# A balanced lever has two weights on it, the first with mass 42 kg  and the second with mass 14 kg. If the first weight is  2 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 3, 2016

$\textcolor{b l u e}{x = 6 m}$

Did you know you can treat units of measurement the same way as numbers? I have shown this in my solution.

#### Explanation:

For the system to be in equilibrium (not rotating or moving in any way) all the 'moments' have to cancel each other out.

$\textcolor{b l u e}{\text{Taking moment about the fulcrum.}}$

$\textcolor{b r o w n}{\text{Consider the LHS of the fulcrum}}$

42 kg xx 2m = 84 kgm -> "force "xx" distance")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Consider the RHS of the fulcrum}}$

$14 k g \times x m = 14 x k g m$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Equating LHS to RHS for equilibrium gives:

$42 k g \times 2 m = 14 k g \times x m$

$\implies 84 k g m = 14 x k g m$

Remember that the units for $x$ is m

Divide both sides by $14 k g$ (leaves $x m$ on the RHS)

$\implies \frac{84}{14} \frac{k g m}{k g} = x \textcolor{w h i t e}{.} m$

$6 \textcolor{w h i t e}{.} \frac{\cancel{k g} m}{\cancel{k g}} = x \textcolor{w h i t e}{.} m$

So $x = 6 m$