HW3S - Problem 2 Problem *3.97 [4] NEW PROBLEM STATEMENT...

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Problem 2
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Problem 3 ( L + c )/2 L c F BB W B F BR W R L /2 a ! Given: Geometry of block and rod Find: Angle for equilibrium 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 c ρ A $ L 2 c 2 % $ 2M B ρ V B $ % 1 2 M R $ ( ) + , . $ L $ # cL 2 2L $ ρ A $ M B ρ V B $ % 1 2 M R $ ( ) + , . $ % # c 5 m $ 2 25 / m $ m 3 1000 kg $ / 1 25 / 1 cm 2 $ 100 cm $ 1m $ ) + , . 2 / 30 kg $ 1000 kg m 3 $ 0.025 / m 3 $ ) * + , - . % 1 2 1.25 / kg $ ( 0 1 2 3 4 5 / % # c 1.58m # Then sin θ a c # with a 0.25 m $ # θ asin a c ) + , . # θ 9.1 deg $ #
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Problem 4
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Problem 5 Given: $%&’’&’( *+,-./ 0/12/3 1, 4’/ /’3 Find: 516&7-7 1’(-218 0%//3 948 ’4 :1;&,1,&4’ Solution: <10&: /=-1,&4’ >’ :47%4’/’,0 8 % ! ! " ρ 1 8 # $ ρ " ? " 8 # $ ρ " ω " # 8 # $ @ % ! ! ρ " ( # $ </,A//’ B 1’3 CD 8 E :4’0,1’,D 04 @ % !
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HW3S - Problem 2 Problem *3.97 [4] NEW PROBLEM STATEMENT...

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