Question

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar form its left end.

Answer 1

Here, α = 36.9°, β = 53.1°

If T₁, T₂ are tensions in the string. Then for equilibrium along the horizontal

T₁ sin α = T₂ sin β

or, T₁/T₂ = sin β/sin α = sin 53.1°/sin 36.9°

= 0.747/0.5477 = 1.3523

d is the distance of the centre of gravity G of the bar from the left end.

For rotational equilibrium about c.

T₁ cos α x d = T₂ cos β(2 - d)

or, T₁ cos 36.9° x d = T₂ cos 53.1°(2 - d)

or, T₁ 0.8366 d = T₂ x 0.6718(2 - d)

Putting T₁ = 1.3523 T₂

We get d = 0.745 m = 74.5 cm