How large is the normal force? (applying Newtons laws phys )

[natty210]

how large is the normal force? (applying Newtons laws phys 1111)

A 23 kg child goes down a straight slide inclined 38 degrees above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30 N force as shown in the figure.

How large is the normal force of the slide on the child?
Express your answer using two significant figures.

attachment below***

 

[tiny-tim]

Hi natty210!

Since there is (obviously!) no acceleration perpendicular to the slide, good ol' Newton's second law means that the sum of the components, in that direction, of all the forces must be zero.

Show us what you get.

 

[TwoTruths]

Might I chime in with a quick question about this problem? The normal force obviously provides an equal and opposite force to the forces pointing down perpendicular to the slide (whatever those may be, natty210). This would create a net force of 0 (equivalent to a pencil resting on a desk). However, the horizontal rope has a component that is upwards perpendicular to the slide (equivalent to pulling on a rope attached to my pencil that is angled a bit - this causes my pencil to rise). What causes the child to stay on the slide?

 

[tiny-tim]

Hi TwoTruths!

Because the weight of the child is (I hope! ) more than enough to balance that …

the normal force will be reduced, but not enough for the child to lose contact.

(This is similar to the question of whether a car will lose contact with a roller-coaster …the other forces, less the acceleration, must be enough to keep the normal force greater than zero)