241_Physics ProblemsTechnical Physics

241_Physics ProblemsTechnical Physics - Chapter 8*P8.64(a...

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Chapter 8 243 *P8.64 (a) The length of string between glider and pulley is given by A 22 0 2 =+ xh . Then 220 A A d dt x dx dt . Now d dt A is the rate at which string goes over the pulley: d dt v x vv yx x A A == = cos θ af . (b) KKU ABg i f ++ = ej 00 1 2 1 2 30 45 = + mgy y mv BA x B y bg Now yy 30 45 is the amount of string that has gone over the pulley, AA 30 45 . We have sin30 0 30 °= h A and sin45 0 45 h A , so 30 45 30 45 040 2 2 0234 −= ° ° =− = hh sin sin .. m m . From the energy equation 05 98 1 2 100 1 2 0500 45 115 135 2 . . . c o s . . kg m s m kg kg J 0.625 kg ms 2 ° == v xx x (c) ° = cos . cos . 0958 (d) The acceleration of neither glider is constant, so knowing distance and acceleration at one point is not sufficient to find speed at another point. P8.65 The geometry reveals DL L sin sin θφ , 50 0 40 0 50 s i n s i n m + φ , 28 9 . (a) From takeoff to alighting for the Jane-Earth system KU W mv mg L FD mg L v v v g i g f i i i i = + +− = ° ° −× × = a f bg a f a f a f wind kg kg m s m N m
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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