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This cheat sheet covers torque and rotational equilibrium through worked-example thinking for high school physics. Students use it to decide where forces act, how far they are from the pivot, and which direction each torque turns an object. It is especially useful for beams, seesaws, ladders, hinges, and hanging signs.

The layout is designed as a clean printable reference with three color-coded sections for quick review.

Key Facts

  • Torque is calculated with τ=rFsinθ\tau = rF\sin\theta, where rr is the distance from the pivot to the force and θ\theta is the angle between r\vec{r} and F\vec{F}.
  • When a force is perpendicular to a lever arm, the torque simplifies to τ=rF\tau = rF.
  • For rotational equilibrium, the net torque must be zero, so τ=0\sum \tau = 0.
  • For complete static equilibrium, both force balance and torque balance must hold: Fx=0\sum F_x = 0, Fy=0\sum F_y = 0, and τ=0\sum \tau = 0.
  • Choosing the pivot at an unknown force can remove that force from the torque equation because its lever arm is r=0r = 0.
  • Clockwise and counterclockwise torques must have opposite signs, such as τccw>0\tau_{\text{ccw}} > 0 and τcw<0\tau_{\text{cw}} < 0.
  • The lever arm is the perpendicular distance from the pivot to the force line of action, so r=rsinθr_{\perp} = r\sin\theta.
  • A balanced beam with two perpendicular forces often uses F1r1=F2r2F_1r_1 = F_2r_2 when the torques are equal and opposite.

Vocabulary

Torque
Torque is the turning effect of a force and is measured in newton-meters, Nm\text{N}\cdot\text{m}.
Pivot
A pivot is the point or axis around which an object can rotate.
Lever Arm
The lever arm is the perpendicular distance from the pivot to the force line of action.
Line of Action
The line of action is the straight line in the direction of a force extending through the point where the force is applied.
Rotational Equilibrium
Rotational equilibrium occurs when the net torque on an object is zero, written as τ=0\sum \tau = 0.
Static Equilibrium
Static equilibrium occurs when an object has no linear acceleration and no angular acceleration, so F=0\sum \vec{F} = 0 and τ=0\sum \tau = 0.

Common Mistakes to Avoid

  • Using the full distance instead of the perpendicular lever arm is wrong because torque depends on rr_{\perp}, not just the distance along the object.
  • Forgetting the sign of torque is wrong because clockwise and counterclockwise torques must cancel in τ=0\sum \tau = 0.
  • Choosing a pivot randomly can make the algebra harder because a smart pivot can eliminate an unknown force with r=0r = 0.
  • Including a force that acts through the pivot is wrong in a torque equation because its torque is τ=0\tau = 0.
  • Mixing force balance with torque balance is wrong because Fy=0\sum F_y = 0 and τ=0\sum \tau = 0 are separate equations with different meanings.

Practice Questions

  1. 1 A student pushes perpendicular to a door with a force of 35N35\,\text{N} at a distance of 0.80m0.80\,\text{m} from the hinge. What is the torque about the hinge?
  2. 2 A 4.0m4.0\,\text{m} uniform beam weighs 120N120\,\text{N} and is supported at its left end. What counterclockwise torque is needed at the right end to hold it in rotational equilibrium?
  3. 3 A seesaw has a 300N300\,\text{N} student sitting 1.5m1.5\,\text{m} from the pivot. How far from the pivot should a 450N450\,\text{N} student sit on the opposite side to balance it?
  4. 4 When solving a ladder or beam problem, why is it often useful to choose the pivot at a point where an unknown support force acts?

Understanding Torque and Rotational Equilibrium Worked Examples

The most important geometric idea is the line of action of a force. Imagine extending the force arrow into a long straight line. The turning effect depends on the shortest distance from the pivot to that line, not simply on the distance to the point where the arrow begins.

This explains why pushing a door near its hinge feels ineffective. A push near the handle works better because its line of action is farther from the hinge.

A force aimed directly toward or away from the pivot has no turning effect. Its line of action passes through the pivot, so the perpendicular distance is zero.

A good diagram separates the object from everything touching it. This is called a free body diagram. Include the object’s weight, support forces, tensions, friction forces, and any applied loads.

Weight acts through the center of mass. For a uniform beam, that point is at its middle. For an uneven object, it may not be.

A hinge can exert both horizontal and vertical forces, while a rope can pull only along its length. Choosing the hinge as the pivot often makes the torque calculation shorter. Its support forces still belong in the force balance equations, even though they produce no torque about that chosen point.

Use one sign convention for every torque in a problem. Many students choose counterclockwise as positive and clockwise as negative. The choice itself does not matter, but changing it halfway through a calculation causes errors.

Write each torque with its sign before combining terms. Keep distances in metres and forces in newtons, giving torque in newton metres. Check the angle carefully when a force is slanted.

The relevant angle is between the position direction from pivot to force point and the force direction. It is often safer to find the perpendicular lever arm from the diagram first.

A quick check helps. A larger force or a larger perpendicular distance should create a larger turning effect.

Real structures show why torque balance needs force balance too. A wall sign can have zero net turning effect while still moving downward if the supporting cable force is too small. A ladder against a wall has weight, wall contact forces, floor contact forces, and often friction at the floor.

Static friction is not always at its maximum value. It adjusts to whatever value is needed up to its limit. Tipping begins when the line of action of the total weight falls outside the support area.

When practicing worked examples, label every distance from a clearly chosen pivot, identify the center of mass, and state the assumptions. Common assumptions include a massless rope, a uniform beam, and contacts that do not slip. These details decide which forces belong in the model.