{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MidtermSolutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3)
Background image of page 2
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.
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}