United StatesPhysics C: MechanicsSyllabus dot point

What does it mean for a rigid body to be in rotational equilibrium, and how do we use zero net force and zero net torque to solve statics problems?

Topic 5.5 Rotational Equilibrium and Newton's First Law: state the two conditions for static equilibrium (zero net force and zero net torque) and apply them to find unknown forces on rigid bodies.

A focused answer to AP Physics C: Mechanics Topic 5.5, covering the two conditions for static equilibrium of a rigid body (zero net force and zero net torque), choosing a convenient pivot, the role of the center of gravity, and solving for unknown support and tension forces on beams and ladders, with worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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What this topic is asking
The two conditions for equilibrium
Choosing the pivot
The center of gravity
A method for statics problems
Try this

What this topic is asking

The College Board (Topic 5.5) wants you to state the two conditions for the static equilibrium of a rigid body, zero net force and zero net torque, and to apply them to find unknown forces. This is the rotational extension of Newton's first law: a body can have balanced forces yet still rotate, so equilibrium of an extended object requires balancing torques as well as forces. Beam, ladder and bridge problems all rest on this.

The two conditions for equilibrium

For a point particle, Newton's first law (zero net force) is enough. But an extended body can have a zero net force and still rotate: two equal and opposite forces applied at different points (a "couple") sum to zero force yet produce a net torque. So equilibrium of a rigid body needs the torque condition as well. Together, zero net force and zero net torque guarantee the body neither translates nor rotates, the state of static equilibrium examined in statics problems.

Choosing the pivot

A powerful simplification: because the body is in equilibrium, the net torque is zero about every point, so you may choose the axis wherever it is most convenient. The smart choice is to put the axis at an unknown force, so that force has zero lever arm and drops out of the torque equation. This leaves one torque equation in the remaining unknowns, often a single unknown you can solve immediately. After finding it, return to the force-balance equations for the eliminated force. This pivot-at-an-unknown trick is the key technique for beam and ladder problems.

The center of gravity

A rigid body's weight acts as though concentrated at its center of gravity, which coincides with the center of mass in a uniform gravitational field. When you draw the free-body diagram of a beam or plank, place the single weight force at this point (the geometric center for a uniform object). This lets you treat the distributed weight as one force with a definite lever arm in the torque equation. For non-uniform bodies, the weight acts at the (mass-weighted) center of gravity, which may be off-center.

A method for statics problems

The reliable procedure: draw the free-body diagram with every force at its point of application; choose axes and write $\sum F_x = 0$ and $\sum F_y = 0$ ; choose a pivot (ideally at an unknown) and write $\sum\tau = 0$ , with a consistent sign convention (counterclockwise positive); then solve the equations. Three equations (two force, one torque, in a plane) determine up to three unknowns. Ladders against walls, beams on supports, and signs on brackets all follow this template.

A plank on two supports

A uniform $12$ kg, $3.0$ m plank rests on two supports, one at the left end and one $2.0$ m from the left end. A $30$ kg child stands at the right end. Take $g = 9.8$ m/s squared. Find the upward force from each support.

step 1 Draw the diagram and place the weights

Plank weight $W_p = (12)(9.8) = 118$ N at the center ( $1.5$ m). Child weight $W_c = (30)(9.8) = 294$ N at the right end ( $3.0$ m). Support forces $N_1$ (left end, $0$ m) and $N_2$ (at $2.0$ m), both up.

step 2 Take torques about the left end to eliminate $N_1$

$\sum\tau = 0$ : $N_2(2.0) - W_p(1.5) - W_c(3.0) = 0 \Rightarrow N_2 = \dfrac{118(1.5) + 294(3.0)}{2.0} = \dfrac{177 + 882}{2.0} = 530$ N.

step 3 Use force balance for $N_1$

$\sum F_y = 0$ : $N_1 + N_2 - W_p - W_c = 0 \Rightarrow N_1 = 118 + 294 - 530 = -118$ N.

step 4 Interpret the sign

$N_1$ is negative, meaning the left support must pull down $118$ N (or the plank would tip about the second support); the child at the far end overbalances it. A real plank resting freely would tip.

Try this

Q1. A uniform $8.0$ kg beam $2.0$ m long is held horizontal by a single support at its center. State the net torque about the support and explain. [2 points]

Cue. Zero: the weight acts at the center, directly above the support, so its lever arm is zero and there is no net torque.

Q2. State the two conditions a ladder leaning against a wall must satisfy to remain in static equilibrium. [2 points]

Cue. Zero net force ( $\sum F_x = 0$ , $\sum F_y = 0$ ) and zero net torque about any axis.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)6 marksSection II (FRQ). A uniform beam of mass

20

kg and length

4.0

m is supported by a pivot at its left end and a vertical cable at its right end. A

50

kg load sits

1.0

m from the left end. Take

g = 9.8

m/s squared. (a) Draw a free-body diagram. (b) Write the two equilibrium conditions. (c) Determine the cable tension. (d) Determine the force the pivot exerts on the beam.

Show worked answer →

A 6-point static-equilibrium FRQ.

(a) Diagram (1 point): pivot force $P$ (up) at the left end, cable tension $T$ (up) at the right end, beam weight $W_b = (20)(9.8) = 196$ N at the center ( $2.0$ m), load $W_L = (50)(9.8) = 490$ N at $1.0$ m.
(b) Conditions (1 point): $\sum F_y = 0$ and $\sum\tau = 0$ .
(c) Tension (2 points): take torques about the left pivot (eliminating $P$ ): $T(4.0) - W_b(2.0) - W_L(1.0) = 0 \Rightarrow T = \dfrac{196(2.0) + 490(1.0)}{4.0} = \dfrac{392 + 490}{4.0} = 220$ N.
(d) Pivot force (2 points): $\sum F_y = 0$ : $P + T - W_b - W_L = 0 \Rightarrow P = 196 + 490 - 220 = 466$ N up.

Markers reward choosing the pivot at an unknown to eliminate it from the torque equation, then using force balance for the pivot force.

AP 2021 (style)1 marksSection I (multiple choice). A rigid body is in static equilibrium. Which condition(s) must hold? (A) zero net force only (B) zero net torque only (C) both zero net force and zero net torque (D) the body must be symmetric. Justify your reasoning.

Show worked answer →

A 1-point conceptual MCQ. The answer is (C).

Static equilibrium of a rigid body requires both the net force and the net torque to be zero: zero net force prevents linear acceleration of the center of mass, and zero net torque prevents angular acceleration. A body can have zero net force yet still rotate (a couple), so both conditions are needed. The trap (A) or (B) supplies only one of the two.

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Sources & how we know this

AP Physics C: Mechanics Course and Exam Description — College Board (2024)