Calculation of Total Pressure on a Square Submerged in Water

When dealing with fluid mechanics, the problem of finding the total pressure on a submerged object is a fundamental concept. This article explores the detailed steps to calculate the total force exerted by water on a square plate, with one corner at the surface and the opposite corner submerged. We will also clarify the difference between pressure and force, and how to use the principles of fluid statics to solve such a problem.

Understanding Fluid Pressure and Force

In fluid mechanics, the pressure at a point in a fluid is defined as the force per unit area at that point. Depending on the reference point chosen, pressure can be classified as gauge pressure or absolute pressure. Gauge pressure is the pressure relative to atmospheric pressure, while absolute pressure includes the atmospheric pressure as well.

Calculation of Total Force on a Submerged Object

Consider a square plate with side length of 2 feet, submerged in water such that one corner is at the surface, and the opposite corner is directly below. The specific weight of water is 62.4 pounds per cubic foot, and the pressure at the surface is 0. The pressure at the deepest point of the square can be calculated using the formula:

Ph 62.4h (psf)

Centroid Calculation

The centroid of the square is at a distance of L_C √2/2 ft 0.707 ft from the surface. Using the formula for the resultant force due to fluid pressure:

F_R γL_C A

For water, γ 62.4 lb/ft^3, and the area A 4 ft^2. Substituting the values, we get:

F_R 62.4 * 0.707 * 4 176.5 lb (force on the centroid of the square)

Calculating the Total Force on Each Face

To find the total force on each face of the square, we need to calculate the force on thin horizontal strips of the square. The pressure at a depth h is given by:

Ph 62.35h (psf)

The force on an infinitesimal strip of width 2h and height Δh is:

dF Ph * 2h * Δh 124.7h^2 Δh

For each strip at depth h, there is an identical strip at depth 2.828 - h, where the pressure is:

P2.828 - h 62.35(2.828 - h) (psf)

The total force on both sides of the 2 ft square is then calculated by integrating over the range of h from 0 to 1.414:

F integrate 249.4h(2.828 - h) dh from 0 to 1.414

This integral simplifies to:

F 705.3h^2/2 evaluated from 0 to 1.414

So, the total force on both sides is:

F 705.3/2 * [1.414^2 - 0^2] 705.3 lb

Dividing by 2 to find the force on each face, we get:

Force on each face 353 lb

Conclusion

The calculation of the total force on a submerged object involves understanding the principles of fluid statics, including the concept of pressure and force. By breaking down the problem into infinitesimal strips and integrating, we can accurately determine the total force on each face of the submerged square plate.

This detailed explanation not only answers the specific problem but also provides insight into how to approach similar fluid mechanics problems. Whether you are an engineer, a student, or simply curious about fluid dynamics, this knowledge is valuable and widely applicable.