Question

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 (take, $g = 10\, m /s^2)$

• 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,

$T \sin 30^{\circ}=T_{1}$
$\Rightarrow T_{1}=\frac{mg}{2 \cos30^{\circ}}\times \sin 30^{\circ}$
$= \frac{mg \tan 30^{\circ}}{2}=\frac{5\times10\times1\sqrt{3}}{2}$
$14.41\,N \approx 14\,N$