# A balanced lever has two weights on it, the first with mass 8 kg  and the second with mass 30 kg. If the first weight is  7 m from the fulcrum, how far is the second weight from the fulcrum?

Dec 26, 2015

$1.867 m$

#### Explanation:

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

So in this particular case,

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

$\therefore \left(7 \times 8 g\right) - \left({r}_{2} \times 30 g\right) = 0$

$\therefore {r}_{2} = \frac{7 \times 8 \times 9 , 8}{30 \times 9 , 8}$

$= 1.867 m$.