A balanced lever has two weights on it, the first with mass 8 kg 8kg and the second with mass 24 kg24kg. If the first weight is 2 m2m from the fulcrum, how far is the second weight from the fulcrum?

1 Answer
Mar 20, 2016

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

Answer is:

r_2=0.bar(66)mr2=0.¯¯¯¯66m

Explanation:

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

Στ=0Στ=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