Problem 3.101 - Problem *3.101 [Difficulty: 3] Given:...

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Problem *3.101 [Difficulty: 3] ( L + c )/2 L c F BB W B F BR W R L /2 a θ F hinge,y F hinge,x Given: Geometry of block and rod Find: Angle for equilibrium Assumptions: Water is static and incompressible Solution: Basic equations Σ M Hinge 0 = F B ρ g V = (Buoyancy) The free body diagram is as shown. F BB and F BR are the buoyancy of the block and rod, respectively; c is the (unknown) exposed length of the rod Taking moments about the hinge W B F BB () L cos θ F BR Lc + 2 cos θ W R L 2 cos θ + 0 = with W B M B g = F BB ρ g V B = F BR ρ g A = W R M R g = Combining equations M B ρ V B L ρ A + 2 M R L 2 + 0 = We can solve for M B ρ A L 2 c 2 2M B ρ V B 1 2 M R + L = M B ρ A 2L
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