# A balanced lever has two weights on it, the first with mass 3 kg  and the second with mass 24 kg. If the first weight is  9 m from the fulcrum, how far is the second weight from the fulcrum?

Dec 30, 2015

$r = 1.125 m$

#### Explanation:

Let ${m}_{1} = 3 k g$ ${m}_{2}$=24kg
Let ${W}_{1}$ and ${W}_{2}$ be the weights of ${m}_{1}$ and ${m}_{2}$ respectively.
$\implies {W}_{1} = {m}_{1} \cdot g = 3 \cdot 9.8 = 29.4 N$
$\implies {W}_{2} = {m}_{2} \cdot g = 24 \cdot 9.8 = 235.2 N$

Since the lever is balanced so both the torques produced by ${W}_{1}$ and ${W}_{2}$ should be equal.

Let the mass ${m}_{2}$ be at a distance from $r$ from the fulcrum.

Let ${\tau}_{1}$ and ${\tau}_{2}$ be the torques produced by ${W}_{1}$ and ${W}_{2}$ respectively.
${\tau}_{1} = {\tau}_{2}$
$\implies {W}_{1} \cdot 9 = {W}_{2} \cdot r$
$\implies 29.4 \cdot 9 = 235.2 \cdot r$
$\implies r = \frac{29.4 \cdot 9}{235.2} = 1.125 m$
$\implies r = 1.125 m$
Hence the second weight is at a distance of 1.125m from the fulcrum.