Torque and rotational statics — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Definition of torque, cross product and lever arm formulas for torque, sign conventions, rotational equilibrium conditions, and problem-solving for pivoted beams, leaning ladders, and suspended rigid bodies.

You should already know: Newton’s first law for translational equilibrium, vector cross products for 3D vectors, rigid body center of mass fundamentals.

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 C: Mechanics 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 and rotational statics?

Torque is the rotational equivalent of force in linear mechanics: just as a net force causes linear acceleration, a net torque causes angular acceleration. Rotational statics specifically describes systems where both net force and net torque are zero, meaning the object has no linear or angular acceleration—it is either at rest or moving at constant translational and angular velocity, though most AP problems focus on stationary systems. This topic is a core component of Unit 5 (Rotation), which makes up 14-20% of the total AP Physics C: Mechanics exam score. Torque concepts regularly appear in multiple-choice questions (MCQ) both individually and as part of larger rotational motion sets, and rotational statics is a common topic for 3-5 point parts of free-response questions (FRQ), often combined with friction or force equilibrium. Mastery of this topic is foundational for all subsequent rotational dynamics concepts, as torque is the starting point for angular acceleration and rotational work-energy.

2. Definition and Calculation of Torque

Torque ($\vec{\tau }$) is a vector quantity that measures the tendency of a force to cause rotation about a given pivot point. The general definition comes from the cross product of the position vector $\vec{r}$ (from the pivot to the point of force application) and the applied force $\vec{F}$: $$\vec{\tau }=\vec{r}\times \vec{F}$$ The magnitude of torque is given by $\tau =rF\sin \theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ when placed tail to tail. An equivalent, often simpler, formulation for static problems uses the lever arm (or moment arm) $d=r\sin \theta$, the perpendicular distance from the pivot to the line of action of the force, giving $\tau =Fd$. The standard AP sign convention uses counterclockwise (CCW) torque as positive and clockwise (CW) torque as negative, which matches the right-hand rule for cross products: thumb pointing out of the page = positive torque, thumb pointing into the page = negative torque.

Worked Example

A 2.0 m tall vertical utility pole is anchored at its base (pivot point). A cable pulls on the top of the pole with a force of 500 N, running from the top of the pole down to the ground at an angle of 30° from the vertical pole. Calculate the magnitude and sign of the torque exerted by the cable about the base of the pole.

1. Identify the position vector $\vec{r}$ points straight up along the pole from the base (pivot) to the top, with magnitude $r=2.0$ m. The force vector $\vec{F}$ points down along the cable, 30° from the pole.
2. The angle between $\vec{r}$ and $\vec{F}$ is $180^{\circ }-30^{\circ }=150^{\circ }$, and $\sin (150^{\circ })=\sin (30^{\circ })=0.5$.
3. Using the magnitude formula: $\tau =rF\sin \theta =(2.0\text{m})(500\text{N})(0.5)=500\text{Ncdotpm}$. Using the lever arm method gives the same result: $d=r\sin (30^{\circ })=1.0$ m, so $\tau =Fd=500\text{Ncdotpm}$.
4. The cable pulls the top of the pole sideways, causing clockwise rotation about the base, so the torque is negative: $\tau =-500\text{Ncdotpm}$.

Exam tip: Always confirm your value of $\theta$ before calculating $\tau$—it is the angle between the position vector and force vector, not the angle between the force and the rod. Many students incorrectly use $\cos \theta$ instead of $\sin \theta$ from misidentifying $\theta$.

3. Equilibrium Conditions for Rotational Statics

For a rigid body to be completely static (no acceleration of any kind), two conditions must hold: translational equilibrium and rotational equilibrium. Translational equilibrium, which you already used for linear statics, means the net force on the body in all directions is zero: $$\sum \vec{F}=0⟹\sum F_x=0,\sum F_y=0$$ The new core condition for rotational statics is that the net torque about any pivot point is zero: $$\sum \tau =0$$ A critical simplification for problem-solving: if a system is already in translational equilibrium, the net torque is identical about any pivot point, so you can choose any pivot to simplify your calculation. The most strategic choice is almost always to pick a pivot at the location of an unknown force, because that force has $r=0$, so its torque is zero, eliminating the unknown from the equation immediately.

Worked Example

A uniform 4.0 m long beam of mass 10 kg is pivoted at its left end, and held horizontal by a rope pulling up at the right end at an angle of 30° above the horizontal. What is the tension in the rope?

1. Draw the free-body diagram: forces are tension $T$ at the right end, weight $mg$ acting at the center of the uniform beam (2.0 m from the pivot), and unknown pivot force at the left end.
2. Choose pivot at the left end, so torque from the unknown pivot force is zero, eliminating it from the calculation.
3. Assign signs: weight creates a clockwise (negative) torque, tension creates a counterclockwise (positive) torque. Calculate torques: ${\tau }_{mg}=-(2.0\text{m})(mg)\sin (90^{\circ })=-196\text{Ncdotpm}$, ${\tau }_T=(4.0\text{m})(T)\sin (30^{\circ })=2T$.
4. Set net torque to zero: $2T-196=0⟹T=98\text{N}$.

Exam tip: Always choose your pivot to eliminate as many unknown forces as possible from your torque equation—this will cut your work in half and reduce the chance of algebra errors.

4. Complex Static Systems: Leaning Ladders

Most AP C FRQ problems on rotational statics involve multi-force systems with multiple unknowns, so you need to use both force and torque equilibrium together to solve for all unknowns. The most common example is a uniform ladder leaning against a frictionless vertical wall, standing on a rough horizontal floor. Four forces act on the ladder: weight at the center of mass, normal force from the wall (purely horizontal, since the wall is frictionless), normal force from the floor (purely vertical), and static friction from the floor (horizontal, opposing the tendency of the ladder to slip inward). Three unknowns require one torque equation and two force equilibrium equations for a full solution.

Worked Example

A uniform 5.0 m long ladder of mass 15 kg leans against a frictionless vertical wall, with its base 3.0 m from the wall on a horizontal rough floor. What is the minimum coefficient of static friction between the ladder and floor needed to keep the ladder from slipping?

1. Use the Pythagorean theorem to find the height of the top of the ladder: $h=\sqrt{5^2-3^2}=4.0$ m, forming a 3-4-5 right triangle.
2. List all forces: $N_w$ (normal from wall, horizontal right at the top), $N_f$ (normal from floor, vertical up at the base), $f_s$ (static friction, horizontal left at the base), $mg$ (weight, vertical down at the center of the ladder, 2.5 m along the ladder from the base).
3. Choose pivot at the base of the ladder, so torque from $N_f$ and $f_s$ is zero, eliminating both unknowns.
4. Calculate torques: ${\tau }_{N_w}=+N_{w}h=4N_w$ (CCW positive), ${\tau }_{mg}=-mg(1.5\text{m})=-15(9.8)(1.5)=-220.5\text{Ncdotpm}$ (CW negative).
5. Set net torque to zero: $4N_w-220.5=0⟹N_w=55.125\text{N}$. Use force equilibrium: horizontal $\sum F_x=f_s-N_w=0⟹f_s=55.125\text{N}$, vertical $\sum F_y=N_f-mg=0⟹N_f=147\text{N}$. Minimum ${\mu }_s=f_s/N_f=55.125/147\approx 0.38$.

Exam tip: For leaning ladder problems, always remember that frictionless walls can only exert normal forces perpendicular to the wall—they cannot exert any vertical force, so all vertical force support comes from the floor.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $\cos \theta$ instead of $\sin \theta$ in $\tau =rF\sin \theta$ because you use the angle between the force and the rod and default to cosine from force component problems. Why: Most problems give the angle between the force and the rod, so students confuse torque formulas with linear force component formulas. Correct move: Always confirm that $\sin \theta$ gives the perpendicular component of force (the only component that creates torque), and use $\sin \theta$ for the angle between $\vec{r}$ and $\vec{F}$.
• Wrong move: Choosing the center of mass as the pivot by default, leaving multiple unknown forces in the torque equation that can't be eliminated. Why: Students associate center of mass with all rigid body problems, so they default to it without strategic thinking. Correct move: Always pick the pivot at the location of the largest number of unknown forces, so their torques go to zero immediately.
• Wrong move: Placing the weight of a uniform rigid body at one end instead of the center of mass. Why: Students forget that weight is distributed evenly across the rigid body. Correct move: For any uniform rigid body, mark weight at the geometric midpoint of the object, unless the problem explicitly states the center of mass is elsewhere.
• Wrong move: Solving only with torque equilibrium and forgetting to use force equilibrium to find all unknowns. Why: Students focus on the new rotational condition and forget that static systems require both equilibria. Correct move: Always write down $\sum F_x=0$, $\sum F_y=0$, and $\sum \tau =0$ for all full static problems.
• Wrong move: Mixing up torque signs, leading to negative magnitudes for unknown forces. Why: Students don't write their sign convention explicitly, so they reverse directions when adding torques. Correct move: State your sign convention (CCW positive, CW negative) at the start of every problem, and check the rotation direction of each force's torque before assigning a sign.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A uniform 4.0 m long seesaw is pivoted at its center. A 30 kg child sits 1.5 m to the left of the pivot. A second 40 kg child sits somewhere to the right of the pivot to balance the seesaw. How far from the pivot must the second child sit to balance the system? (A) 1.125 m (B) 1.25 m (C) 1.5 m (D) 1.875 m

Worked Solution: The seesaw is uniform, so its weight acts at the pivot and contributes zero torque. Let $m_1=30\text{kg}$, $d_1=1.5\text{m}$, $m_2=40\text{kg}$, $d_2$ = unknown distance. Set net torque to zero: $m_{1}g{d}_1-m_{2}g{d}_2=0$. Cancel $g$ from both sides, so $d_2=\frac{m_1{d}_1}{m_2}=\frac{30\times 1.5}{40}=1.125\text{m}$. The correct answer is (A).

Question 2 (Free Response)

A non-uniform 3.0 m long beam of total mass 12 kg is suspended horizontally from two vertical ropes: one attached at the left end of the beam, and one attached 0.5 m from the right end. The tension in the left rope is measured to be 50 N. (a) What is the tension in the right rope? (b) How far is the center of mass of the beam from the left end? (c) If the rope at the right end is cut, what is the instantaneous torque about the left end immediately after the cut?

Worked Solution: (a) Use vertical translational equilibrium: $\sum F_y=T_L+T_R-mg=0$. Substitute values: $T_R=(12\times 9.8)-50=117.6-50=67.6\text{N}$. (b) Take pivot at the left end, so torque from $T_L$ is zero. The right rope is $3.0-0.5=2.5\text{m}$ from the left end. Let $x$ = distance of center of mass from the left end. Set net torque to zero: $T_R(2.5)-mgx=0$. Solve for $x$: $x=\frac{67.6\times 2.5}{12\times 9.8}\approx 1.44\text{m}$. (c) After cutting, only weight exerts torque about the left end (tension at the left end acts at the pivot, so torque is zero). $\tau =-mgx=-(117.6)(1.44)\approx -169\text{Ncdotpm}$ (169 N·m clockwise).

Question 3 (Application / Real-World Style)

A construction worker wants to lift a uniform 3.6 m long steel I-beam that weighs 1200 N, pivoted at one end on the ground. The worker can pull with a maximum force of 400 N, and pulls perpendicular to the beam at the free end. What is the maximum angle above the horizontal that the worker can hold the beam in static equilibrium?

Worked Solution: Choose pivot at the ground end, eliminating the unknown pivot force. Let $\theta$ = maximum angle of the beam above the horizontal. Tension $T=400\text{N}$ pulls perpendicular to the beam, so torque from tension is ${\tau }_T=+TL=400\times 3.6=1440\text{Ncdotpm}$ (CCW positive). Weight acts at the center of the beam (1.8 m from the pivot), and the lever arm for weight is $1.8\cos \theta$, so torque from weight is ${\tau }_{mg}=-1200\times 1.8\cos \theta =-2160\cos \theta$. Set net torque to zero: $1440-2160\cos \theta =0⟹\cos \theta =\frac{2}{3}⟹\theta \approx 48^{\circ }$. This means the worker cannot hold the beam at any angle steeper than 48° above horizontal, because the required tension would exceed their maximum pulling strength.

7. Quick Reference Cheatsheet

8. What's Next

This chapter lays the foundational understanding of torque that you will apply immediately to rotational dynamics, the next major topic in Unit 5 Rotation. Just as Newton's second law relates net force to linear acceleration, Newton's second law for rotation relates net torque to angular acceleration—without a solid understanding of how to calculate net torque and identify torque directions, you will not be able to correctly set up rotational dynamics problems for rolling motion, rotating rigid bodies, or pulley systems with massive pulleys. Beyond Unit 5, torque is also a key concept for angular momentum, which relies on the relationship between net torque and change in angular momentum. This topic also extends the equilibrium concepts you learned in Newton's laws to the rotational domain.

• Rotational dynamics
• Angular momentum and energy
• Rolling motion