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At the airport, you pull a 20 kg suitcase across the floor at constant velocity with a strap that is at an angle of 45° above the horizontal. Find the normal force on the suitcase, given that the coefficient of kinetic friction between the suitcase and the floor is 0.25
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|dw:1392429026406:dw| u is coefficient of friction Since it moves with constant velocity, net Force=0. in horizontal direction, Fcos45-umg=0 gives \[F=50\sqrt{2}\] In vertical direction, Fsin45 + Normal reaction= mg solving gives Normal reaction = 150N
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