1. Practice with Inclined Planes
Renate Fiora

2. A 23 kg child goes down a straight slide inclined at 38° above horizontal. The child is acted on by his weight, the normal force from the slide, and kinetic friction (m=0.20). What is the net force on the child? q
Try it on your own, then advance to the next slide to see the solution.

3. A 23 kg child goes down a straight slide inclined at 38° above horizontal. The child is acted on by his weight, the normal force from the slide, and kinetic friction (m=0.20). What is the net force on the child? q FW FN Ff
Direction perpendicular to the slide’s surface: The normal force is the reaction force to the perpendicular component of the child’s weight. In this case, there is no motion in the perpendicular direction, so we can write Newton’s second law as SFnet  = FN – FW = 0 Which tells us that FN = FW +x +y FW FN
Fnet = 0 q Ff

4. A 23 kg child goes down a straight slide inclined at 38° above horizontal. The child is acted on by his weight, the normal force from the slide, and kinetic friction (m=0.20). What is the net force on the child? q FW FN Ff
Direction parallel to the slide’s surface: The only parallel force we see on the diagram is the force of kinetic friction. Remember, however, that there is also a parallel component to the child’s weight. So, Newton’s second law can be written as SFnet = FW – Ff = FW sin q – mFN But FN = FW cos q So SFnet = FW sin q – mFW cos q SFnet = FW (sin q – mcos q) SFnet = mg (sin q – mcos q) Plugging in our values and calculating, we get SFnet = (23 kg)(9.8 m/s2) (sin 38 – 0.20(cos 38)) Fnet = 103 N +x +y FW FN
Fnet = 0 q Ff