† † margin: y y x - 2 - 1 1 2 - 2 - 1 1 2 50 water line not to scale d ( y ) = 50 - y Figure 6.5.8: Measuring the fluid force on an underwater porthole in Example 6.5.4.\( \newcommand\). (The weight-density of water is 62.4 pounds. Find the fluid force on the vertical side of the tank, where the dimensions are given in feet. The truth is that it is not, hence the survival tips mentioned at the beginning of this section. (The weight-density of water is 62.4 pounds per cubic foot.) XIb Trapezoid 4 7 2. This is counter-intuitive as most assume that the door would be relatively easy to open. Most adults would find it very difficult to apply over 500 lb of force to a car door while seated inside, making the door effectively impossible to open. Suppose that the tank is full to a depth of 7 feet with water of weight density 62.4 pounds/ft3. = ∫ - 2.25 0 62.4 ( - y ) 10 / 3 d y What is the total force exerted by water against a dam. (The weight-density of water is 62.4 pounds per cubic foot. 3 ¯, - 2.25 ) ( 0, - 2.25 ) ( 0, 0 ) y y x Figure 6.5.7: Sketching a submerged car door in Example 6.5.3. Find the fluid force on the vertical side of the tank, where the dimensions are given in feet. ![]() Using the weight-density of water of 62.4 lb/ft 3, we have the total force as Example2: Consider a swimming pool 5 m wide, 10 m long and 3m deep. We adopt the convention that the top of the door is at the surface of the water, both of which are at y = 0. So the force exerted by a fluid of constant weight density w, against a submerged vertical plane from ya to yb is given as, Where h(y) is depth of fluid at y and L(y) is the horizontal length of region at y. ![]() Its length is 10 / 3 ft and its height is 2.25 ft. ![]() SolutionThe car door, as a rectangle, is drawn in Figure 6.5.7.
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