# A balanced lever has two weights on it, the first with mass 8 kg  and the second with mass 24 kg. If the first weight is  2 m from the fulcrum, how far is the second weight from the fulcrum?

Mar 20, 2016

Since the lever is balanced, the sum of torques is equal to 0

${r}_{2} = 0. \overline{66} m$

#### Explanation:

Since the lever is balanced, the sum of torques is equal to 0:

Στ=0

About the sign, obviously for the lever to be balanced if the first weight tends to rotate the object with a certain torque, the other weight will have opposite torque. Let the masses be:

${m}_{1} = 8 k g$

${m}_{2} = 24 k g$

τ_(m_1)-τ_(m_2)=0

τ_(m_1)=τ_(m_2)

${F}_{1} \cdot {r}_{1} = {F}_{2} \cdot {r}_{2}$

${m}_{1} \cdot \cancel{g} \cdot {r}_{1} = {m}_{2} \cdot \cancel{g} \cdot {r}_{2}$

${r}_{2} = {m}_{1} / {m}_{2} \cdot {r}_{1}$

${r}_{2} = \frac{8}{24} \cdot 2$ $\cancel{\frac{k g}{k g}} \cdot m$

${r}_{2} = \frac{2}{3} m$ or ${r}_{2} = 0. \overline{66} m$