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(D) At the point of support,the tension $T$ acts at an angle of $30^{\circ}$ with the vertical. For the vertical equilibrium of the entire rope,the ve

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$14$

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(D) At the point of support,the tension $T$ acts at an angle of $30^{\circ}$ with the vertical. For the vertical equilibrium of the entire rope,the vertical components of the tension at both ends must balance the weight of the rope:$2T \cos 30^{\circ} = mg$$T = \frac{mg}{2 \cos 30^{\circ}}$Let $T_1$ be the tension at the lowest point of the rope. At this point,the tension is purely horizontal. Considering the equilibrium of half the rope,the horizontal component of the tension at the support must be equal to the tension at the lowest point:$T_1 = T \sin 30^{\circ}$Substituting the value of $T$:$T_1 = \left( \frac{mg}{2 \cos 30^{\circ}} ight) \sin 30^{\circ} = \frac{mg}{2} 	an 30^{\circ}$Given $m = 5 \,kg$,$g = 10 \,m/s^2$,and $	an 30^{\circ} = \frac{1}{\sqrt{3}}$:$T_1 = \frac{5 	imes 10}{2} 	imes \frac{1}{\sqrt{3}} = \frac{25}{\sqrt{3}} \approx \frac{25}{1.732} \approx 14.43 \,N$Thus,the tension at the lowest point is approximately $14 \,N$.

$A$ rope of mass $5 \,kg$ is hanging between two supports as shown. The tension at the lowest point of the rope is close to ........... $N$ (take $g=10 \,m/s^2$).

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