MidtermSolutions

MidtermSolutions - With h CG = h/2 then F = h CG A = 62.4 h...

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MAE 101A Summer I 2008 Midterm Solutions 1) a. Given = 0.4 lb s ft 2 and applying Newton's relation = du dy = u 2 u 1 y 2 y 1 = u 2 0 y 2 0 = V h = 0.4 lb s ft 2 0.7 ft s 0.18 12 ft = 18.67 lb ft 2 . b. Applying Newton's law F = ma then F = ma = A 18.67 lb ft 2 9 ft 2 = 420 lb 32.2 ft s 2 a a =− 12.88 ft s 2 . 2) dy dx = v u = y 2x dy y = dx 2x dy y = dx 2x ln y = 1 2 ln x ln C y = Cx 1 2 from given point (2,3) then C = 3 2 .
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3)
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4) Keep in mind that the wet portion of the gate is considered, and there is no buoyancy force since the other side is the ambient air. The weight of the gate W = (7.85)(62.4 lbf/ft 3 )(15 ft)(1/12 ft)(6 ft) = 3674 lbf.
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Unformatted text preview: With h CG = h/2 then F = h CG A = 62.4 h 2 6 h sin 60 o = 216.16 h 2. y cp = 1 / 12 6 h sin 60 o 3 sin 60 o h 2 h sin 60 o = h 6 1 sin 60 o . Taking sum of moments around B then M B = 10000 15 216.16 h 2 h 2sin 60 o h 6sin 60 o 3674 7.5cos60 o = which gives 83.2 h 3 = 136222.5 h = 1637.3 1 3 = 11.8 ft....
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This note was uploaded on 03/30/2011 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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MidtermSolutions - With h CG = h/2 then F = h CG A = 62.4 h...

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