Question

"A 23kg kg child goes down a straight slide inclined 38∘ above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30N force as shown in the figure.(Figure 1)
How large is the normal force of the slide on the child?"

Answer 1

Here is the solution for the given problem above:

Normal force = component of child's weight that acts normal to slide - 30 sin 38° 
Normal force = mg(cos 38°) - 30(sin 38°) 
Normal force = (23)(9.8)(cos 38) - 30(sin 38)
                      = 178 - 18.5
                      = 159.5 N

159.5 N is the answer.