244_Physics ProblemsTechnical Physics

244_Physics ProblemsTechnical Physics - 246 Potential...

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246 Potential Energy (c) Call θ the equilibrium angle with the vertical. FTF FTm g x y =⇒ = = 0 0 sin cos , and Dividing: tan . . == = F mg 14 7 0750 N 19.6 N , or 36 9 . Therefore, HL eq m m =− = ° = 12 0 0 1 3 6 9 0 4 0 0 cos . cos . . a f a fa f (d) As F →∞ , tan , →° 90 0 . and eq A very strong wind pulls the string out horizontal, parallel to the ground. Thus, eq ej max = . P8.70 Call φθ 180 the angle between the upward vertical and the radius to the release point. Call v r the speed here. By conservation of energy KU EKU mv mgR mv mgR gR gR v gR vg R g R ii rr ir r r ++ =+ ++ = + += + 1 2 0 1 2 22 32 2 cos cos cos φ The components of velocity at release are vv xr = cos and yr = sin so for the projectile motion we have The path after string is cut i v = Rg R θ C FIG. P8.70 xv t x = 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 .

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