Physics Levers Question

Question

A uniform meter stick of weight 10N is loaded as follows:

An 80N weight attached at 10cm mark,

A 30N weight attached at 20cm mark,

A 50N weight attached at 60cm mark.

What is the center of gravity of the loaded meterstick?

Answer 1

The center of gravity is a point where the moments of force (also called Torque) on either side of that point are in balance.

When the stick is not loaded and has uniformly distributed mass, the center of gravity (CG) is exactly in the middle (i.e. at 0.5m). At that point, the torques generated by natural mass of the lever are balanced, and the lever experiences no rotational movement.

When you load that stick, that point of balance (or the CG) shifts. The stick can no longer be balanced at the stick's natural CG.

Assume that the new CG is now located at a distance 'x' from the zero-mark. We will assume that the new CG is shifted to the left of 50cm (the natural CG of the unloaded meter-stick), and to the right of the 30N force at the 20cm mark. You can confirm later that it doesn't matter exactly where you assume the new CG to lie.

Draw a picture on your sheet showing the points where different forces are applied. The weight of the unloaded stick will be considered to act at the 50cm mark.

The torques on two sides of the CG must be equal. To use the torque balance equation, you will need to calculate distances from the new CG, not from the zero-mark on the stick.

(x-10).80 + (x-20).30 = (50-x).10 + (60-x).50

Solving for x

x= 28.82 cm (approx).

To verify that the assumption about the location of new CG does not matter to the final result, you can repeat this calculation by assuming a different position for the new CG (say, to the right of natural CG of the stick), and re-writing the torque balance equation.

Hope this helps.