A uniform metre stick of mass 200 g is suspended from the ceiling by two vertical strings of equal lengths attached at its ends. A small object of mass 20 g is positioned on the stick at a distance of 70 cm from the left end. Determine the tensions in the two strings.

Step-by-step solution

Let the length of the metre stick be L = 100 cm . Mass of the stick, M = 200 g = 0.2 kg . Weight of the stick, W₁ = Mg = 0.2 9.8 = 1.96 N , acting at its centre of gravity ( 50 cm from the left end). Mass of the small object, m = 20 g = 0.02 kg . Weight of the object, W₂ = mg = 0.02 9.8 = 0.196 N , acting at 70 cm from the left end. Let T₁ and T₂ be the tensions in the left and right strings respectively. For rotational equilibrium, taking torque about the left end: T₂ 100 = W₁ 50 + W₂ 70 T₂ 100 = (1.96 50) + (0.19