# A balanced lever has two weights on it, the first with mass 5 kg  and the second with mass 8 kg. If the first weight is  4 m from the fulcrum, how far is the second weight from the fulcrum?

Mar 2, 2017

$F d = F d$

$5 \cdot 4 = 8 d$

$d = 2.5 m$

#### Explanation:

This is a question about moments. Moments are the product of force and distance. For a lever to be balanced, the moments on either side must be equal.

We can describe this mathematically, using $M$ for moment, $F$ for force and $d$ for distance.

In a balanced lever, the moments are equal, so

${M}_{1} = {M}_{2}$

therefore

${F}_{1} {d}_{1} = {F}_{2} {d}_{2}$

We are not given the force either side, only the mass. Weight is the force that we need to work out, which is given by the product of mass and gravity, which we know is always $g = 9.8 m {s}^{-} 2$, so

(Note that mass is not a force, but weight is. Your mass doesn't change as you move between planets, but your weight might increase or decrease. In a question like this, assume that the force is the weight, which you can find from the mass but is not the mass.)

$F = W = m g$

We can put this into our equation for balanced moments,

$F = m g$

${m}_{1} g {d}_{1} = {m}_{2} g {d}_{2}$

from which we can eliminate $g$, which is a constant, so

${m}_{1} {d}_{1} = {m}_{2} {d}_{2}$

(Note that this isn't always the case, sometimes there are forces other than gravity acting on a lever.)

We know the first mass ($5 k g$) and the first distance ($4 m$) and the second mass ($8 k g$), so

$5 \cdot 4 = 8 {d}_{2}$

${d}_{2} = \frac{20}{8} = 2.5 m$