# A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The b - Physics

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A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. 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.

#### Solution

The free body diagram of the bar is shown in the following figure.

Length of the bar, l = 2 m

Tand T2 are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have:

T_1 sin 36.9^@ = T_2 sin 53.1

T_1/T_2 = (sin 53.1^@)/(sin 36.9)

= (0.800)/0.600 = 4/3

=> T_1 = 4/3 T_2

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

T_1 cos 36.9 xx d = T_2 cos 53.1 (2-d)

T_1 xx 0.800d = T_2 0.600(2-d)

4/3xxT_2xx0.800d = T_2[0.600xx2- 0.600d]

1.067 d + 0.6d = 1.2

∴ d = 1.2/1.67

= 0.72 m

Therefore, the center of gravity (C.G.) of the specified bar is located 0.72 meters from its left end.

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Chapter 7: System of Particles and Rotational Motion - Exercises [Page 178]

#### APPEARS IN

NCERT Physics Class 11
Chapter 7 System of Particles and Rotational Motion
Exercises | Q 8 | Page 178
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