Practice with Inclined Planes Renate Fiora A 23

Practice with Inclined Planes Renate Fiora

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.

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? Direction perpendicular to the slide’s surface: FN Ff The normal force is the reaction force to the perpendicular component of the child’s weight. q FW In this case, there is no motion in the perpendicular direction, so we can write Newton’s second law as +y Fnet = 0 FN Ff +x q FW SFnet = FN – FW = 0 Which tells us that FN = FW

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? FN Ff q FW +y Fnet = 0 FN Ff +x q FW 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 – m. FN But FN = FW cos q So SFnet = FW sin q – m. FW 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/s 2) (sin 38 – 0. 20(cos 38 )) Fnet = 103 N