Torque — Pushing a Door (Worked Example)

Easy

Problem

You push a door perpendicularly at a point 0.90 m from the hinges with a force of 30 N. What torque do you exert about the hinge axis?

Given

r = 0.90 m
F = 30 N (perpendicular)

Find

The torque τ.

Principle

The hinge is the pivot; torque τ = rF measures the turning effect, which is why handles sit far from hinges.

Formula

\tau = rF

τ = rF

Solution

Write the formula: $\tau = rF$
Substitute: $\tau = (0.90)(30)$
Multiply: $\tau = 27\ \mathrm{N·m}$

27 N·m

Unit check

m × N = N·m — correct.

What this means physically

27 N·m swings the door easily. Push near the hinge (small r) and the same 30 N produces far less torque — the everyday reason door handles are at the far edge.

Common mistake

Measuring r from the wrong point. Torque needs the distance from the pivot (hinge), not from the door's edge or your hand's starting position.