Question

A 1.10-kg wrench is acting on a nut trying to turn it. The length of the wrench lies directly to the east of the nut. A force 150.0 N acts on the wrench at a position 15.0 cm from the center of the nut in a direction 30.0° north of east. What is the magnitude of the torque about the center of the nut

Answer 1

Answer:

τ = 11.25 Nm

Explanation:

Given,

Mass of wrench, m = 1.10 Kg

Force on the wrench, F = 150 N

position of force, r = 15 cm = 0.15 m

angle of nut = 30°

We know,

torque = Force x distance

F = 150 sin 30°

F = 75 N

τ = 75 x r

τ = 75 x 0.15

τ = 11.25 Nm

Torque at the center of the nut is equal to 11.25 N.