Torque Physics Example 5

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

Example 5

hard
A 40 N40 \text{ N} force is applied to a 0.8 m0.8 \text{ m} lever arm at an angle of 60°60° to the lever. What is the torque? How does this compare to the maximum possible torque with the same force?

Solution

  1. 1
    Torque: τ=rFsin⁡θ=0.8×40×sin⁡60°=32×0.866=27.7 N m\tau = rF\sin\theta = 0.8 \times 40 \times \sin 60° = 32 \times 0.866 = 27.7 \text{ N m}.
  2. 2
    Maximum torque occurs at 90°90°: τmax⁡=rF=0.8×40=32 N m\tau_{\max} = rF = 0.8 \times 40 = 32 \text{ N m}.
  3. 3
    Ratio: ττmax⁡=sin⁡60°=0.866\frac{\tau}{\tau_{\max}} = \sin 60° = 0.866, so the torque is 86.6%86.6\% of the maximum.

Answer

τ≈27.7 N m(86.6% of maximum)\tau \approx 27.7 \text{ N m} \quad (86.6\% \text{ of maximum})
Torque is maximized when the force is perpendicular to the lever arm. At any other angle, only the perpendicular component of the force contributes to torque, reducing it by a factor of sin⁡θ\sin\theta.

About Torque

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

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