A rope of mass 5 ,kg is hanging between two s | vedclass

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Question

$(d)$ At point of contact $P$ tension $T$ can be resolved into vertical and horizontal components.

For equilibrium,

$2 T \cos 30^{\circ}=m g$

Vedclass · Accepted Answer

$14$

Vedclass · Answer

$(d)$ At point of contact $P$ tension $T$ can be resolved into vertical and horizontal components.

For equilibrium,

$2 T \cos 30^{\circ}=m g$

$	herefore \quad T=\frac{m g}{2 \cos 30^{\circ}}$

If tension at lowest point is $T_{1}$, then for horizontal equilibrium,

$T \sin 30^{\circ}=T_{1}$

$\Rightarrow T_{1} =\frac{m g}{2 \cos 30^{\circ}} 	imes \sin 30^{\circ}$

$=\frac{m g 	an 30^{\circ}	imes 5 	imes 10 	imes 1 / \sqrt{3}}{2}$

$=14.41 \,N \approx 14 \,N$

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

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