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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. F T F F T mg x y = = = = 0 0 sin cos θ θ , and Dividing: tan . . θ = = = F mg 14 7 0 750 N 19.6 N , or θ = ° 36 9 . Therefore, H L eq m m = = ° = 1 2 00 1 36 9 0 400 cos . cos . . θ a f a fa f (d) As F → ∞ , tan θ → ∞ , θ ° 90 0 . and H L eq A very strong wind pulls the string out horizontal, parallel to the ground. Thus, H L eq e j max = . P8.70 Call φ θ = °− 180 the angle between the upward vertical and the radius to the release point. Call v r the speed here. By conservation of energy K U E K U mv mgR mv mgR gR gR v gR v gR gR i i r r i r r r + + = + + + = + + = + = 1 2 0 1 2 2 2 3 2 2 2 2 cos cos cos φ φ φ The components of velocity at release are v v x r = cos φ and v v y r = sin φ so for the projectile motion we have The path after string is cut i v =
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