Torque — Pushing a Door Open (Worked Example)

Easy

Problem

You push perpendicularly on a door handle 0.80 m from the hinges with a force of 55 N. What torque do you exert on the door?

Given

r = 0.80 m
F = 55 N
θ = 90°

Find

The torque τ about the hinges.

Principle

Torque equals lever arm times force for a perpendicular push; pushing farther from the hinge gives more rotation for the same effort.

Formula

\tau = rF

τ = rF

Solution

Since the push is perpendicular: $\tau = rF$
Substitute: $\tau = 0.80 \times 55$
Multiply: $\tau = 44\ \mathrm{N\cdot m}$

44 N·m

Unit check

m × N = N·m — correct for torque.

What this means physically

44 N·m swings the door easily. Push at 0.20 m from the hinge instead and you'd need 220 N for the same torque — four times the effort, which is why handles sit far from hinges.

Common mistake

Thinking the door's weight matters for this torque. Gravity acts at the center of mass straight down through the hinge axis region — it contributes no torque about the vertical hinge axis.