A uniform wooden metre rule of mass 90g is pivoted at the 40cm mark.if the rule is in equilibrium with an unknown mass M placed at the 10cm mark and 72g mass at the 70cm mark ,determine M.(WAEC 2000)

Answer 1

This is not possible to solve. Unless we know how long the rod is, we have no idea how far from the fulcrum the mass of the rod is distributed. Just for simplicity, let's assume that the pivot is at the center, so the rod is 80cm long.

Now consider the rod as massless, and having two extra masses of 40g at the 20cm mark, and 50g at the 60cm mark. For the forces to balance, we need

30M + 20*40 = 20*50 + 30*72

Now just solve for M.

Answer 2

oops. I was dividing the mass 40:50 instead of evenly. Try
30M + 20*45 = 20*45 + 30*72