# A balanced lever has two weights on it, the first with mass 4 kg  and the second with mass 6 kg. If the first weight is  4 m from the fulcrum, how far is the second weight from the fulcrum?

Dec 29, 2015

$2.667 m$

#### Explanation:

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

Application thereof in this particular case gives

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

$\therefore \left(4 \times 4 g\right) - \left({r}_{2} \times 6 g\right) = 0$

$\therefore {r}_{2} = \frac{4 \times 4 \times 9.8}{6 \times 9 , 8} = 2.667 m$.