# A balanced lever has two weights on it, the first with mass 5 kg  and the second with mass 35 kg. If the first weight is  9 m from the fulcrum, how far is the second weight from the fulcrum?

Dec 12, 2016

$= 1.29 m \left(3 s . f .\right)$

#### Explanation:

use turning moments:

${M}_{1} \cdot {D}_{1} = {M}_{2} \cdot {D}_{2}$

($M$ being mass of weight on lever, $D$ being distance from fulcrum)

${M}_{1} = 5 k g$
${M}_{2} = 35 k g$

${D}_{1} = 9 m$

find ${D}_{2}$:

${M}_{1} \cdot {D}_{1} = 5 \cdot 9 = 45$
${M}_{2} \cdot {D}_{2} = 35 \cdot {D}_{2} = 45$

${D}_{2} = \left(\frac{45}{35}\right) m$

$= \left(\frac{9}{7}\right) m$

$= 1.29 m \left(3 s . f .\right)$