Question

Video Example EXAMPLE 1 A dam has the shape of a trapezoid shown in the top figure. The height is 20 m and the width is 40 m at the top and 20 m at the bottom. Find the force on the dam due to hydrostatic pressure if the water level is 6 m from the top of the dam. 40 m SOLUTION We choose a vertical x-axis with origin at the surface of the water and directed downward as in the middle figure. The depth of the water is 14 m, so we divide the interval [0, 14] into subintervals of equal length with endpoints x and we choose x' [x - 1, X]. The Ith horizontal strip of the dam is approximated by a rectangle with height Ax and width w, where, from similar triangles in the bottom figure, 20 m and so -3x)-_ -xi w; = 2(10 + a) = 2(10 + _ If A is the area of the ith strip, then A; = wdx=(O - x^)ex 20 m If Ax is small, then the pressure P, on the ith strip is almost constant and we can use this equation to write P = 1000gx The hydrostatic force F, acting on the ith strip is the product of the pressure and the area: 10 10 F, = PA; = 1000gx"( -xi" Ax Adding these forces and taking the limit as n 0 , we obtain the total hydrostatic force on the dam: 10 AN == . lim§ 1000gxi ( D - x*)ex - (*10009~( -x) dx = 1000(9.8)["" (C ) ox - 98001 N (round to the nearest Integer) 14-X

Answer 1

how 10 [u;-) 12]. rom ith height an horizontal Strip from similar triangle of figure So, width of Strip - d (10+2) = a (10+ 177-**) - 134 - x Area of th Strip A, = W; An = ( 134 x) Aa if An is small, the pressure of 1th Strip is P = booget Effore = P; A = 1000grt (134) - x) an. now sotal force f = lim & 1000g art (134) Dar no ha

In [34) –n) du F- 1000 (9.8) | £ 200 - 22 ) in F = 9000 [31) - 1 F = 9800 [1172 - F = 9.800 [ 17(60)- (Ly] F = 1800 (13" (51=1"] f = 9800 (196) (39) (F= 23689866.667 N | ox/F = 23689867 N