·m} [/katex]. Both turn the beam the same way (clockwise, downward), so the total is about [katex] 1370 \text{ N·m} [/katex] clockwise.
AP trap: Putting the weight of a uniform beam at its end. It acts at the middle, [katex] \tfrac{L}{2} [/katex] from either end.
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Lesson 5.3 Preview
In the next lesson we look at what happens when the torques cancel: rotational equilibrium, the idea behind seesaws, shelves, cranes and ladders.
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This article was written by Jason