A uniform rod AB is suspended by two light strings, as shown in figure. The tension in the left string just after the right string is cut is 100 $α$ N. Find ‘ $α$ ’. Take Mass of rod as $70 \sqrt{3} k g$ and $g = 9.8 m / s^{2}$ .

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Detailed Solution

$\frac{T}{2} = m a_{c x} (i); m g - \frac{\sqrt{3}}{2} T = m a_{c y} (i i) \frac{T}{2} \times \frac{l}{2} = \frac{m l^{2} α}{l^{2} 3} \Rightarrow T = \frac{m l α}{3} (i i)$

$\frac{a_{c x}}{2} + \frac{l α}{4} = \frac{\sqrt{3}}{2} a_{c y} (i v) \Rightarrow \frac{T}{4 m} + \frac{1}{4} \times \frac{3 T}{m} = \frac{\sqrt{3}}{2} (g - \frac{\sqrt{3}}{2 m}) \Rightarrow \frac{T}{4} {1 + 3 + 3} = \frac{\sqrt{3}}{2} \times 70 \sqrt{3} \times g \Rightarrow \frac{T}{4} \times 7 = \frac{3}{2} \times 70 \times 10 g \Rightarrow T = 58 N$