which yields fk= 47 N. We also apply Newton’s second law to the yaxis (perpendicular to the incline surface), where the acceleration-component is zero: 140cos250 127 N.NNFF−°=¡=Therefore,μk= fk/FN= 0.37. (b) Returning to our first equation in part (a), we see that if the downhill component of the weight force were insufficient to overcome static friction, the child would not slide at all. Therefore, we require 140 sin 25° > fs,max= sFN, which leads to tan 25° = 0.47 > s.The minimum value of sequals kand is more subtle; reference to §6-1 is recommended. Ifkexceeded sthen when static friction were overcome (as the incline is raised) then it
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.