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

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Problem 2
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
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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