Question

A vertical dam retains 15 m of water. It is pierced at its base with a semi-circular door 8 m in diameter (the base of the door is its diameter), as indicated in the figure below. a) Let y denote the height in meters measured from the base of the dam. The hydrostatic force exerted by the water on the portion of the door comprised between y m and y + 4y m is approximately p(y) delta y N, what is p(y)? Note that the density of water is rho = 1000 Kg/m^3 and the acceleration due to gravity on the Earth's surface is g = 9.8 m/s^2. Express your answer as a formula. Answer: b) In Newtons, what is the total hydrostatic force exerted on the door? Give the answer correct to three significant digits.

Answer 1

Given Vertical dam retains 15 m of water. Diameter of the semi-circular door = 8 m rho = 1,000 kg/m^3 g = 9.8 m/s^2 (a) Since the portion of the door comprised between y m and y + delta y m. Implies Area = 2 squareroot 16 - y^2 delta y m^2 At height y, the column of water is at 18 - y m. Implies Pressure = rho g (15 - y) = 1000 times 9.8 times (15 - y) = 9800 (15 - y) N/m^2 The hydrostic force acting on the portion of the door is (Pressure) times (Area) congruence (9800 (15 - y)) times (2 squareroot 16 - y^2 delta y) = 19600 (15 - y) squareroot 16 - y^2 (b) Since the door has a height 8/2 = 4 m The total force exerted on it is: integral^4_0 p (y) dy = integral^4_0 19600 (15 - y) squareroot 16 - y^2 dy = 19600 integral^4_0 (15 - y^2) squareroot 16 - y^2 dy ...(1) Let y = 4 sin (t) then dy = 4 cos (t) dt. t