# A balanced lever has two weights on it, the first with mass 6 kg  and the second with mass 7 kg. If the first weight is  4 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 19, 2016

$x \approx 3.429 m e t r e s$

#### Explanation:

$\textcolor{b l u e}{\text{Comment}}$
As we are determining the distance it does not matter what units of weight we use as long as they are consistent.

I am also going to demonstrate how to deal with the units of measurement. If you deal with 'Statics' to any extent then manipulating units is exceptionally important!

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ If the moments do not match then it means that the system is in motion. We can use this fact to determine the moment arms needed to establish equilibrium.

A Moment is the number resulting from an applied force multiplied by its distance to a point of rotation ( actual or theoretical).
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Taking Moments about the fulcrum B}}$

Let clockwise moment be positive.
Let anticlockwise moments be negative

Then for equilibrium to exist

$\left(+ \left(6 \times 4\right) K g m\right) + \left(- \left(7 \times x\right) K g m\right) = 0$

$\textcolor{b r o w n}{\text{:::::::::::: Units are very important:::::::::::::::}}$

color(brown)("Note that weight"xx"distance"->Kgxxm->Kgm

$\textcolor{b r o w n}{\text{::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::}}$

For this to be true then:

$24 K g m = 7 x K g m$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To find the value of } x}$

To find $x$ divide both sides by $7 K g$: Note I have included the units

$\frac{24}{7} \textcolor{w h i t e}{.} \frac{\cancel{K g} m}{\cancel{K g}} = \frac{\cancel{7 K g}}{\cancel{7 K g}} \times x m$

$x \approx 3.429 m e t r e s$ to 3 decimal places