Engineering Mechanics - DynamicsChapter 14Solution:NcMgcosθ()=12MvA2Mga()sin+μkNca−12B2=vBvA22ga(+2kgcosa−=vB4.52ms=*Problem 14-28When the skier of weight Wis at point A he has a speed vA. Determine his speed when he reachespoint B on the smooth slope. For this distance the slope follows the cosine curve shown. Also, what isthe normal force on his skis at B and his rate of increase in speed? Neglect friction and air resistance.Given:W150 lb=vA5fts=a50 ft=b100 ft=d35 ft=Solution:yxa( )cosπxb⎛
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