A balanced lever has two weights on it, the first with mass 3 kg 3kg and the second with mass 24 kg24kg. If the first weight is 9 m9m from the fulcrum, how far is the second weight from the fulcrum?

1 Answer
Dec 30, 2015

r=1.125mr=1.125m

Explanation:

Let m_1=3kgm1=3kg m_2m2=24kg
Let W_1W1 and W_2W2 be the weights of m_1m1 and m_2m2 respectively.
implies W_1=m_1*g=3*9.8=29.4NW1=m1g=39.8=29.4N
implies W_2=m_2*g=24*9.8=235.2NW2=m2g=249.8=235.2N

Since the lever is balanced so both the torques produced by W_1W1 and W_2W2 should be equal.

Let the mass m_2m2 be at a distance from rr from the fulcrum.

Let tau_1τ1 and tau_2τ2 be the torques produced by W_1W1 and W_2W2 respectively.
tau_1=tau_2τ1=τ2
implies W_1*9=W_2*rW19=W2r
implies 29.4*9=235.2*r29.49=235.2r
implies r=(29.4*9)/235.2=1.125mr=29.49235.2=1.125m
implies r=1.125mr=1.125m
Hence the second weight is at a distance of 1.125m from the fulcrum.