A balanced lever has two weights on it, the first with mass $4 k g$ $4 kg$ and the second with mass $30 k g$ $30 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2017-10-25T07:07:29

The distance is $= 0.93 m$ $=0.93m$

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The mass $M_{2} = 4 k g$ $M_2=4kg$

The mass $M_{1} = 30 k g$ $M_1=30kg$

The distance $b = 7 m$ $b=7m$

Taking moments about the fulcrum

$M_{1} \times a = M_{2} \times b$ $M_1xxa=M_2xxb$

$a = \frac{M_{2} \times b}{M_{1}} = \frac{4 \cdot 7}{30} = \frac{14}{15} = 0.93 m$ $a=(M_2xxb)/(M_1)=(4*7)/30=14/15=0.93m$

A balanced lever has two weights on it, the first with mass 4kg4 kg and the second with mass 30kg30 kg. If the first weight is 7m 7 m from the fulcrum, how far is the second weight from the fulcrum?