Question

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

• A. 22 N
• B. 44 N
• C. 28 N
• D. 14 N

Answer 1

Answer: (D)

At point of contact $P$ tension $T$ can be resolved into vertical and horizontal components.

$\therefore T=\frac{mg}{2\cos 30^{\circ }}$
If tension at lowest point is $T_1$ then for horizontal equilibrium,

$T\sin 30^{\circ }=T_1$
$\Rightarrow T_1=\frac{mg}{2\cos 30^{\circ }}\times \sin 30^{\circ }$
$=\frac{mg\tan 30^{\circ }}{2}=\frac{5\times 10\times 1\sqrt{3}}{2}$
$14.41N\approx 14N$