Question

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?

Answer 1

Answer:

Angle between force exerted and handle of the wrench is 29.56 degree

Explanation:

We have given force exerted at the end of the wrench  F = 95 N m

Length of the wrench d = 0.32 m

Torque is given as $\tau =15Nm$

Let the angle between wrench handle and force exerted is $\Theta$

Torque is equal to $\tau =Fsin\Theta \times d$

$15=95\times sin\Theta \times 0.32$

$sin\Theta =0.4932$

$\Theta =29.56$

So angle between force exerted and handle of the wrench is 29.56 degree