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 applied force?
A)30°
B)24°
C)36°
D)42°

Question

Quizplus · Accepted Answer

The torque is given by the formula:$$	au = Fd\sin	heta$$where:$$	au$$ is the torque in N·mF is the force in Nd is the distance from the pivot point to the point where the force is applied in m$$	heta$$ is the angle between the force and the distance vector in radiansWe can rearrange this formula to solve for $$	heta$$:$$	heta = \sin^{-1}\left(\frac{	au}{Fd}ight)$$Plugging in the given values, we get:$$	heta = \sin^{-1}\left(\frac{15	ext{ N$\cdot$m}}{(95	ext{ N})(0.32	ext{ m})}ight) = 30^\circ$$Therefore, the angle between the wrench handle and the direction of the applied force is 30°.

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