# A balanced lever has two weights on it, the first with mass 3 kg  and the second with mass 4 kg. If the first weight is  25 m from the fulcrum, how far is the second weight from the fulcrum?

Aug 13, 2016

${l}_{2} = 18.75 \text{ m}$

#### Explanation:

$\text{Torque for the first weight:}$

${L}_{1} = m \cdot g \cdot {l}_{1}$

${m}_{1} = 3 \text{ kg";g=9.81 N/(kg);l_1=25 "m}$

${L}_{1} = 3 \cdot g \cdot 25 = 75 \cdot g \text{ } N \cdot m$

$\text{Torque for the second weight:}$

${L}_{2} = {m}_{2} \cdot g \cdot {l}_{2}$

m_2=4" kg" " ; "l_2=?

${L}_{2} = 4 \cdot g \cdot {l}_{2}$

${L}_{1} = {L}_{2} \text{ so the system is balanced}$

$75 \cdot \cancel{g} = 4 \cdot \cancel{g} \cdot {l}_{2}$

${l}_{2} = \frac{75}{4}$

${l}_{2} = 18.75 \text{ m}$