A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. 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 from its left end.
Given, the length of bar is 2 m and the angle made by the strings with vertical are 36.9 ° and 53.2 ° .
The following figure shows the tensions T 1 and T 2 , the corresponding angles of string with the vertical direction θ 1 = 36.9 ° and θ 2 = 53.1 ° also the forces resolved by T 1 and T 2 .
The bar is in equilibrium hence,
T 1 sin θ 1 = T 2 sin θ 2 T 1 T 2 = sin θ 2 sin θ 1
Substitute the values in the above direction,
T 1 T 2 = sin 53.1 ° sin 36.9 ° = 1.332 (1)
Let the centre of gravity of the bar is at a distance d from the left. Then balance the moment at C.
T 1 cos θ 1 × d = T 2 cos θ 2 ( 2 − d ) T 1 cos 36.9 ° × d = T 2 × cos 53.1 ° ( 2 − d ) T 1 T 2 = cos 53.1 ° ( 2 − d ) cos 36.9 ° d (2)
From equations (1) and (2),
1.332 = cos 53.1 ° ( 2 − d ) cos 36.9 ° d d = ( 0.72 m ) ( 1 00 cm 1 m ) ≈ 72 cm
Hence, the centre of gravity is at a distance of 72 cm from the left end.
Given, the length of bar is
The following figure shows the tensions
The bar is in equilibrium hence,
Substitute the values in the above direction,
Let the centre of gravity of the bar is at a distance
From equations (1) and (2),
Hence, the centre of gravity is at a distance of
