Torque Physics Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

medium
A 30 N force is applied at the end of a 0.4 m wrench at 60\u00b0 to the handle. Calculate the torque.

Solution

  1. 1
    Recall the torque formula: τ=Frsinθ\tau = F \cdot r \cdot \sin\theta, where FF is the applied force, rr is the distance from the pivot, and θ\theta is the angle between the force and the lever arm.
  2. 2
    Substitute the given values: F=30NF = 30\,\text{N}, r=0.4mr = 0.4\,\text{m}, θ=60°\theta = 60°.
  3. 3
    Calculate: τ=30×0.4×sin60°=12×32=12×0.866=10.4Nm\tau = 30 \times 0.4 \times \sin 60° = 12 \times \frac{\sqrt{3}}{2} = 12 \times 0.866 = 10.4\,\text{N\,m}

Answer

The torque is τ10.4Nm\tau \approx 10.4\,\text{N\,m}.
Torque depends on the component of force perpendicular to the lever arm. The sinθ\sin\theta factor accounts for the angle: maximum torque occurs at 90\u00b0 (force perpendicular to wrench), and zero torque at 0\u00b0 or 180\u00b0 (force along the wrench).

About Torque

The rotational equivalent of force; a measure of how much a force tends to cause an object to rotate about an axis.

Learn more about Torque →

More Torque Examples