You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass m = 95 g on a pivot. stacked on the 0-cm end of the meter stick are n identical coins, each with mass m = 3.1 g, so that the center of mass of the coins is directly over the end of the meter stick. the pivot point is a distance d from the 0-cm end of the meter stick.
part (a): determine the distance d = d1, in centimeters, if there is only one coin o the 0 end of the meter stick and the system is in static equilibrium

d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

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