# A wrench with a length of 42 cm is being used to unscrew a 7/2 cm bolt. If a torque of 32 Nm is needed to overcome the friction keeping the bolt in place, what is the minimum torque that must be applied to the wrench to unscrew the bolt?

Aug 10, 2017

$F \approx 76 \text{N}$

#### Explanation:

I assume the question is meant to ask what is the minimum force required to unscrew the bolt.

Torque is given by the equation:

$\textcolor{\mathrm{da} r k b l u e}{\tau = r F \sin \phi}$

where $F$ is the applied force, $r$ is the distance from the pivot (axis of rotation) that the force is applied, and $\phi$ is the angle at which the force is applied.

We can rearrange the equation for torque to find $F$.

$\textcolor{c r i m s o n}{F = \frac{\tau}{r \sin \phi}}$

We are provided the following information:

• $\mapsto \text{r"=42"cm"=0.42"m}$

• $\mapsto \tau = 32 \text{Nm}$

And it is implied that $\phi = {90}^{o}$ since no angle—nor the necessary information to calculate one—is provided.

Note that I am not including the radius of the bolt in the measurement of the total radius because I am assuming that this is the radius of the body of the bolt and not the head of the bolt.

Therefore, we have:

$F = \left(32 \text{Nm")/(0.42"m}\right)$

$= 76.19 \text{N}$

$\approx 76 \text{N}$