![]() ![]() Why? How far down is that from the top? Let that be y and we left the thickness of this part, be delta y, and then it's also going to depend on the length of that section or the area of that rectangle the length times its width. So if we had a thin layer of water swimming, that's filled with water and there's a here's gonna be a thin layer of that water to find the force the hydrostatic force on that section. We have these 4 lines containing the this area and that's representing the flat vertical side of a tank. There is y equals 0 point, so we can see. Minus 4 and then the line y equals 0 would just be the x axis there. So here we have the line y equals 14 point. Each unit in the coordinate plane represents 1 foot and water has a density of 62.4 per cubic foot. ![]() Minus 4 y equals negative 2 x minus 4 and y equals 14 point. † † 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.Find the hydrostatic force on the flat vertical side of a tank that is bounded by the lines y equals 0 y equals 2 x. ![]() The truth is that it is not, hence the survival tips mentioned at the beginning of this section. 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. Using the weight-density of water of 62.4 lb/ft 3, we have the total force as We adopt the convention that the top of the door is at the surface of the water, both of which are at y = 0. 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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