# A balanced lever has two weights on it, the first with mass 66 kg  and the second with mass 9 kg. If the first weight is  1 m from the fulcrum, how far is the second weight from the fulcrum?

Jun 21, 2016

The $9 k g$ mass should be $7.33 m$ from the fulcrum.

#### Explanation:

For this solution we will use the formula ${F}_{1} {d}_{1} = {F}_{2} {d}_{2}$

Where Force times the distance of one side of the lever must equal the combination of the Force and distance applied on the opposite side of the lever.

Force equals mass time gravity

$F = m \cdot g$

${F}_{1} = 66 k g \cdot 9.81 \frac{m}{s} ^ 2 = 647.46 N$

${F}_{2} = 9 k g \cdot 9.81 \frac{m}{s} ^ 2 = 88.29 N$

${d}_{1} = 1 m$

d_2 = ?

${F}_{1} {d}_{1} = {F}_{2} {d}_{2}$

$\left(647.46 N\right) \left(1 m\right) = \left(88.29 N\right) \left({d}_{2}\right)$

(647.46cancelN*m)/(88.29 cancelN) = ((cancel(88.29N))(d_2))/(cancel(88.29N)

${d}_{2} = 7.33 m$