A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0 = 30. what is the torque on the bar?

9.375Nm

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given F = 25N, r = 0.75m and ∅ = 30°

torque on the bar τ = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm