A hemispherical plate with diameter 10 ft is submerged vertically 2 ft below the surface of the water. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Recall that the weight density of water is 62.5 lb/ft^3.)
To express the hydrostatic force against one side of the plate as an integral, we need to consider the pressure at each infinitesimally small area on the surface of the plate and integrate it over the entire surface.
The pressure at a certain depth in a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth below the surface.
In this case, the depth below the surface is given as 2 ft, and the density of water is given as 62.5 lb/ft^3. The acceleration due to gravity can be considered as 32.2 ft/s^2.
We can consider an infinitesimally small area on the surface of the plate. Since the plate is a hemisphere, the area differential can be represented as dA = 2πr ds, where r is the radius of the plate and ds is an infinitesimally small segment on the surface of the plate.
The radius of the plate is half of the diameter, so r = 10 ft / 2 = 5 ft.
The hydrostatic force on the infinitesimally small area dF is given by dF = P dA.
Substituting the values, dF = (ρgh) * (2πr ds) = (62.5 lb/ft^3) * (32.2 ft/s^2) * (2π * 5 ft * ds)
Now, we need to integrate this expression over the entire surface of the plate.
The limits of integration for ds can be taken as 0 to πr, as we need to cover the entire surface of the plate.
So, the integral for the hydrostatic force F becomes:
F = ∫[0,πr] (62.5 lb/ft^3) * (32.2 ft/s^2) * (2π * 5 ft * ds)
Simplifying this expression gives us:
F = (62.5 lb/ft^3) * (32.2 ft/s^2) * (2π * 5 ft) * ∫[0,πr] ds
F = 62.5 * 32.2 * 2 * π * 5 * r
Now, substituting the value of r as 5 ft:
F = 62.5 * 32.2 * 2 * π * 5 * 5
Evaluating this expression gives:
F ≈ 12500 * 32.2 * 10 * π
F ≈ 12500 * 322 * π
F ≈ 12500 * 1004.48
F ≈ 12,555,000 lb
Rounding this to the nearest whole number, the hydrostatic force against one side of the plate is approximately 12,555,000 lb.
To express the hydrostatic force against one side of the plate as an integral, we need to consider the pressure at each infinitesimally small element of the surface and integrate it over the entire surface area of the plate.
Here's what we can do step by step:
1. Let's first determine the radius of the hemisphere. The diameter is given as 10 ft, so the radius (r) is half of that, which is 10/2 = 5 ft.
2. We need to find the pressure at each point on the surface of the plate. The pressure at any depth in a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
Since the plate is submerged vertically 2 ft below the water's surface, the depth h will vary from 2 ft at the bottom of the plate to 0 ft at the top. We can express this as h = 0 to h = 2.
3. The weight density of water is given as 62.5 lb/ft^3. Therefore, the density (ρ) of water is equal to the weight density, which is 62.5 lb/ft^3.
4. Now, let's determine the pressure at each infinitesimally small element of the surface. This pressure can be expressed as P = ρgh.
Since the plate is a hemisphere, we can consider a differential element on the curved surface of the hemisphere as a thin circular strip with thickness dx. The height of this strip can be expressed as h = 2 - x, where x is the distance from the top of the plate.
Using P = ρgh, we have P = (62.5 lb/ft^3) * (32.2 ft/s^2) * (2 - x) lb/ft^2.
5. The differential area of this infinitesimally small strip on the curved surface of the hemisphere is given by dA = 2πrx dx.
6. Now, we can express the hydrostatic force (dF) acting on this infinitesimally small strip as dF = P * dA.
Substituting the expressions for P and dA, we have dF = (62.5 lb/ft^3) * (32.2 ft/s^2) * (2 - x) lb/ft^2 * 2πrx dx.
7. Finally, to evaluate the total hydrostatic force, we need to integrate the expression for dF over the entire surface area of the hemisphere.
The limits of integration for x are from 0 to 2, since x varies from the top to the bottom of the hemisphere.
The integral for the hydrostatic force (F) is given by F = ∫[0 to 2] (62.5 lb/ft^3) * (32.2 ft/s^2) * (2 - x) lb/ft^2 * 2πrx dx.
To evaluate this integral, we can use any mathematical software or tool capable of solving integrals. Upon evaluating the integral, you should get an approximate value for the hydrostatic force against one side of the plate.