a vertical gate in a dam has the shape of an isosceles trapezoid 8 feet across the top and 6 feet across the…

Explanation:

Step1: Find the depth - function

The depth (d(y)) of a horizontal strip of the gate at a distance (y) from the top of the gate is (d(y)=4 + y). The height of the gate is (h = 5) feet.

Step2: Find the width - function

We use similar - triangles to find the width (w(y)) of the horizontal strip. The change in width over the height of the gate is (\frac{8 - 6}{5}=0.4). The width of the strip at height (y) above the bottom of the gate is (w(y)=6+0.4y).

Step3: Set up the pressure integral

The pressure on a small horizontal strip of area (dA=w(y)dy) at depth (d(y)) is (dP=\rho g d(y)dA), where (\rho = 62.4\ lb/ft^{3}) and (g = 32.2\ ft/s^{2}) (but since we are using the density (\rho) directly, we don't need (g) separately). The total pressure (P) is given by the integral (P=\int_{0}^{5}\rho d(y)w(y)dy=\int_{0}^{5}62.4(4 + y)(6 + 0.4y)dy). First, expand the integrand: ((4 + y)(6 + 0.4y)=24+1.6y+6y + 0.4y^{2}=24 + 7.6y+0.4y^{2}). Then, (P = 62.4\int_{0}^{5}(24 + 7.6y+0.4y^{2})dy).

Step4: Integrate term - by - term

(\int_{0}^{5}(24 + 7.6y+0.4y^{2})dy=\int_{0}^{5}24dy+\int_{0}^{5}7.6ydy+\int_{0}^{5}0.4y^{2}dy). (\int_{0}^{5}24dy=24y\big|{0}^{5}=24\times5 - 24\times0 = 120). (\int{0}^{5}7.6ydy=7.6\times\frac{y^{2}}{2}\big|{0}^{5}=3.8y^{2}\big|{0}^{5}=3.8\times25-3.8\times0 = 95). (\int_{0}^{5}0.4y^{2}dy=0.4\times\frac{y^{3}}{3}\big|_{0}^{5}=\frac{0.4}{3}\times125=\frac{50}{3}\approx16.67). The sum of the integrals is (120 + 95+\frac{50}{3}=\frac{360 + 285+50}{3}=\frac{695}{3}).

Step5: Calculate the pressure

(P = 62.4\times\frac{695}{3}). (P=62.4\times\frac{695}{3}=62.4\times231.67 = 14456.2\approx14456\ lb).

Answer:

14456