A balanced lever has two weights on it, the first with mass $7 k g$ $7 kg$ and the second with mass $4 k g$ $4 kg$ . If the first weight is $8 m$ $8 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2017-12-08T07:21:29

The distance is $= 14 m$ $=14m$

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The first mass is $M_{1} = 7 k g$ $M_1=7kg$

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

The distance $a = 8 m$ $a=8m$

Taking moments about the fulcrum

$M_{1} \times a = M_{2} \times b$ $M_1 xx a=M_2 xx b$

Therefore,

$b = \frac{M_{1} \cdot a}{M_{2}} = \frac{7 \cdot 8}{4} = 14 m$ $b=(M_1*a)/(M_2)=(7*8)/4=14m$

A balanced lever has two weights on it, the first with mass 7kg7 kg and the second with mass 4kg4 kg. If the first weight is 8m 8 m from the fulcrum, how far is the second weight from the fulcrum?