The trick t this problem is to realize that the suitcase iss ON THE VERGE OF slipping.
The static coefficient of friction detremines the maximum possible friction force wthout slipping. The actual force can be much less.
Based upon the observation that the suitcase slipped part way down but not all the way down the sloped carousel, you can infer that the friction force is the maximum value, which in this case is M g cos 37 * mus. The net inward horizontal force equals the centripetal force. Both friction and gravity contribute to the inward force.
M g cos^2 37 mus + M g sin 37 = M V^2/R
"mus" is the static coefficient of friction
I had to square the cosine term because the friction force does not act directly inward.
Cancel the M's, and solve for V. Then get the time T for one trip around the carousel from V T 2 pi R
The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle (r = 11.2 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.760, and the angle in the drawing is 37.0°. How much time is required for your suitcase to go around once?
2 answers
21.6 seconds
0 answers