Problem 3.85

# Problem 3.85 - Problem 3.85 Given[Difficulty 3 Model cross...

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Problem 3.85 [Difficulty: 3] Given: Model cross section of canoe as a parabola. Assume constant width W over entire length L y a x 2 = a 1.2 ft 1 = W 2 ft = L 18 ft = Find: Expression relating the total mass of canoe and contents to distance d. Determine maximum allowable total mass without swamping the canoe. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g = (Hydrostatic Pressure - h is positive downwards from free surface) F v A y p d = (Vertical Hydrostatic Force) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the water and inner surface of the canoe. At any value of d the weight of the canoe and its contents is balanced by the net vertical force of the water on the canoe. Integrating the hydrostatic pressure equation: p ρ g h = F v A y p d = x ρ g h L d = where h H d ( ) y = To determine the upper limit of integreation we remember that y a x 2 = At the surface y H d = Therefore, x H d a = and so the vertical force is: F v 2 0 H d a x ρ g H d ( ) a x 2
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