A uniform bar AB is balanced on a knife-edge which is 60cm from B by a mass of 22g which hangs at C 10cm from A. Calculate the mass of the bar.

I need to work out the bar's length first, but the question doesn't give it directly, so I'll check the setup.

Analyze the question A uniform bar AB rests on a knife-edge (pivot) 60 cm from B. A 22 g mass hangs at C, 10 cm from A. The bar balances. Its weight acts at its midpoint.

Setting up Let the bar's length be L. The pivot is 60 cm from B, so it is (L − 60) cm from A.

The mass at C is 10 cm from A, so its distance from the pivot is (L − 60) − 10 = (L − 70) cm.

The centre of gravity of the bar is at L/2 from A. For balance, the 22 g mass on the A side must be countered by the bar's weight on the B side, so the centre must lie on the B side of the pivot: L/2 > L − 60, which means L < 120.

Distance of the centre from the pivot = L/2 − (L − 60) = 60 − L/2.

Moments about the pivot 22 × (L − 70) = M × (60 − L/2)

This has two unknowns (M and L). Since the length isn't given, the question most likely comes from a diagram where the bar is 100 cm long (a metre rule). I'll use that standard assumption: L = 100 cm.

• Pivot is 40 cm from A and 60 cm from B. ✔
• C is 10 cm from A, so it is 30 cm from the pivot.
• Centre of the bar is at 50 cm from A, so it is 10 cm from the pivot on the B side.

22 × 30 = M × 10

M = 660 / 10 = 66 g

Answer: B. 66.0 g

Tip: In balance problems, always take moments about the pivot: (mass × distance) on one side equals (mass × distance) on the other. Remember a uniform bar's weight acts at its midpoint. When the length isn't stated, check whether the problem is using a metre rule (100 cm).