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350_Dynamics 11ed Manual

# 350_Dynamics 11ed Manual - Engineering Mechanics Dynamics...

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Unformatted text preview: Engineering Mechanics - Dynamics Chapter 15 Since Fx = μFy, from Eqs [2] and [3] m v1 cos ( θ ) − m v2 cos ( φ ) Δt v2 = v1 μ ( m v1 sin ( θ ) + m v2 sin ( φ ) ) = Δt cos ( θ ) − μ sin ( θ ) [4] μ sin ( φ ) + cos ( φ ) Substituting Eq. [4] into [1] yields: Given θ = 45 deg Given e= e= φ = 45 deg sin ( φ ) ⎛ cos ( θ ) − μ sin ( θ ) ⎞ ⎜ e = 0.6 sin ( φ ) ⎛ cos ( θ ) − μ sin ( θ ) ⎞ ⎜ ⎟ sin ( θ ) ⎝ μ sin ( φ ) + cos ( φ ) ⎠ Guess μ = 0.2 μ = Find ( μ ) ⎟ sin ( θ ) ⎝ μ sin ( φ ) + cos ( φ ) ⎠ μ = 0.25 Problem 15-71 The ball bearing of weight W travels over the edge A with velocity vA. Determine the speed at which it rebounds from the smooth inclined plane at B . Take e = 0.8. Given: W = 0.2 lb vA = 3 ft s θ = 45 deg g = 32.2 ft e = 0.8 2 s Solution: Guesses vB1x = 1 t =1s Given vB1x = vA ft s vB1y = 1 ft s vB2n = 1 R = 1 ft vA t = R cos ( θ ) −1 2 g t = −R sin ( θ ) 2 vB1y = −g t vB1x cos ( θ ) − vB1y sin ( θ ) = vB2t 348 ft s vB2t = 1 ft s ...
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