A balanced lever has two weights on it, the first with mass $32 k g$ $32 kg$ and the second with mass $24 k g$ $24 kg$ . 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 · 2017-12-01T07:13:48

The distance is $= \frac{28}{3} m$ $=28/3m$

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The masses are

$M_{1} = 32 k g$ $M_1=32kg$

and

$M_{2} = 24 k g$ $M_2=24kg$

The distance $a = 7 m$ $a=7m$

Taking moments about the fulcrum

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

$32 \times 7 = 24 \times b$ $32xx7=24xxb$

$b = \frac{32 \times 7}{24} = \frac{28}{3} m$ $b=(32xx7)/24=28/3m$

A balanced lever has two weights on it, the first with mass 32kg32 kg and the second with mass 24kg24 kg. If the first weight is 7m 7 m from the fulcrum, how far is the second weight from the fulcrum?