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Numerical · Q21

Q.A force of 25 N25\ \text{N} is applied at the end of a wrench of length 0.3 m0.3\ \text{m} to turn a bolt.

(a) Find the torque produced when the force is applied exactly perpendicular to the wrench.
(b) Find the torque produced when the same force is instead applied at an angle of 30∘30^\circ to the length of the wrench.
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✓ Free question

(a) Perpendicular force (θ=90∘\theta = 90^\circ, sin⁡90∘=1\sin90^\circ=1):

τ=rFsin⁡θ=(0.3)(25)(1)=7.5 N m\tau = rF\sin\theta = (0.3)(25)(1) = 7.5\ \text{N}\,\text{m}

(b) Force at 30∘30^\circ to the wrench (sin⁡30∘=0.5\sin30^\circ = 0.5):

τ=rFsin⁡θ=(0.3)(25)(0.5)=3.75 N m\tau = rF\sin\theta = (0.3)(25)(0.5) = 3.75\ \text{N}\,\text{m}

Applying the same force at 30∘30^\circ instead of perpendicular produces exactly HALF the torque, directly reflecting that sin⁡30∘=0.5=12(1)=12sin⁡90∘\sin30^\circ = 0.5 = \tfrac12(1) = \tfrac12\sin90^\circ -- this is exactly why a wrench, spanner, or door handle is always most effective when pushed or pulled as close to perpendicular to it as possible.

✓Final answer

  1. τ=7.5 N m\tau = 7.5\ \text{N}\,\text{m} (perpendicular);
  2. τ=3.75 N m\tau = 3.75\ \text{N}\,\text{m} (at 30∘30^\circ).

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