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Oscillations

Question

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A body of mass 60 kg is suspended by means of three strings P Q and R as shown in the figure is in equilibrium. The tension in the string P is

130.9 g N

60 g N

50 g N

103.9 g N

Solution

D.

103.9 g N

R cos 60^o = Q = 60 g

⇒ R = $\frac{60 g}{\cos 60^{o}}$ $(∵ \cos 60^{o} = \frac{1}{2})$

P = R sin60^o

= 120 g × $\frac{\sqrt{3}}{2}$

⇒ P = 103.9 g N

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