A balanced lever has two weights on it, the first with mass $14 k g$ $14 kg$ and the second with mass $9 k g$ $9 kg$ . 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 · 2017-03-07T07:16:09

$3.1 m$ $3.1m$

A moment is the product of force and the distance over which the force acts.

$M = F d$ $M = Fd$

In a balanced lever, both moments are equal to each other, so

$M_{1} = M_{2}$ $M_1 = M_2$

$F_{1} d_{1} = F_{2} d_{2}$ $F_1d_1 = F_2d_2$

We can work out the force either side by $F = m a$ $F=ma$ , where $a \approx 10 m s^{- 2}$ $a~~10ms^-2$ , so

$F_{1} = 14 \times 10 = 140 N$ $F_1 = 14 xx 10 = 140N$

$F_{2} = 9 \times 10 = 90 N$ $F_2 = 9 xx 10 = 90N$

and put these values, along with $d_{1} = 2 m$ $d_1 = 2m$ (from the question) into the equation:

$2 m \times 140 N = d \times 90 N$ $2m xx 140N = d xx 90N$

$d = \frac{280}{90} = 3.1 m$ $d = 280/90 = 3.1m$

A balanced lever has two weights on it, the first with mass 14kg14 kg and the second with mass 9kg9 kg. If the first weight is 2m 2 m from the fulcrum, how far is the second weight from the fulcrum?