Question

A uniform rope of length $20m$ and mass $20kg$ is shown in the figure. The hanging part on each side is $6m$. $C$ is the middle point of $ACB$ part of rope. Find the tension at point $C$. Assume that the rope is in equilibrium.

Answer 1

Since, this body is in equilibrium tension at point $A$ and $B$ will be same, let it be $T$.

Then,

$T=6g$

$T=60N$

For tension at point $C$ consider its half part,

Since, there is no force to balance the weight $50$ a component of tension in each element will balance its weight.

Total tension=$60N+40N$

$T=100N$