determine the force due to hydrostatic pressure on the flat vertical side of a tank

† † 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. 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. So our force becomes 62.4 and then we need to figure out how to find the area of that layer.SolutionThe car door, as a rectangle, is drawn in Figure 6.5.7. So we have square feet times feet and if we divide that by cubic feet, that would cancel out our feet units and we're going to lift with pounds, which is what our force is going to be measured in. Consider the density of the oil to be 3 900 kg/m. Also, determine the location of this resultant, measured from the top of the tank. Determine the resultant force that both of these liquids exert on the side ABC of the tank if the side has a width of b 1.25 m. We would need to multiply that by the area of that layer right there, which should be measured in square feet, and then we would need to multiply that by the depth of that which is y and that we measured in feet. A storage tank contains oil and water acting at the depths shown. So the tank is bounded by these curves, so this thing is filled with water, so we're just going to take a very thin layer of water here and we're gonna, let the thickness of that thickness of that be delta y, okay and then it's going to be At a depth of y units, that's gonna be the depth of that layer of water, so the hydrostatic force of the water on that plate just on this thin layer would be the weight density of water, which is 62.4 pounds per cubic foot. That'S y equals 3 times of natural log of 17 point okay. That'S just the x axis, and then we have this constant flat line. This on the left side would be y equals 3 times the natural log of negative x, and we got y equals 0. This curve here will be y equals 3 times the natural log of x. Has a density of 62.4 per cubic foot, so we have this taint shaped like this bounded by these curves. Find the hydrostatic force on the flat vertical side of a tank that has the shape of the region bounded by y equals 0 y equals 3 l n x y equals 3 l n of negative x and y equals 3 l n of 17 point water.

Determine the force due to hydrostatic pressure on the flat vertical side of a tank