# A balanced lever has two weights on it, the first with mass 7 kg  and the second with mass 25 kg. If the first weight is  6 m from the fulcrum, how far is the second weight from the fulcrum?

Dec 24, 2015

$1 , 68 m$

#### Explanation:

We may use the second condition for static equilibrium to solve this problem, that is, that the resultant torque of the system must be zero.
ie, $\sum \vec{\tau} = 0$, where $\vec{\tau} = \vec{r} \times \vec{F}$ is the vector cross product of the position vector from the axis of rotation (fulcrum) to the applied force and the applied force.

So in this case,

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

$\therefore \left(6 \times 7 g\right) - \left({r}_{2} \times 25 g\right) = 0$

$\therefore {r}_{2} = \frac{6 \times 7 \times 9 , 8}{25 \times 9 , 8} = 1 , 68 m$.