A child, who has a mass of 41 kg, tries to swing from a 10.0 m long rope.

Question

Its speed at the bottom of the swing is 5.0 m/s. The rope is only rated to handle 3000 N of tensile force. The maximum amount of force will be put on the rope at the bottom of the swing. What is the amount of force the child exerts on the rope at the bottom of the swing?

Answer 1

Let

m = 41 kg be the mass of the child,

r = 10.0 m be the length of the rope,

v = 5.0 m/s be the child's velocity at the bottom of the swing,

g = 9.81 m/s² be gravitational acceleration.

As long as the rope does not snap, there are two forces acting on the child at the bottom of the swing.

1) the force of gravity acting downward of magnitude mg,

2) unknown force of tension acting upwards of some magnitude T.

Therefore the net force is T - mg acting upward.

The child experiences upward centripetal acceleration of magnitude v²/r.

By Newton's Second Law

T - mg = mv²/r

From that

T = m(g + v²/r)

This tension T is what the rope needs to provide to avoid breaking.

Substitute actual numbers

T = 41×(9.81 + 5²/10) = 505 N

That is well below 3000N, so the rope will not break.