Two masses of mass 10.0 and 6.0 kg are hung from massless strings at the end of a light rod. The rod is virtually weightless. A pivot (fulcrum) is placed off center and the system is free to rotate. If the 6.0 kg mass is 4.0 m away from the pivot, how far away is the 10.0 kg mass if the system is not rotating?

If the 6.0 kg mass is 4.0 m away from the pivot, the 10kg mass is far away from the pivot by 2.4 m

Moment: Moment of a force is the product of force and its perpendicular distance. The unit of moment is Nm

In other to solve this question, we need to make use of the principle of moment.

Principle of moment: For a body to be at equilibrium,  the sum of clockwise moment =  the sum of anticlock wise moment.

For the system to be at equilibrium (not rotating),

The sum of clockwise moment must be equal to the  sum of anticlock wise moment. (Principle of moment)

From the question,

moment = mass×garvity×distance

M = mgd Equation 1

where m = mass, d = distance g = gravity = 10 m/s²

Clockwise moment = 6×10×4 = 240 Nm.

Anticlock wise moment = 10×10×y = 100y

Where y = distance of the 10 kg mass from the point of pivot

therefore,

240 = 100y

y = 240/100

y = 2.4 m.

Hence, the 10 kg mass is 2.4 m away from the pivot

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