Question

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?

Answer 1

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

Answer is:

`r_2=0.bar(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=8kg`

`m_2=24kg`

`τ_(m_1)-τ_(m_2)=0`

`τ_(m_1)=τ_(m_2)`

`F_1*r_1=F_2*r_2`

`m_1*cancel(g)*r_1=m_2*cancel(g)*r_2`

`r_2=m_1/m_2*r_1`

`r_2=8/24*2` `cancel((kg)/(kg))*m`

`r_2=2/3 m` or `r_2=0.bar(66)m`