# A balanced lever has two weights on it, one with mass 1 kg and one with mass 8 kg. If the first weight is 9 m from the fulcrum, how far is the second weight from the fulcrum?

May 20, 2017

The $8$ $k g$ mass will be $1.125$ $m$ from the fulcrum in order to balance a $1$ $k g$ mass placed $9$ $m$ from the fulcrum.

#### Explanation:

For the lever to be balanced, the torque on each side of the fulcrum must be balanced.

The torque due to the first weight is given by $\setminus \tau = F r = m g r$ where $r$ is the distance from the fulcrum, since the weight force on the mass is given by $F = m g$.

$\setminus \tau = F r = m g r = 1 \times 9.8 \times 9 = 88.2$ $N m$

The torque due to the second mass must have the same value:

$\setminus \tau = m g r$

Rearranging to make $r$ the subject:

$r = \setminus \frac{\tau}{m g} = \frac{88.2}{8 \times 9.8} = 1.125$ $m$

This is as we would expect: a larger mass needs to be closer to the fulcrum to balance a smaller mass further from the fulcrum.