# A balanced lever has two weights on it, the first with mass 16 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?

Jul 31, 2016

$\text{The second weight } = 2 \frac{1}{7} m e t r e s$ from the fulcrum as an exact value

$\text{The second weight } \approx 2.143 m e t r e s$ from the fulcrum to 3 decimal places

#### Explanation:

$\textcolor{red}{\text{Assumption 1}}$
The beam ends at the points of loading. It is not stated as such

$\textcolor{red}{\text{Assumption 2}}$
The weight of the beam is discounted. No uniformly distributed load given.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For the beam to be in equilibrium (balanced and not moving)

All forces and moments cancel out.

$\textcolor{b l u e}{\text{Taking moments about point B}}$

Let clockwise moments be positive
Let anticlockwise moments be negative

A moment is $\text{force "xx" length of moment arm}$

The force of the 'Reaction' has moment arm length of 0. So this cancels itself out giving only:

So for the system to be in equilibrium (not moving)

$\text{(clockwise moment) + (anticlockwise moment)} = 0$
$\text{ } \textcolor{b r o w n}{\uparrow}$
$\textcolor{b r o w n}{\text{We have chosen that anticlockwise rotation is negative}}$

$\left(16 \times 2\right) + \left(- 14 \times x\right) = 0$

$32 = + 14 x$

$x = \frac{32}{14} \equiv \frac{16}{7} m e t r e s \text{ " ->" } 2 \frac{1}{7} m e t r e s$ as an exact value

$x \approx 2.143 m e t r e s$ to 3 decimal places