The tree branch in the preceding problem sags, and the child's feet rest on the ground. If the tension in the rope is reduced to 220 N, what is the value of the normal force being exerted on the child's feet?

The complete question is

A 45-kg child sits on a 3.2-kg tire swing. The tension on the rope is 470 N. The tree branch sags and the child's feet rest on the ground. If the tension in the rope is reduced to 220 N, what is the value of the normal force being exerted on the child's feet?

Therefore the Normal force is being exerted on the child's feet is 252.36( 250N).

Explanation:

We are given

mass of child = 45kg

mass of the tire = 3.2kg

g = 9.8 m/s^2

We have the first case in which the tire is in swing so tension in that case will be which is the resultant force.

Fr = Resultant force

Ft =  Tension force

Fg= Gravitational force

Fr = Fg + Ft

0 = -mg + Ft

Ft = mg

   = (45+3.2)*9.8

   = 48.2 * 9.8

Ft = 472.36N  ( which is close to 470)

In the second case the child is in the rest position, The resultant force will be zero.

Fr = Ft + Fn + Fg  ( Normal force is denoted by Fn )

0 = Ft + Fn + Fg

Fn =  -Fg -Ft

Fn = mg -Ft

    = 472.36 - 220

Fn  = 252.36N  ( close to 250N)

Therefore the Normal force being exerted on the child's feet will be 252.36N.  

answer; equivalent to a one(1) horsepower motor;