Chapter 8243*P8.64(a)The length of string between glider and pulley is given by A2202=+xh. Then 220AAddtxdxdt.Now ddtAis the rate at which string goes over the pulley: ddtvxvvyxxAA== =cosθaf.(b)KKUABgif++=ej0012123045−=+mgyymvBAxBybgNow yy3045−is the amount of string that has gone over the pulley, AA3045−. We havesin30030°=hAand sin45045hA, so 30453045040220234−=°−°=−=hhsinsin..m m.From the energy equation059812100120500451151352...cos..kg m s mkgkgJ0.625 kgms2°==vxxx(c)°=cos.cos.0958(d)The acceleration of neither glider is constant, so knowing distance and acceleration at onepoint is not sufficient to find speed at another point.P8.65The geometry reveals DLLsinsinθφ, 50 040 050sinsinm=°+φ, 28 9.(a)From takeoff to alighting for the Jane-Earth systemKUWmvmgLFDmgLvvvgigfiiii=++−=°−°−×−×=af bgafafafwindkgkgm smNm
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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 .