Torque — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Definition of torque, the lever arm formula, sign conventions, torque from center of mass weight, rotational equilibrium, and problem-solving for static rigid body systems on the AP exam.

You should already know: Newton's first law for translational motion, center of mass calculation for uniform rigid bodies, vector components of force.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Torque?

Torque is the rotational equivalent of force in linear motion: where force causes translational acceleration, torque causes angular acceleration of a rigid body around a fixed pivot point. It is a core concept of Unit 6 Rotational Motion, which makes up 14-18% of the total AP Physics 1 exam score per the official College Board CED. Torque appears in both multiple-choice (MCQ) and free-response (FRQ) sections, most commonly paired with static equilibrium, rotational kinematics, or angular momentum problems. 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, which is the core principle of levers and wrenches. Torque is typically denoted with the Greek letter $\tau$ (tau), and is sometimes called the "moment of force" or "moment" in older texts.

2. Calculating Torque: Magnitude and Sign

There are two equivalent, widely used formulas to calculate torque for AP Physics 1, both rooted in the definition that torque depends on force, distance from pivot, and the angle between them. The standard formula is: $$\tau =rF\sin \theta$$ where $r$ is the straight-line distance from the pivot point 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 application) 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, we use a standard AP Physics 1 sign convention: 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 the torque from that force is zero regardless of force magnitude: pushing directly through the pivot cannot make an object rotate.

Worked Example

Problem: 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: $r=3.0\text{m}$, $F=10\text{N}$, $\theta =30^{\circ }$ between $\vec{r}$ (which points right along the rod from pivot to force) and $\vec{F}$.
2. Confirm rotation direction: the force pulls up and left at the right end, so it causes counterclockwise rotation around the pivot, meaning torque is positive by convention.
3. Substitute into the torque formula: $\tau =rF\sin \theta =(3.0\text{m})(10\text{N})(\sin 30^{\circ })$
4. Calculate: $\sin 30^{\circ }=0.5$, so $\tau =+15\text{Ncdotpm}$.

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.

3. 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 is a major simplification for problem-solving: you do not need to calculate torque for every tiny segment of the object, just 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, so the distance from the pivot to the center of mass is just the distance between the pivot and the midpoint of the beam. Torque from weight is calculated exactly like torque from any other force: ${\tau }_g=r_{CM}mg\sin \theta$, 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

Problem: 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: $F_g=mg=(12\text{kg})(10{\text{m/s}}^2)=120\text{N}$, acting straight down.
3. The beam is horizontal, so the angle between $r_{CM}$ (pointing right horizontally from pivot to CM) and weight (pointing 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.
4. Calculate final torque: ${\tau }_g=-r_{CM}{F}_g\sin 90^{\circ }=-(1.0\text{m})(120\text{N})(1)=-120\text{Ncdotpm}$.

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.

4. 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. The mathematical condition for rotational equilibrium is: $$\sum \tau =0$$ This can be rewritten as $\sum {\tau }_{CCW}=\sum {\tau }_{CW}$, meaning the total of all positive counterclockwise torques equals the total of all 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 (since $r=0$ for that force), letting you solve for other unknowns without force balance first.

Worked Example

Problem: 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, so positive; torque from the right child is clockwise, so negative.
3. Write the rotational equilibrium condition: $\sum \tau ={\tau }_1+{\tau }_2=0\to r_1{m}_{1}g-r_2{m}_{2}g=0$. The $g$ term cancels out, leaving $r_1{m}_1=r_2{m}_2$.
4. Solve for $r_2$: $r_2=\frac{r_1{m}_1}{m_2}=\frac{(2.0\text{m})(40\text{kg})}{30\text{kg}}\approx 2.7\text{m}$.

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.

Common Pitfalls (and how to avoid them)

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

Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

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 $g=10{\text{m/s}}^2$. A) $0\text{Ncdotpm}$ B) $25\text{Ncdotpm}$ C) $50\text{Ncdotpm}$ D) $100\text{Ncdotpm}$

Worked Solution: 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, so the entire ladder forms a right triangle with horizontal leg 1 m and hypotenuse 2 m, meaning the horizontal distance from the base to the midpoint of the ladder is 0.5 m. Since weight acts straight down, the perpendicular lever arm for weight is exactly this horizontal distance. Weight magnitude is $mg=10\text{kg}\ast 10{\text{m/s}}^2=100\text{N}$. Torque magnitude is $\tau =F{d}_{\perp }=100\text{N}\ast 0.5\text{m}=50\text{Ncdotpm}$. The correct answer is C.

Question 2 (Free Response)

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.

Worked Solution: (a) Forces: 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. (b) The bridge is 6.0 m long 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: $$\sum \tau =T(6.0\text{m})\sin 45^{\circ }-mg(3.0\text{m})\sin 90^{\circ }=0$$ Substitute values: $T\ast 6\ast (\sqrt{2}/2)=500\ast 10\ast 3\to 3\sqrt{2}T=15000\to T=\frac{5000}{\sqrt{2}}\approx 3500\text{N}$. (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 for the horizontal component directly without force balance.

Question 3 (Application / Real-World Style)

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?

Worked Solution: Use the standard torque formula $\tau =rF\sin \theta$. Rearrange to solve for $\sin \theta$: $$\sin \theta =\frac{\tau }{rF}=\frac{160\text{Ncdotpm}}{(0.5\text{m})(400\text{N})}=0.8$$ Take the inverse sine to find $\theta =\arcsin (0.8)\approx 53^{\circ }$. This means that the mechanic must pull at an angle of at least 53° between the wrench handle and the direction of their pull to produce enough torque to loosen the bolt; pulling at a smaller angle will not produce enough torque.

Quick Reference Cheatsheet

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 Newton's Second Law for Rotation Angular Momentum and Conservation Static Equilibrium of Rigid Bodies Rolling Motion