Verified Solution • 2025–26 Session moderate

Q6.8: NCERT Physics - Part 1 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.8 Solution: A non-uniform bar of weight W is suspended at rest by two…

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. 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.

Step-by-Step Verified Solution

**Answer:** d=0.72d = 0.72 m from the left end. **Step-by-Step Explanation:** Let T1T_1 and T2T_2 be tensions. For equilibrium:
T1sin⁡36.9�=T2sin⁡53.1�T_1 \sin 36.9� = T_2 \sin 53.1�
T1cos⁡36.9�+T2cos⁡53.1�=WT_1 \cos 36.9� + T_2 \cos 53.1� = W
Solving: T1=0.8WT_1 = 0.8 W, T2=0.6WT_2 = 0.6 W. Taking torques about the left end:
T2cos⁡53.1�×2=W×dT_2 \cos 53.1� \times 2 = W \times d
0.6W×0.6×2=Wd  ⟹  d=0.72 m0.6 W \times 0.6 \times 2 = W d \implies d = 0.72 \text{ m}

All Questions in this Chapter

Q6.1 Q6.2 Q6.3 Q6.4 Q6.5 Q6.6 Q6.7 Q6.8 Q6.9 Q6.10 Q6.11 Q6.12 Q6.13 Q6.14 Q6.15 Q6.16 Q6.17

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