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ID: 1696403 • Letter: H
Question
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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?
Explanation / Answer
always eager to do a problem when the DIAGRAM is supplied ! :>) 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 ANS
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