# AP Physics 1 Torque

> AP Physics 1 · Rotational Motion
> Source: https://www.owlsprep.com/study/ap-physics-1-u6-torque/

This module covers torque definition, magnitude/sign calculation, torque from object weight, and rotational equilibrium for static rigid body problems aligned to AP Physics 1 Unit 6.

**Prerequisites:** Newton's first law for translational motion; Center of mass calculation for uniform rigid bodies; Vector components of force

## Learning objectives

- Define torque as the rotational equivalent of force
- Calculate torque magnitude and sign using two equivalent formulas
- Calculate torque from an object's weight acting at its center of mass
- Solve rotational equilibrium problems for static rigid bodies
- Avoid common exam pitfalls in torque problems

## What Is Torque?

Torque is the rotational equivalent of force in linear motion: while force causes translational acceleration, torque causes angular acceleration of a rigid body around a fixed pivot. It is a core concept of Unit 6 Rotational Motion, which makes up 14-18% of the total AP Physics 1 exam score, appearing in both multiple-choice and free-response questions.

Unlike force, torque depends not just on the magnitude of the applied force, but also on where the force is applied relative to the pivot, and the direction of the force. A small force applied far from the pivot can produce the same torque as a large force applied very close to the pivot, the core principle behind levers and wrenches. Torque is denoted by the Greek letter $\tau$ (tau), and is sometimes called the *moment of force*.

## Calculating Torque: Magnitude and Sign

There are two equivalent formulas for calculating torque that you will use on the AP Physics 1 exam. The standard definition formula is:

$$\tau = r F \sin\theta$$

where $r$ is the straight-line distance from the pivot to the point where the force is applied, $F$ is the magnitude of the applied force, and $\theta$ is the angle between the position vector $\vec{r}$ (from pivot to force) and the force vector $\vec{F}$.

An equivalent form using the lever arm (or moment arm) concept is:

$$\tau = F d_\perp$$

where $d_\perp = r \sin\theta$ is the perpendicular distance from the pivot to the line of action of the force. This form is often easier for visual problem-solving on the exam.

For fixed-axis rotation, the standard AP Physics 1 sign convention is: counterclockwise (CCW) rotation is caused by positive torque, and clockwise (CW) rotation is caused by negative torque. If a force's line of action passes directly through the pivot, $d_\perp = 0$, so torque is zero regardless of force magnitude.

**Worked example:** A 3.0 m long uniform rod is pivoted at its left end. A 10 N force is applied at the right end, at an angle of 30° above the rod (pointing up and to the left). What is the magnitude and sign of the torque from this force around the pivot?

1. Identify known values:
2. $$r = 3.0 \, \text{m}, F = 10 \, \text{N}, \theta = 30^\circ$$
3. Confirm rotation direction: the force pulls up and left at the right end, causing counterclockwise rotation around the pivot, so torque is positive by convention.
4. Substitute into the torque formula:
5. $$\tau = r F \sin\theta = (3.0 \, \text{m})(10 \, \text{N})(\sin 30^\circ)$$
6. Calculate, using $\sin 30^\circ = 0.5$:
7. $$\tau = +15 \, \text{N·m}$$

> **Exam tip:** If you get confused about which angle to use for $\theta$, always draw the full line of the force and measure the perpendicular distance from the pivot to that line—you can never go wrong with the lever arm method on the AP exam.

## Torque from an Object's Weight

Any rigid body has weight distributed evenly across its mass, but the total torque from gravity around any pivot is equal to the torque produced by the entire weight of the object acting at its center of mass. This simplifies problem-solving significantly: you only need to locate the center of mass and treat weight as a single point force acting there.

For a uniform rigid body (constant density), the center of mass is always at the geometric center of the object. For example, a uniform beam of total length $L$ has its center of mass at $L/2$ from either end. Torque from weight is calculated the same way as any other torque:

$$\tau_g = r_{CM} mg \sin\theta$$

where $r_{CM}$ is distance from pivot to center of mass, $m$ is total mass of the object, and $g = 9.8 \, \text{m/s}^2$.

**Worked example:** A uniform 4.0 m long beam has a mass of 12 kg. The beam is pivoted 1.0 m from its left end, and held horizontal. What is the torque from the beam's own weight around the pivot? Use $g = 10 \, \text{m/s}^2$ for simplicity.

1. Find the center of mass of the uniform beam: it sits 2.0 m from the left end, so $r_{CM} = 2.0 \, \text{m} - 1.0 \, \text{m} = 1.0 \, \text{m}$ to the right of the pivot.
2. Calculate total weight:
3. $$F_g = mg = (12 \, \text{kg})(10 \, \text{m/s}^2) = 120 \, \text{N}$$
4. The beam is horizontal, so the angle between $r_{CM}$ (right horizontally from pivot to CM) and weight (straight down) is 90°, so $\sin 90^\circ = 1$. Weight pulling down on the right side of the pivot causes clockwise rotation, so torque is negative.
5. Calculate final torque:
6. $$\tau_g = - (1.0 \, \text{m})(120 \, \text{N})(1) = -120 \, \text{N·m}$$

> **Exam tip:** Never forget that the object's own weight produces torque unless the pivot is exactly at the center of mass. This is the most frequently omitted term in AP Physics 1 torque equilibrium problems.

## Rotational Equilibrium

Newton's first law extended to rotation states that if the net torque on a rigid body around a pivot is zero, the body has zero angular acceleration. This condition is called rotational equilibrium. If an object is also in translational equilibrium (net force on the object is zero), the entire system is in static equilibrium, meaning it is completely stationary—this is the most common type of torque problem tested on the AP exam.

$$\sum \tau = 0$$

This can also be written as $\sum \tau_{CCW} = \sum \tau_{CW}$, meaning total positive counterclockwise torques equal total negative clockwise torques. A key problem-solving trick for static equilibrium: you can choose any point as your pivot for calculating net torque, because net torque is zero around every point for a static system. Choosing the pivot at the location of an unknown force eliminates that force from the torque equation, letting you solve for other unknowns without force balance first.

**Worked example:** A uniform 5.0 m long seesaw has a total mass of 20 kg. It is pivoted at its center. A 40 kg child sits 2.0 m to the left of the pivot. How far to the right of the pivot must a 30 kg child sit to balance the seesaw? Use $g=10 \, \text{m/s}^2$.

1. The seesaw is uniform, so its center of mass is at the pivot. This means torque from the seesaw's weight is zero ($r=0$), so we can ignore it entirely.
2. We choose the pivot at the center of the seesaw, so torque from the pivot's normal force is also zero. Assign signs: torque from the left child (weight pulling down on the left of the pivot) is counterclockwise (positive), torque from the right child is clockwise (negative).
3. Write the rotational equilibrium condition:
4. $$\sum \tau = \tau_1 + \tau_2 = 0 \rightarrow r_1 m_1 g - r_2 m_2 g = 0$$
5. The $g$ term cancels out, leaving $r_1 m_1 = r_2 m_2$.
6. Solve for $r_2$:
7. $$r_2 = \frac{r_1 m_1}{m_2} = \frac{(2.0 \, \text{m})(40 \, \text{kg})}{30 \, \text{kg}} \approx 2.7 \, \text{m}$$

> **Exam tip:** Always choose your pivot at the location of an unknown force to eliminate that variable from your torque equation. This saves time and reduces algebra errors on FRQs.

## Additional AP-Style Worked Examples

**Worked example:** A uniform 2 m long ladder of mass 10 kg leans against a frictionless wall, with the base of the ladder 1 m from the wall, pivoted at the base. What is the magnitude of the torque from the ladder's weight around the base pivot? Use $g = 10 \, \text{m/s}^2$.

Options: A) $0 \, \text{N·m}$, B) $25 \, \text{N·m}$, C) $50 \, \text{N·m}$, D) $100 \, \text{N·m}$

1. The uniform ladder has its center of mass at its midpoint, 1.0 m along the ladder from the base pivot. The base is 1.0 m from the wall, forming a right triangle with horizontal leg 1 m and hypotenuse 2 m, so the horizontal distance from the base to the midpoint is 0.5 m.
2. Since weight acts straight down, the perpendicular lever arm for weight is exactly this horizontal distance. Calculate weight:
3. $$mg = (10 \, \text{kg})(10 \, \text{m/s}^2) = 100 \, \text{N}$$
4. Calculate torque magnitude:
5. $$\tau = F d_\perp = (100 \, \text{N})(0.5 \, \text{m}) = 50 \, \text{N·m}$$
6. The correct answer is C.

**Worked example:** A uniform 6.0 m long drawbridge has a mass of 500 kg. It is held at rest, horizontal, by a cable attached to the far end of the bridge, connected to the castle wall 6.0 m above the pivot of the bridge. The pivot is at the end of the bridge attached to the wall.

(a) Label all forces acting on the bridge, identifying the point each force acts on.
(b) Write the equation for net torque around the pivot of the bridge, and solve for the tension in the cable. Use $g = 10 \, \text{m/s}^2$.
(c) Explain why choosing the pivot at the bridge's center of mass would allow you to solve for the horizontal component of the force from the pivot on the bridge, without using force balance.

1. (a) Forces are: 1) Weight of the bridge $mg$, acts straight down at the center of the bridge (3.0 m from the pivot). 2) Tension $T$, acts along the cable from the far end of the bridge up to the wall attachment. 3) Pivot force $\vec{F}_p$, acts at the pivot (0 m from the pivot) with both horizontal and vertical components.
2. (b) The bridge is 6.0 m long and horizontal, so the cable forms a 45-45-90 right triangle, meaning the angle between the bridge and the cable is 45°. For equilibrium, net torque is zero:
3. $$\sum \tau = T(6.0 \, \text{m})\sin 45^\circ - mg(3.0 \, \text{m})\sin 90^\circ = 0$$
4. Substitute values and solve:
5. $$T * 6 * (\sqrt{2}/2) = 500 * 10 * 3 \rightarrow 3\sqrt{2} T = 15000 \rightarrow T = \frac{5000}{\sqrt{2}} \approx 3500 \, \text{N}$$
6. (c) If the pivot is chosen at the center of mass, torque from the bridge's weight is zero (it acts at the new pivot). Only tension and the pivot force produce non-zero torque around the new pivot. The torque from tension is known, so the torque from the horizontal component of the pivot force must balance it, allowing us to solve directly for the horizontal component without force balance.

**Worked example:** A mechanic uses a 0.5 m long wrench to loosen a stuck bolt. The mechanic can apply a maximum force of 400 N with their hand. To loosen the bolt, a total torque of 160 N·m is required. What is the minimum angle between the wrench handle and the direction the mechanic is pulling that will allow the mechanic to loosen the bolt?

1. Use the standard torque formula:
2. $$\tau = r F \sin\theta$$
3. Rearrange to solve for $\sin\theta$:
4. $$\sin\theta = \frac{\tau}{rF} = \frac{160 \, \text{N·m}}{(0.5 \, \text{m})(400 \, \text{N})} = 0.8$$
5. Take the inverse sine to find the angle:
6. $$\theta = \arcsin(0.8) \approx 53^\circ$$
7. The mechanic must pull at an angle of at least 53° between the wrench handle and the direction of pull to produce enough torque to loosen the bolt.

## Common pitfalls

- **Wrong:** Using $\cos\theta$ instead of $\sin\theta$ in the torque formula when the given angle is between the force and the rod
  - Why it fails: Students mix up trigonometric terms from memorization instead of checking the formula definition
  - Correct: Always draw the line of action of the force, then calculate the perpendicular lever arm $d_\perp$ directly to confirm your trigonometry
- **Wrong:** Forgetting to include torque from the object's own weight when the pivot is not at the center of mass
  - Why it fails: Students focus on external applied forces and ignore the weight of the object itself
  - Correct: After listing all external torques, always add a line for torque from the object's weight acting at its center of mass before writing the net torque equation
- **Wrong:** Switching sign conventions mid-problem, calling some clockwise torques positive and others negative based on which side of the pivot they sit
  - Why it fails: Students do not set the convention at the start of the problem
  - Correct: Explicitly set "counterclockwise = positive, clockwise = negative" at the top of your work for every torque problem, and check every torque against this rule
- **Wrong:** Using the total length of the object for $r$, instead of the distance from the pivot to the force application point
  - Why it fails: Students default to the only distance given in the problem, which is often the total object length
  - Correct: Always label $r$ on your diagram for every force, measuring directly from the pivot to the point the force is applied
- **Wrong:** Including torque from the pivot's normal force in the equation when the pivot is chosen at that point
  - Why it fails: Students forget that torque depends on distance from the pivot
  - Correct: Any force acting at your chosen pivot has $r=0$, so its torque is zero and can be omitted from the calculation

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| General Torque | $\tau = r F \sin\theta$ | $r$ = distance from pivot to force, $\theta$ = angle between $\vec{r}$ and $\vec{F}$ |
| Torque (Lever Arm Form) | $\tau = F d_\perp$ | $d_\perp = r \sin\theta$ = perpendicular distance from pivot to force line of action |
| Torque from Weight | $\tau_g = r_{CM} mg \sin\theta$ | Total weight acts at the object's center of mass |
| Rotational Equilibrium | $\sum \tau = 0$ | Applies to any rigid body with zero angular acceleration |
| Standard AP Sign Convention | Counterclockwise = +, Clockwise = - | Be consistent in every problem; reverse is acceptable if you stay consistent |
| Static Equilibrium | $\sum \tau = 0$, $\sum F_x = 0$, $\sum F_y = 0$ | All three conditions hold for non-moving rigid bodies |
| Torque at Pivot | $\tau = 0$ | Any force acting through your chosen pivot has zero torque |

## What's next

Torque is the foundational concept for all of rotational motion, so mastering it is required for every upcoming topic in Unit 6. Next, you will use torque to derive and apply Newton's second law for rotation, which relates net torque to angular acceleration and moment of inertia. Without a solid understanding of how to calculate net torque correctly, you will not be able to solve multi-concept problems involving rotating pulleys, rolling wheels, or rotating systems. Torque also connects to angular momentum, where net torque equals the rate of change of angular momentum, just as net force equals the rate of change of linear momentum. On the AP exam, torque is almost always combined with these concepts in multi-point FRQs.

- [Rotational Inertia and Rotational Newton's Second Law](https://www.owlsprep.com/study/ap-physics-1-u6-rotational-inertia-and-rotational-newton/)
- [Angular Momentum](https://www.owlsprep.com/study/ap-physics-1-u6-angular-momentum/)
- [Conservation of Angular Momentum](https://www.owlsprep.com/study/ap-physics-1-u6-conservation-of-angular-momentum/)

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