Problem 2.50

# Problem 2.50 - Problem 2.50 [Difficulty: 3] Given: Block on...

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Problem 2.50 [Difficulty: 3] Given: Block on oil layer pulled by hanging weight N Mg x y F v F t mg F t Find: Expression for viscous force at speed V; differential equation for motion; block speed as function of time; oil viscosity Solution: Governing equations: τ yx μ du dy = Σ F x Ma x = Assumptions: Laminar flow; linear velocity profile in oil layer The given data is M5 k g = Wm g = 9.81 N = A2 5 c m 2 = h 0.05 mm = Equation of motion (block) Σ F x x = so F t F v M dV dt = 1 () Equation of motion (block) Σ F y ma y = so mg F t m dV dt = 2 Adding Eqs. (1) and (2) F v Mm + dV dt = The friction force is F v τ yx A = μ du dy A = μ V h A = Hence μ A h V + dV dt = To solve separate variables dt + μ A h V dV = t + h μ A ln m g μ A h V ln m g = + h μ A ln 1 μ A h V = Hence taking antilogarithms 1 μ A h V e μ A + h t =

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Finally V mg h μ A 1e μ A Mm + () h t = The maximum velocity is V h μ A = In Excel: The data is M = 5.00
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## This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 2.50 - Problem 2.50 [Difficulty: 3] Given: Block on...

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