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?

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?

Olivia Askew · Accepted Answer

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

Answer is:

#r_2=0.bar(66)m# 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#

Adam Arnold · Answer

undefined To find the distance of the second weight from the fulcrum, use the formula for a balanced lever: $ \text{mass}_1 \times \text{distance}_1 = \text{mass}_2 \times \text{distance}_2 $. Substitute the given values and solve for $ \text{distance}_2 $.

$ 8 \times 2 = 24 \times \text{distance}_2 $

$ \text{distance}_2 = \frac{8 \times 2}{24} $

$ \text{distance}_2 = \frac{16}{24} $

$ \text{distance}_2 = \frac{2}{3} $ meters from the fulcrum.