How to Calculate Torque and Solve Rotational Equilibrium

To calculate torque, multiply the force by its distance from the pivot and by the sine of the angle between them: torque = rF sin theta, in newton meters. For rotational equilibrium, the counterclockwise torques must balance the clockwise torques so the net torque about any pivot is zero.

AP Physics: Unit 5 (topics 5.3 Torque, 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form). Covers Topics 5.3 and 5.5 in Unit 5 of AP Physics 1, which carries 10-15% of the multiple-choice section. The same torque and equilibrium ideas appear in Unit 5 of AP Physics C: Mechanics at the same 10-15% weight.

Start with the torque formula

To calculate torque, multiply the force by the distance from the pivot to the point where the force is applied, then by the sine of the angle between the position vector and the force:

\tau = r_{\perp} F = rF\sin\theta

Here $r$ is the distance from the axis of rotation to where the force acts, $F$ is the magnitude of the force, and $\theta$ is the angle between them. Torque is measured in newton meters (N·m). This is exactly how the formula appears on the AP Physics 1 equation sheet, with $r_{\perp}$ standing for the lever arm.

Two quick sanity checks: a force applied at the pivot ( $r = 0$ ) produces no torque, and a force pointing straight along the object toward the pivot ( $\theta = 0$ ) produces no torque either. Only the perpendicular part of the force makes something rotate. You can test numbers quickly with the torque calculator.

What the lever arm actually means

The lever arm $r_{\perp}$ is the perpendicular distance from the axis of rotation to the line of action of the force. Slide a force anywhere along its own line of action and the torque stays the same. What matters is how far that line misses the pivot.

Both groupings of the formula give the same number:

$\tau = (r\sin\theta)F$ : the lever arm times the full force.
$\tau = r(F\sin\theta)$ : the full distance times the perpendicular component of the force.

Use whichever is easier to picture in a given problem. Think about a door: pushing perpendicular to the door at the handle gives the largest lever arm, so it swings easily. Pushing at the hinge gives $r = 0$ , and pushing along the door straight toward the hinge gives a lever arm of zero. In both cases the door does not rotate, no matter how hard you push.

Pick a sign convention and stay with it

Torque has a direction. In AP Physics 1, rotation problems stay in a plane, so all you need is clockwise versus counterclockwise. The usual convention takes counterclockwise as positive and clockwise as negative, which matches the direction of increasing angle used in rotational kinematics.

The net torque is the signed sum of every individual torque:

\tau_{net} = \Sigma\tau

If $\tau_{net}$ is positive under your convention, the object gains counterclockwise rotation. If it is negative, it gains clockwise rotation. If it is zero, the rotation does not change at all.

A common error is adding torque magnitudes that actually oppose each other. Assign each torque its sign the moment you write it down, before any algebra. One consistent convention per problem, applied to every single force, prevents nearly all sign disasters.

Rotational equilibrium: net torque equals zero

An object is in rotational equilibrium when the net torque on it is zero. This is Newton's first law in rotational form: with zero net torque, angular velocity stays constant, and in a statics problem that constant is zero.

Full static equilibrium requires two conditions at the same time:

$\Sigma F = 0$ : no translational acceleration.
$\Sigma\tau = 0$ : no angular acceleration.

Here is the trick that makes these problems manageable: when an object is in equilibrium, the net torque is zero about every axis, so you may put the pivot anywhere you want. Choose it at the point where an unknown force acts. That force then has $r = 0$ , contributes zero torque, and vanishes from the torque equation. You are left with one equation and one unknown instead of a messy system.

A five-step method for equilibrium problems

Draw an extended free body diagram. Sketch the object as a bar and place each force at the exact point where it acts, not all at the center. The basics are covered in how to draw a free body diagram.
Choose the pivot at the location of an unknown force so that force drops out of the torque equation.
Write each torque as $rF\sin\theta$ with a sign: counterclockwise positive, clockwise negative. For horizontal beams with vertical forces, $\theta = 90^\circ$ and $\sin\theta = 1$ .
Set $\Sigma\tau = 0$ and solve for the unknown.
If the problem asks for a second force, use $\Sigma F = 0$ on the whole object.

For a uniform beam or plank, treat its entire weight $mg$ as acting at the midpoint. That one substitution handles the beam's own weight in almost every AP equilibrium problem.

Mistakes that cost points

The most common torque errors are all avoidable:

Dropping $\sin\theta$ when the force is not perpendicular. If the angle between $r$ and $F$ is $30^\circ$ , you keep only half the force's turning effect, since $\sin 30^\circ = 0.5$ .
Forgetting the object's own weight. A uniform beam's weight acts at its center and usually contributes a torque.
Mixing units. Convert centimeters to meters before multiplying, so torques come out in N·m.
Writing torque in joules. N·m has the same base units as the joule, but torque is not energy, so keep it as N·m.
Assuming zero net torque means nothing moves. It means angular velocity is constant; a wheel spinning at a steady rate also has zero net torque.

When the net torque is not zero, the object has angular acceleration, which is where Newton's second law in rotational form picks up later in Unit 5.

Torque on a wrench at an angle

You pull on the end of a wrench 0.25 m from the bolt with a 40.0 N force, at an angle of $60^\circ$ to the wrench handle. Find the magnitude of the torque about the bolt.

List the values: $r = 0.25$ m, $F = 40.0$ N, $\theta = 60^\circ$ .
Apply the formula: $\tau = rF\sin\theta = (0.25\ \text{m})(40.0\ \text{N})(\sin 60^\circ)$ .
Evaluate the sine: $\sin 60^\circ = 0.866$ .
Multiply: $\tau = (0.25)(40.0)(0.866) = 8.66$ N·m.
Check with the lever arm form: $r_{\perp} = r\sin\theta = (0.25\ \text{m})(0.866) = 0.217$ m, and $(0.217\ \text{m})(40.0\ \text{N}) = 8.66$ N·m. Same answer, as it must be.

The torque is $\tau = 8.66$ N·m, about 8.7 N·m. Pulling at $60^\circ$ instead of $90^\circ$ costs you about 13 percent of the maximum possible torque of 10.0 N·m.

Balancing a seesaw

A 30.0 kg child sits 2.0 m to the left of a seesaw's pivot. The uniform board is pivoted at its center. How far from the pivot must a 40.0 kg child sit on the right side to balance the seesaw?

The board is uniform and pivoted at its center, so its weight acts at the pivot with $r = 0$ and contributes zero torque. Only the two children matter.
Left child's weight: $F_1 = m_1 g = (30.0\ \text{kg})(9.8\ \text{m/s}^2) = 294$ N. Both weights act perpendicular to the horizontal board, so $\sin\theta = 1$ for each.
Left child's torque (counterclockwise, positive): $\tau_1 = +(2.0\ \text{m})(294\ \text{N}) = +588$ N·m.
Right child's weight: $F_2 = (40.0\ \text{kg})(9.8\ \text{m/s}^2) = 392$ N. Torque (clockwise, negative): $\tau_2 = -(d)(392\ \text{N})$ .
Set the net torque to zero: $588 - 392d = 0$ , so $d = 588 / 392 = 1.5$ m.
Sanity check: the heavier child sits closer to the pivot, which is what everyday seesaw experience predicts.

The 40.0 kg child must sit $d = 1.5$ m from the pivot on the right side.

Person standing on a supported plank

A uniform 15.0 kg plank of length 6.0 m rests on supports at its two ends. A 55.0 kg person stands 2.0 m from the left support. Find the upward force from each support.

Identify the forces: left support pushes up with $N_L$ at $x = 0$ , right support pushes up with $N_R$ at $x = 6.0$ m, the person's weight $(55.0\ \text{kg})(9.8\ \text{m/s}^2) = 539$ N acts at $x = 2.0$ m, and the plank's weight $(15.0\ \text{kg})(9.8\ \text{m/s}^2) = 147$ N acts at the midpoint, $x = 3.0$ m.
Choose the pivot at the left support. Then $N_L$ has $r = 0$ and drops out of the torque equation.
Write the torques with signs. $N_R$ turns the plank counterclockwise: $+(6.0)N_R$ . The two weights turn it clockwise: $-(2.0)(539) = -1078$ N·m and $-(3.0)(147) = -441$ N·m.
Set $\Sigma\tau = 0$ : $6.0N_R - 1078 - 441 = 0$ , so $6.0N_R = 1519$ and $N_R = 1519 / 6.0 = 253$ N.
Use $\Sigma F = 0$ for the vertical direction: $N_L + N_R = 539 + 147 = 686$ N, so $N_L = 686 - 253 = 433$ N.
Check: the person stands closer to the left support, so the left support should carry more of the load, and 433 N is indeed larger than 253 N.

The left support pushes up with about $N_L = 433$ N and the right support with about $N_R = 253$ N. Together they equal the total weight of 686 N.

Frequently asked questions

What units does torque use?

Torque is measured in newton meters (N·m). The base units match the joule, but torque is not energy: it measures a force's turning effect about an axis, not work done along a path. Keep torque in N·m and energy in joules so the two never get confused.

What angle goes into the torque formula?

Theta is the angle between the position vector (from the pivot to where the force acts) and the force itself. When the force is perpendicular to the object, the angle is 90 degrees, sin theta = 1, and torque is simply rF. That is the case in most horizontal beam and seesaw problems, since weights point straight down while the beam is horizontal.

Does choosing a different pivot point change the answer?

No. When an object is in equilibrium, the net torque is zero about every axis, so any pivot gives the same physical answer. Pick the pivot where an unknown force acts: that force gets a moment arm of zero, drops out of the torque equation, and leaves you with fewer unknowns to juggle.

Is zero net torque the same as not rotating?

Not quite. Zero net torque means the angular velocity is constant. A stationary beam stays stationary, but a wheel already spinning at a steady rate also has zero net torque. This mirrors Newton's first law for straight-line motion, just written for rotation.

What happens when the net torque is not zero?

The object has an angular acceleration, and its angular velocity changes over time. Predicting that motion uses the rotational versions of the kinematic equations, covered in the rotational kinematics guide, plus Newton's second law in rotational form from Topic 5.6.