A balanced lever has two weights on it, the first with mass $3 k g$ $3 kg$ and the second with mass $27 k g$ $27 kg$ . If the first weight is $6 m$ $6 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2017-11-30T06:55:36

The distance is $= \frac{2}{3} m$ $=2/3m$

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The masses are

$M_{1} = 27 k g$ $M_1=27kg$

and

$M_{2} = 3 k g$ $M_2=3kg$

The distance $b = 6 m$ $b=6m$

Taking moments about the fulcrum

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

$27 \times a = 3 \times 6$ $27xxa=3xx6$

$a = \frac{3 \times 6}{27} = \frac{2}{3} m$ $a=(3xx6)/27=2/3m$

A balanced lever has two weights on it, the first with mass 3kg3 kg and the second with mass 27kg27 kg. If the first weight is 6m 6 m from the fulcrum, how far is the second weight from the fulcrum?