# A mechanic can exert 113Nm of torque on his wrench. What is the torque exerted if the wrench were 7 times shorter AND the man could exert 5 times more force?

Dec 4, 2015

$\frac{5}{7} \mathrm{df} = \frac{5}{7} \times 113 N m \approx 80.72 N m \text{ to 2 decimal places}$

#### Explanation:

It is a matter of $\text{Force"xx"distance}$ equated to a constant

Let Force be f
Let distance be d

Given: $f d = 113 N m \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

Consider the question parts:

Wrench were 7 times shorter $\to \frac{d}{7}$

The man could exert 5 times more force$\to 5 f$

Consider equation (1)

By applying these changes to the left the equation becomes unbalanced. To maintain balance we have to change the right hand side in the same manor.

By the new conditions we have for the LHS $\frac{d}{7} \times 5 f$

To make the LHS of (1) the same as this we apply

$\frac{5}{7} \times \mathrm{df} \to L H S \ldots \ldots \ldots \ldots \ldots \left(2\right)$

We have to do the same to the right hand side (RHS)

$\frac{5}{7} \times 113 N m \to R H S \ldots \ldots \ldots \ldots . \left(3\right)$

Combining (2) and (3) gives

$\frac{5}{7} \mathrm{df} = \frac{5}{7} \times 113 N m \approx 80.72 N m \text{ to 2 decimal places}$