Question

A mechanic pushes perpendicularly on the end of a 0.15m wrench. If the

mechanic applies a force of 230N counter clockwise, what is the torque applied
on the bolt?

Answer 1

τ = 34.5 N.m

Step-by-step explanation:

The force applied by the mechanic is 230 N.

Thus, F = 230 N

Now, this force is applied in a counterclockwise manner while pushing perpendicularly at the end of the 0.15m long wrench.

Thus, r = 0.15 m

Formula for the torque in this case is;

τ = Fr sin θ

θ = 90° since mechanic pushed perpendicularly.

Thus;

τ = 230 × 0.15 × sin 90

τ = 34.5 N.m