# A wrench with a length of 42 cm is being used to unscrew a 7/2 cm bolt. If a torque of 42 N*m 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 7, 2017

$F = 100 N$

#### 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:

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

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 $F$.

$\textcolor{red}{F = \frac{\tau}{r \sin \phi}}$

We are provided the following information:

$\text{r"=42"cm"=0.42"m}$

$\tau = 42 \text{Nm}$

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 screw and not the head of the screw.

Therefore, we have:

$F = \left(42 \text{Nm")/(0.42"m}\right)$

$= 100 \text{N}$