A balanced lever has two weights on it, the first with mass $15 k g$ $15 kg$ and the second with mass $5 k g$ $5 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2017-03-05T08:26:37

The distance is $= 6 m$ $=6m$

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Here, we have

$M_{1} = 15 g N$ $M_1=15gN$

$M_{2} = 5 g N$ $M_2=5gN$

$a = 2 m$ $a=2m$

Taking moments about the fulcrum

$M_{1} \cdot a = M_{2} \cdot b$ $M_1*a=M_2*b$

$b = \frac{M_{1}}{M_{2}} \cdot a$ $b=M_1/M_2*a$

$= \frac{15 g}{5} \cdot 2 = 6 m$ $=(15g)/(5)*2=6m$

A balanced lever has two weights on it, the first with mass 15 kg 15kg15 kg and the second with mass 5 kg5kg5 kg. If the first weight is 2 m2m 2 m from the fulcrum, how far is the second weight from the fulcrum?