A balanced lever has two weights on it, the first with mass $8 k g$ $8 kg$ and the second with mass $35 k g$ $35 kg$ . If the first weight is $5 m$ $5 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2016-03-03T09:44:19

It is at a distance of $1.143$ $1.143$ $m .$ $m.$ from fulcrum

For a balance lever moments of the two forces must be equal. Moment is given by the product of weight and its distance from fulcrum.

Moment of first weight is $8 \times 5 = 40$ $8xx5=40$ $k g . m$ $kg.m$ . as the second weight weighs $35$ $35$ $k g .$ $kg.$ ,

it is at a distance of $\frac{40}{35} = 1.143$ $40/35=1.143$ $m$ $m$ from fulcrum.

A balanced lever has two weights on it, the first with mass 8kg8 kg and the second with mass 35kg35 kg. If the first weight is 5m 5 m from the fulcrum, how far is the second weight from the fulcrum?