The fluid force on the window is about 1608.5 pounds. Observing the first integral is a semicircle's area with radius 1 produces The second integral is 0 because the integrand is odd and the integration limits are symmetric to the origin. The fluid force on a submerged horizontal surface with area \(A\) is Figure 7.7.2 show the pressure at each depth is the same for all three objects. Pascal’s Principle states that the pressure exerted by a fluid at a depth \(h\) is transmitted equally in all directions. When calculating fluid pressure, you can use an important physical law called Pascal’s Principle, named after the French mathematician Blaise Pascal. Where \(w\) is the liquid's weight-density per volume unit.īelow are weight-densities for common fluids in pounds per cubic foot. The pressure on an object at depth \(h\) in a liquid is At 30 feet below sea level the pressure is 30 pounds per square inch. At sea level the atmospheric pressure is 15 pounds per square inch. It is measured in pounds per square inch or kilograms per square centimeter. The bottom is flat (but not horizontal) and the sides are vertical. Pressure is the force on a body's surface area. Ex1: A swimming pool is 20 m long and 10 m wide. That is, we need to find the integral $\int_0^h 62.4x(y)dy \cdot h$.At sea level 15 pound presses on the one inch square and one kilogram presses on the one centimeter square. To find the total fluid force on the vertical side of the tank, we need to integrate this expression over the entire height of the tank. So, the fluid force on this strip is $62.4x(y)dy \cdot h$.ĥ. The volume of the water above the strip is $x(y)dy \cdot h$, and the weight-density of water is 62.4 pounds per cubic foot. The fluid force on this strip is equal to the weight of the water above it. The width of this strip is $dy$, and its length is a function of $y$. Now, we need to find the fluid force on a small horizontal strip of the vertical side of the tank. Since the tank is full of water, the depth will vary linearly from 0 at the top to the height of the tank at the bottom.ģ. Next, we need to find the depth of the water at any point on the vertical side of the tank. The formula for the area of a trapezoid is $A = \frac(b_1 + b_2)h$, where $b_1$ and $b_2$ are the lengths of the parallel sides and $h$ is the height.Ģ. First, we need to find the area of the trapezoid.
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