161_Physics ProblemsTechnical Physics

# 161_Physics ProblemsTechnical Physics - Chapter 6 b ge j Fg...

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Chapter 6 163 *P6.11 Fm g g == = 49 8 3 9 2 kg m s N 2 bg ej .. sin . . θ = 15 48 6 m 2 m r = 24 8 6 1 3 2 m m af cos . . a mv r TT xx ab °+ °= += ° = 2 2 48 6 48 6 46 132 109 48 6 165 cos . cos . . cos . kg m s m N N b g a yy = −° = −= ° = sin . sin . . . sin . . 48 6 48 6 39 2 0 39 2 48 6 52 3 N N N 39.2 N T a T b forces v a c motion FIG. P6.11 (a) To solve simultaneously, we add the equations in T a and T b : TTTT abab ++−= + 165 52 3 N N . T a 217 108 N 2 N (b) ba =− =−= 165 165 108 56 2 N N N . *P6.12 a v r c = 2 . Let f represent the rotation rate. Each revolution carries each bit of metal through distance 2 π r , so vr f = 2 and a v r rf g c
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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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