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Unformatted text preview: Math 21BB  Homework Set 7 Section 6.7: 1. Calculate the fluid force on one side of the triangular plate: Note that in this coordinate system, we have 5 y  2. We have that strip depth = y and L ( y ) = 2( y + 5) (using the fact that the right leg of the triangle lies on the line y = x 5 and the fact that our triangle is symmetric over the yaxis). Fluid Force = Z b a w ( strip depth ) L ( y ) dy = Z 2 5 62 . 4  y (2 y + 10) dy = Z 2 5 124 . 8 y 2 624 y dy = 41 . 6 y 3 312 y 2 2 5 = [( 41 . 6  8) (312 4)] [( 41 . 6  125) (312 25)] = 1684 . 8 lb 11. a. What is the fluid force on the gate when the liquid is 2 ft deep? We will assume that the vertex of the gate is along the bottom edge of the cubical tank. Using the coordinates given in the illustration in the book, we know that the water level is at the line y = 2. To calculate the force we are concerned with 0 y 1. We have that the strip depth = 2 y and L ( y ) = 2 y (using the symmetry of the gate over the yaxis). Fluid Force = Z 1 50 (2 y ) 2 y dy = Z 1 200 y 100 y 3 / 2 dy = 400 3 y 3 / 2 40 y 5 / 2 1 = 400 3 40 = 280 3 93 . 33 lb 1 b. What is the maximum height to which the container can be filled without exceeding its design limitation? Suppose that the water level is at y = h for some positive number h . Then to calculate the force on the gate we are still just concerned with 0 y 1 and we have that strip depth = h y and L ( y ) = 2 y . Thus treating h as a number we can calculate the force on the gate in terms of h by using our integral formula: Fluid Force = Z 1 50 ( h y ) 2 y dy = Z 1 100 h y 100 y 3 / 2 dy = 200 h 3 y 3 / 2 40 y 5 / 2 1 = 200 h 3 40 We know that the gate is designed to withstand Fluid Force 160 lb. We substitute our above result for the force into this inequalitylb....
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 Spring '07
 MAT21B

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