A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $32 k g$ $32 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?

Question

1476 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-21T09:47:02

Second weight is $3.5$ $3.5$ $m .$ $m.$ from the fulcrum.

In a balanced lever, the moment of each of the weights should be equal.

Moment is mass of the weight multiplied by its distance from fulcrum.

As first weighs $16 k g .$ $16kg.$ and is at a distance of $7 m$ $7m$ frm fulcrum, its moment is $16 \times 7 = 112$ $16xx7=112$ $k g . m$ $kg.m$ .

As lever is balanced, moment of second weight too would be $112$ $112$ $k g . m$ $kg.m$ . But its weight is $32 k g .$ $32kg.$ .

Hence, it should be at a distance of

$\frac{112}{32} = \frac{16 \times 7}{16 \times 2} = \frac{7}{2} = 3.5$ $112/32=(16xx7)/(16xx2)=7/2=3.5$ $m .$ $m.$

A balanced lever has two weights on it, the first with mass 16 kg 16kg16 kg and the second with mass 32 kg32kg32 kg. If the first weight is 7 m7m 7 m from the fulcrum, how far is the second weight from the fulcrum?