# 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  8 m from the fulcrum, how far is the second weight from the fulcrum?

Apr 3, 2016

The equation is ${F}_{L} \times {d}_{L} = {F}_{R} \times {d}_{R}$
Where $F =$force, $d =$distance, $L =$left, $R = r i g h t$

#### Explanation:

When we translate the masses into forces, we have the constant $g$ on both sides, so they cancel. The equation then turns into:
${m}_{L} \times \cancel{g} \times {d}_{L} = {m}_{R} \times \cancel{g} \times {d}_{R}$

$4 k g \times 8 m = 6 k g \times {d}_{R}$

${d}_{R} = \frac{4 \cancel{k g} \times 8 m}{6 \cancel{k g}} = \frac{32}{6} m = 5 \frac{1}{3} m$