A balanced lever has two weights on it, the first with mass #4 kg # and the second with mass #6 kg#. If the first weight is # 8 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 #4 kg # and the second with mass #6 kg#.  If the first weight is # 8 m# from the fulcrum, how far is the second weight from the fulcrum?

Elena Albarran · Accepted Answer

The equation is #F_Lxxd_L=F_Rxxd_R#
Where #F=#force, #d=#distance, #L=#left, #R=right# When we translate the masses into forces, we have the constant #g# on both sides, so they cancel. The equation then turns into: #m_Lxxcancelgxxd_L=m_Rxxcancelgxxd_R# #4kgxx8m=6kgxxd_R# #d_R=(4cancel(kg)xx8m)/(6cancel(kg))=32/6m=5 1/3m#

Charles Aeschliman · Answer

undefined To find the distance of the second weight from the fulcrum, use the principle of moments: $ \text{Moment}_1 = \text{Moment}_2 $.

$ \text{Moment}_1 = \text{force}_1 \times \text{distance}_1 = 4 \, \text{kg} \times 8 \, \text{m} = 32 \, \text{Nm} $.

$ \text{Moment}_2 = \text{force}_2 \times \text{distance}_2 = 6 \, \text{kg} \times \text{distance}_2 $.

Setting the moments equal: $ 32 \, \text{Nm} = 6 \, \text{kg} \times \text{distance}_2 $.

Solve for distance_2: $ \text{distance}_2 = \frac{32 \, \text{Nm}}{6 \, \text{kg}} = 5.33 \, \text{m} $.