246 Potential Energy(c)Call θthe equilibrium angle with the vertical.FTFFTmgxy∑∑=⇒==00sincos, andDividing: tan..===Fmg14 70750N19.6 N, or =°36 9.Therefore, HLeqm m=−=−°=120013690400cos.cos..afafaf(d)As F→∞, tan, →°90 0.and eq→A very strong wind pulls the string out horizontal, parallel to the ground. Thus,eqejmax=.P8.70Call φθ−180the angle between the upward vertical andthe radius to the release point. Call vrthe speed here. Byconservation of energyKU EKUmvmgRmvmgRgRgRvgRvgRgRiirrirrr++ =+++=++=+∆1201222322coscoscosφThe components of velocity at release are vvxr=cosandyr=sinso for the projectile motion we haveThe path after string is cut iv=RgR θ C FIG. P8.70xvtx=Rv
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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 .