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170_Physics ProblemsTechnical Physics

170_Physics ProblemsTechnical Physics - 172 P6.35 Circular...

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172 Circular Motion and Other Applications of Newton’s Laws P6.35 Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward applied force equals the sum of the gravitational and drag forces (both downward): F mg bv = + . The mass of the copper ball is m r = = F H G I K J × × = 4 3 4 3 8 92 10 2 00 10 0 299 3 3 2 3 πρ π . . . kg m m kg 3 e je j . The applied force is then F mg bv = + = + × = 0 299 9 80 0 950 9 00 10 3 01 2 . . . . . a fa f a f e j N . P6.36 F ma T mg T F ma R T R D Av D R Av y y x x = + °− = = ° = × = + °= = × °= × = = = × F H I K = cos . . cos . . sin . . sin . . . . . . . 40 0 0 620 9 80 40 0 7 93 10 40 0 0 7 93 10 40 0 5 10 10 1 2 2 2 5 10 10 1 20 3 80 40 0 1 40 3 3 3 2 2 3 2 kg m s N N N N kg m m m s 2 kg m s N 2 2 2 b g e j e j e j e je j b g ρ ρ FIG. P6.36 P6.37 (a) At terminal velocity, R v b mg T = = = = × × = b mg v T 3 00 10 9 80 2 00 10 1 47
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