Force and Torque on a Dam. A dam has the shape of a rectangular solid. The side facing the lake has area A and height H. The surface of the freshwater lake behind the dam is at the top of the dam. (a) Show that the net horizontal force exerted by the water on the dam equals —that is, the average gauge Pressure across the face of the dam tunes the area (see below Problem). (b) show that the torque exerted by the water about an axis along the bottom of the dam is ?gH2A/6. (c) How do the force and torque depend on the size of the lake? Problem: A swimming pool is 5.0 m long, 4.0 m wide, and 3.0 m deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth h, and integrate this over the end of the pool.) Do not include the force due to air pressure

Solution 55P Step 1: Pressure due to the water column at a height, y, P = g(H-y) Where, - Density of water g - Acceleration due to gravity Suppose we are considering a strip of elemental area dA The force exerted by this elemental area, dF = PdA Consider, dA = L dy Where L is the width of the strip Then, dF = gL(H-y)dy Step 2: a) In order to get the force exerted by the whole area, A, we have to integrate the equation. H dF = gL(H y)dy 0 H That is, F = gL (H y) dy 0 2 Or, F = gL Hy[ y2] 0 to H Substituting the upper and lower limits, H2 F = gL [ 2 We know that, LH = A 1 Therefore, F = 2 gHA