A balanced lever has two weights on it, one with mass $12 k g$ $12 kg$ and one with mass $15 k g$ $15 kg$ . If the first weight is $4 m$ $4 m$ from the fulcrum, how far is the second weight from the fulcrum?

Question

1895 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T09:29:58

The distance is $= 3.2 m$ $=3.2m$

![www.thoughtco.com](useruploads.socratic.org)

The mass $M_{1} = 12 k g$ $M_1=12kg$

The mass $M_{2} = 15 k g$ $M_2=15kg$

The distance $a = 4 m$ $a=4m$

Taking moments about the fulcrum

$M_{1} \times a = M_{2} \times b$ $M_1xxa=M_2xxb$

$b = \frac{M_{1} \times a}{M_{2}} = \frac{12 \cdot 4}{15} = 3.2 m$ $b=(M_1xxa)/(M_2)=(12*4)/(15)=3.2m$

A balanced lever has two weights on it, one with mass 12kg12 kg and one with mass 15kg15 kg. If the first weight is 4m 4 m from the fulcrum, how far is the second weight from the fulcrum?