A balanced lever has two weights on it, the first with mass $15 k g$ $15 kg$ and the second with mass $18 k g$ $18 kg$ . 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 · 2017-04-11T06:40:27

The distance is $= 5 m$ $=5m$

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Here, we have

$M_{2} = 15 g N$ $M_2=15gN$

$b = 6 m$ $b=6m$

$M_{1} = 18 g N$ $M_1=18gN$

Taking moments about the fulcrum

$15 g \cdot 6 = 18 g \cdot a$ $15g*6=18g*a$

$a = \frac{15 \cdot 6}{18} = 5 m$ $a=(15*6)/18=5m$

A balanced lever has two weights on it, the first with mass 15kg15 kg and the second with mass 18kg18 kg. If the first weight is 6m 6 m from the fulcrum, how far is the second weight from the fulcrum?