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Question

A uniform metre stick of mass 200 g is suspended from the ceiling thorough two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.

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Solution

Given:
Mass of the stick = m1=200 g
Mass of the small object = m2=20
Length of the string = l=1 m
As the system is in equilibrium, we have:
Ï„total=0 about O
T1×r1-T2×r2-m1g×r3=0

⇒T1×0.7-T2×0.3-2×0.2×g=0⇒7T1-3T2=3.92 ... (i)
Now, we have:
Total upward force = Total downward force
T1+T2=m1g+m2g =0.2×9.8+0.02×9.8⇒T1+T2=2.156 ...ii



Solving equations (i) and (ii), we get:
T1=1.038 N≈1.04 N;T2=1.18≈1.12 N

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