# A 42-gram weight and a 18-gram weight are placed on opposite sides of a beam, each at a distance of 15 centimeters from the fulcrum. Where should a 60-gram weight be placed to balance the beam?

Jan 28, 2017

The 60g loading should be placed between the 18g loading and the fulcrum. Which is $15 c m - 9 c m = 6 c m$ from the fulcrum.

#### Explanation: NOTE THAT: the 18 gram loading is not sufficient to maintain equilibrium with the 42 grams. So the addition loading must be placed on the 18 gram side of the beam.

$\textcolor{w h i t e}{.}$

Assumption 1: The beam is of uniform weight thus the weight
$\text{ }$influence of each side cancels the other out.

Assumption 2:The fulcrum is located centrally to the length of the
$\text{ }$ beam.

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$\textcolor{g r e e n}{\text{Let clockwise moment be positive (+v)}}$
$\textcolor{p u r p \le}{\text{Let anticlockwise moments be negative (-v)}}$
Let the reaction be R

A moment is the force times its distance to the point under consideration.

All moments must be in equilibrium (balanced) else the beam would be in motion.

The upward reaction of the support must equal the downward loading otherwise the system would fail and collapse.

$R = 42 + 60 + 18 = 120 g$

$\left(18 C\right) \textcolor{p u r p \le}{+ 15 R} \textcolor{g r e e n}{- 42 A - 60 x} = 0$

But the distance of point C to the point of moments is 0.
So $18 C = 18 \times 0 = 0$ Thus it has no influence so we have

$\textcolor{p u r p \le}{+ 15 R} \textcolor{g r e e n}{\text{ "-42A" } - 60 x} = 0$

$\left(15 \times 120\right) - \left(42 \times 30\right) - \left(60 x\right) = 0$

$60 x = 540$

$x = \frac{540}{60} = 9 c m$

The 60g loading should be placed between the 18g loading and the fulcrum such that it is $15 - 9 = 6 c m$ from the fulcrum.