Ncert Exercises & Solutions

Question 7. 8. 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 center of gravity of the bar from its left end.

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

Length of the bar, l = 2 m
T₁ and 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 °

\frac{T_{1}}{T_{2}} = \frac{\sin 53.1 °}{\sin 36.9 °}

= \frac{0.800}{0.600} = \frac{4}{3}

\Rightarrow T_{1} = \frac{4}{3} T_{2}

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

T_{1} \cos 36.9 ° \times d = T_{2} \cos 53.1 ° (2 - d)

T_{1} \times 0.800 d = T_{2} 0.600 (2 - d)

\frac{4}{3} \times T_{2} \times 0.800 d = T_{2} [0.600 \times 2 - 0.600 d]

1.067 d + 0.6 d = 1.2

∴ d = \frac{1.2}{1.67}

= 0.72 m

Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions