Question

A metre rod is half made of copper and half made of iron. If the mass of the copper part is 900 g and the mass of iron is 800 g, then calculate the position at which the rod can remain in equilibrium.

Answer 1

The copper and iron parts are each 50 cm long. Their centres are 25 cm and 75 cm, respectively, from the copper end.

Let the balance point be x cm from the copper end. Using the principle of moments:

900(x − 25) = 800(75 − x)

900x − 22,500 = 60,000 − 800x

1700x = 82,500

x = 48.5 cm (approximately).

Therefore, the rod balances about 48.5 cm from the copper end, or about 1.5 cm towards the copper end from its midpoint.