# A balanced lever has two weights on it, the first with mass 3 kg  and the second with mass 1 kg. If the first weight is  7 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 3, 2016

$21 m$

#### Explanation:

We apply the second condition for static equilibrium which states that the result torque (moments) must be zero,
ie. $\sum \vec{\tau} = 0$, where $\vec{\tau} = \vec{r} \times \vec{F}$ is the cross product of the position vector from the axis of rotation (fulcrum) to the applied force and the applied force.

So in this case :

${r}_{1} {F}_{1} - {r}_{2} {F}_{2} = 0$

$\therefore \left(7 \times 3 g\right) - \left({r}_{2} \times 1 g\right) = 0$

$\therefore {r}_{2} = \frac{7 \times 3 \times 9.8}{1 \times 9.8} = 21 m$.