Question

A 25N force F is applied to a bar that can pivot around its end as shown below. The force is r = 0.75 m away from the end and at an angle θ = 30°.

Answer 1

The value of torque on the bar is τ = 9.375 N.m

Explanation:

We are given that:

F = applied force = 25 N

r = radius = 0.75 m

∅ = angle that the force made with the bar = 30°

Solution:

The formula for calculating torque τ = Frsin∅  

Now put the values in the above formula.

Torque on the bar τ  = 25 x 0.75 x sin(30°)

τ = 25 x 0.75 x 0.5

τ = 9.375 N.m