Question

A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long and, he exerts a 17.0-N force at the end of the handle at $\mathbf{37}\mathbf{°}$with the handle (Fig. E10.7). (a) What torque does the machinist exerts about the center of the nut? (b) What is the maximum torque he could exert with this force, and how should the force be oriented?

Figure E10.7

Short answer

(a) The torque exerted about the center of the nut is, $\tau =2.56 N·m$

(b) The maximum torque exerted with the given force is, ${\tau }_{\max }=4.25 N·m$

Step-by-step solution

To mention the given data

We have the given data:

The length of the wrench $r=25$ cm.

The magnitude of the force exerted at the end of the handle $F=17.0N$.

The angle between the force vector and position vector =$\theta =37°$.

Concept

The magnitude of the torque associated with the force $F$acting on an object at a distance r from the rotation axis is,

$\tau =rF\sin \theta =Fd \cdots \cdots \left(1\right)$

where $\theta$is the angle the force vector and position vector and d is the moment arm of the force, a perpendicular distance from the rotation axis to the line of action of the force.

(a)To find the torque exerted

The torque exerted by the mechanist about the center of the nut is given by,

$\tau =rF\sin \theta$

Substituting the values, we get,

$\tau =\left(0.250\right)·\left(17.0\right)·\sin \left(37°\right)=2.56 N·m$

Hence, the torque exerted about the center of the nut is, $\tau =2.56 N·m$

(b)To find the maximum torque

From $\left(1\right)$, the torque is maximum if the sine term is 1. It is possible if the angle $\theta =90°$.

Therefore, the torque is given by,

${\tau }_{\max }=rF$

Substituting the values, we get,

$$\begin{gathered} {\tau }_{\max }=\left(0.250\right)\left(17.0\right) \\ {\tau }_{\max }=4.25 N·m \end{gathered}$$

Hence, the maximum torque exerted with the given force is, ${\tau }_{\max }=4.25 N·m$