Homework solution. Chapter 14. P14.6 (a) ()()()53201.01310Pa1024 kgm9.80 ms1000 mPPghρ=+=×+71.0110PaP=×(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. 7gauge01.0010PaPPPghρ=−==×The resultant inward force on the porthole is then ()275gauge1.0010Pa0.150 m7.0910NFPAπ⎡⎤==×=×⎣⎦. 2.00 m1.00 m2.00 mP14.14 The air outside and water inside both exert atmospheric pressure, so only the excess water pressure ghρcounts for the net force. Take a strip of hatch between depth hand hdh+. It feels force ()2.00 mdFPdAghdhρ==. (a) The total force is ()2.00 m1.00 m2.00 mhFdFghdρ===∫∫hFIG. P14.14()()()()()()()2.00 m222321.00 m2.00 m2.00 m1000 kgm9.80 ms2.00 m1.00 m2229.4 kN to the righthFgFρ⎡⎤==−⎣⎦=(b) The lever arm of dFis the distance ()1.00 mh−from hinge to strip: ()()()()()()()2.00 m1.00 m2.00 m321.00 m33322.00 m1.00 m2.00 m1.00 m327.00 m3.00 m1000 kgm9.80 ms2.00 m3216.3 kNm counterclockwisehdghhdhhhgττρτρττ===−⎡⎤=−⎢⎥⎣⎦⎛⎞=−⎜⎟⎝⎠=⋅∫∫411
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