A 95-N force exerted at the end of a 0.32-m long torque wrench produces a torque of 15 N ∙ m.What is the angle (less than 90°)between the wrench handle and the direction of the exerted force?
A)30°
B)24°
C)36°
D)42°

Question

Quizplus · Accepted Answer

The torque is given by the product of the force and the lever arm, which is the distance from the point of application of the force to the pivot point. In this case, the lever arm is 0.32 m and the force is 95 N, so the torque is 15 N ∙ m. The angle between the wrench handle and the direction of the exerted force can be found using the formula:torque = force × lever arm × sin(angle)Solving for the angle, we get:angle = sin^-1(torque / (force × lever arm))Plugging in the values we know, we get:angle = sin^-1(15 N ∙ m / (95 N × 0.32 m)) = 30°Therefore, the angle between the wrench handle and the direction of the exerted force is 30°.

A 95-N Force Exerted at the End of a 0