### Q.8:- 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.

**Answer:-**

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

Length of the bar, *l* = 2 m

*T*_{1 }and *T*_{2} are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have:

*T*_{1} *Sin* 36.9^{0} = *T*_{2} *Sin* 53.1^{0}

*T*_{1} / *T*_{2} = 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) × *d* = *T*_{2} *Cos* 53.1 (2 – d)

*T*_{1} × 0.800 *d* = *T*_{2} × 0.600 (2 – *d*)

(4/3) × *T*_{2} × 0.800d = *T*_{2} (0.600 × 2 – 0.600*d*)

1.067*d* + 0.6*d* = 1.2

∴ *d* = 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.