Torque and Rotational Statics

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Rotation· 14 min read

1. Definition and Calculation of Torque★★☆☆☆⏱ 4 min

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📘 Definition

Torque

$τ$

A vector quantity that measures the tendency of a force to cause rotation about a given pivot point, the rotational equivalent of force in linear mechanics.

Example:

A force pulling sideways on the top of a vertical pole exerts a torque about the pole's base.

The general definition of torque comes from the cross product of the position vector $r$ (from the pivot to the point of force application) and the applied force $F$ :

vec{tau} = vec{r} times vec{F}

The magnitude of torque is given by $τ = r F sin θ$ , where $θ$ is the angle between $r$ and $F$ when placed tail to tail. An equivalent formulation using the lever arm (moment arm) $d = r sin θ$ , the perpendicular distance from the pivot to the line of action of the force, simplifies static problems to $τ = F d$ . The standard AP sign convention takes counterclockwise (CCW) torque as positive and clockwise (CW) torque as negative, matching the right-hand rule for cross products.

📐 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 $r$ points straight up along the pole from the base (pivot) to the top, with magnitude $r = 2.0$ m. The force vector $F$ points down along the cable, 30° from the pole.
2
The angle between $r$ and $F$ is $18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ}$ , and $sin (15 0^{\circ}) = sin (3 0^{\circ}) = 0.5$ .
3
Calculate torque magnitude with both methods to confirm:
$tau = rF\sin\theta = (2.0\ \text{m})(500\ \text{N})(0.5) = 500\ \text{N·m}$
4
Using the lever arm method: $d = r sin (3 0^{\circ}) = 1.0$ m, so $τ = F d = 500 N \cdotp m$ , matching the first result.
5
The cable causes clockwise rotation about the base, so the torque is negative:
$tau = -500\ \text{N·m}$

2. Equilibrium Conditions for Rotational Statics★★☆☆☆⏱ 3 min

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For a rigid body to be completely static (no acceleration of any kind), two conditions must hold: translational equilibrium and rotational equilibrium. Translational equilibrium, from linear statics, means net force in all directions is zero:

sum vec{F} = 0 implies sum F_x = 0, quad sum F_y = 0

The new core condition for rotational statics is that net torque about any pivot point is zero:

sum tau = 0

A critical problem-solving simplification: if a system is in translational equilibrium, net torque is identical about any pivot point. You can choose any pivot to simplify calculations, and the most strategic choice is almost always a pivot at the location of an unknown force, since that force has $r = 0$ , so its torque is zero, eliminating the unknown 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 $m g$ 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:
$tau_{mg} = -(2.0\ \text{m})(mg)\sin(90^\circ) = -196\ \text{N·m}, quad tau_T = (4.0\ \text{m})(T)\sin(30^\circ) = 2T$
4
Set net torque to zero and solve for $T$ :
$2T - 196 = 0 implies T = 98\ \text{N}$

3. Complex Static Systems: Leaning Ladders★★★☆☆⏱ 4 min

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Most AP C FRQ problems on rotational statics involve multi-force systems with multiple unknowns, requiring both force and torque equilibrium together to solve. The most common example is a uniform ladder leaning against a frictionless vertical wall, standing on a rough horizontal floor.

📐 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:
$h = sqrt{5^2 - 3^2} = 4.0\ \text{m}$
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), $m g$ (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 (CCW positive):
$tau_{N_w} = +N_w h = 4N_w, quad tau_{mg} = -mg (1.5\ \text{m}) = -15(9.8)(1.5) = -220.5\ \text{N·m}$
5
Set net torque to zero, then use force equilibrium to find remaining unknowns:
$4N_w - 220.5 = 0 implies N_w = 55.125\ \text{N} \\ sum F_x = f_s - N_w = 0 implies f_s = 55.125\ \text{N} \\ sum F_y = N_f - mg = 0 implies N_f = 147\ \text{N}$
6
Calculate minimum coefficient of static friction:
$mu_s = f_s / N_f = 55.125 / 147 approx 0.38$

4. Practice Worked Examples★★★☆☆⏱ 3 min

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📐 Worked Example

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

1
The seesaw is uniform, so its weight acts at the pivot and contributes zero torque. Let $m_{1} = 30 kg$ , $d_{1} = 1.5 m$ , $m_{2} = 40 kg$ , $d_{2}$ = unknown distance.
2
Set net torque to zero for equilibrium:
$m_1 g d_1 - m_2 g d_2 = 0$
3
Cancel $g$ from both sides and solve for $d_{2}$ :
$d_2 = frac{m_1 d_1}{m_2} = frac{30 times 1.5}{40} = 1.125\ \text{m}$
4
The correct answer is (A).

📐 Worked Example

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?

1
Part (a): Use vertical translational equilibrium to solve for $T_{R}$ :
$sum F_y = T_L + T_R - mg = 0 \\ T_R = (12 times 9.8) - 50 = 67.6\ \text{N}$
2
Part (b): Take pivot at the left end, so torque from $T_{L}$ is zero. The right rope is $2.5 m$ from the left end. Let $x$ = distance of center of mass from the left end. Set net torque to zero:
$T_R (2.5) - mg x = 0 \\ x = frac{67.6 times 2.5}{12 times 9.8} approx 1.44\ \text{m}$
3
Part (c): After cutting, only weight exerts torque about the left end (tension at the pivot contributes zero torque):
$tau = -mg x approx -169\ \text{N·m} (169\ \text{N·m clockwise})$

📐 Worked Example

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?

1
Choose pivot at the ground end, eliminating the unknown pivot force. Let $θ$ = maximum angle of the beam above the horizontal.
2
Tension $T = 400 N$ pulls perpendicular to the beam, so torque from tension is:
$tau_T = +TL = 400 times 3.6 = 1440\ \text{N·m}$
3
Weight acts at the center of the beam (1.8 m from the pivot), and the lever arm for weight is $1.8 cos θ$ , so torque from weight is:
$tau_{mg} = -1200 times 1.8 \cos\theta = -2160 \cos\theta$
4
Set net torque to zero and solve for $θ$ :
$1440 - 2160 \cos\theta = 0 implies cos\theta = frac{2}{3} implies theta approx 48^\circ$

5. Common Pitfalls

Wrong move:

Using $cos θ$ instead of $sin θ$ in $τ = r F sin θ$ 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 $sin θ$ gives the perpendicular component of force (the only component that creates torque), and use $sin θ$ for the angle between $r$ and $F$ .

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 $\sum τ = 0$ 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. Quick Reference Cheatsheet

Category	Formula	Notes
General Torque Vector	$τ = r \times F$	Direction given by right-hand rule for cross products
Torque Magnitude	$τ = r F sin θ = F d$	$θ$ = angle between $r$ and $F$ ; $d$ = lever arm (perpendicular distance to line of force)
Translational Equilibrium	$\sum F_{x} = 0, \sum F_{y} = 0$	Required for all static rigid body problems; applies to x/y directions separately
Rotational Equilibrium	$\sum τ = 0$	Net torque zero about any pivot for static systems; any pivot is valid if translation equilibrium holds
Standard Torque Sign Convention	CCW = positive, CW = negative	Matches AP Physics C convention for rotation axes out of the page
Uniform Rigid Body Weight	Acts at geometric center	For beams, ladders, etc., center of mass is at the midpoint of length
Frictionless Wall Force	Purely normal (perpendicular) to wall	No vertical component; all vertical support comes from the ground

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Torque sign and magnitude calculation
2022 · FRQ
Leaning ladder static equilibrium problem

Going deeper

official syllabusAP Physics C: Mechanics CEDOfficial Unit 5 content outline

What's Next

This subtopic 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, extending the equilibrium concepts you learned in Newton's laws to the rotational domain.