Chapter 5149P5.63(1)maA T aTmA11−=⇒= +af(2)MARTATMx==⇒=(3)ma mg T T m g a222=−⇒=−bg(a)Substitute the value for afrom (1) into (3) and solve for T:FIG. P5.63TmgTmA=−+FHGIKJLNMOQP21.Substitute for Afrom (2):TmTMmgmMmM m mM=−+FHGIKJLNMOQP=++LNMOQP2121121.(b)Solve (3) foraand substitute value of
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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 .