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# s14 - Homework solution Chapter 14 P14.6(a P = P0 gh = 1...

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Homework solution. Chapter 14. P14.6 (a) ( ) ( ) ( ) 5 3 2 0 1.013 10 Pa 1024 kg m 9.80 m s 1000 m P P gh ρ = + = × + 7 1.01 10 Pa P = × (b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere. 7 gauge 0 1.00 10 Pa P P P gh ρ = = = × The resultant inward force on the porthole is then ( ) 2 7 5 gauge 1.00 10 Pa 0.150 m 7.09 10 N F P A π = = × = × . 2.00 m 1.00 m 2.00 m P14.14 The air outside and water inside both exert atmospheric pressure, so only the excess water pressure gh ρ counts for the net force. Take a strip of hatch between depth h and h dh + . It feels force ( ) 2.00 m dF PdA gh dh ρ = = . (a) The total force is ( ) 2.00 m 1.00 m 2.00 m h F dF gh d ρ = = = h FIG. P14.14 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2.00 m 2 2 2 3 2 1.00 m 2.00 m 2.00 m 1000 kg m 9.80 m s 2.00 m 1.00 m 2 2 29.4 kN to the right h F g F ρ = = = (b) The lever arm of dF is the distance ( ) 1.00 m h from hinge to strip: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2.00 m 1.00 m 2.00 m 3 2 1.00 m 3 3 3 2 2.00 m 1.00 m 2.00 m 1.00 m 3 2 7.00 m 3.00 m 1000 kg m 9.80 m s 2.00 m 3 2 16.3 kN m counterclockwise h d gh h dh h h g τ τ ρ τ ρ τ τ = = = = = = 411

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