force exrted on the side and circular base of the tank

Force Exerted on the Side and Circular Base of a Tank: A Hydrostatic Problem

In our daily lives, we often encounter problems involving fluid mechanics, particularly hydrostatic force. This article will explore one such problem - calculating the force exerted on the side and circular base of a tank.

The given scenario involves a circular swimming pool with a diameter of 14 meters, sides that are 3 meters high, and water depth of 2.5 meters. To solve this problem, we must first understand the concept of hydrostatic pressure. Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to gravity.

To find the force exerted on the side and circular base of the tank, we need to integrate the density function of the water. The density of water is 1000 kg/m^3, and the acceleration due to gravity is 9.8 m/s^2.

Let's consider a small element of volume in the pool with an area A and height dy. The force exerted on this element by the surrounding water is given by:

dF = ρ g A dy

where ρ is the density of water, g is the acceleration due to gravity, and A is the area of the element.

To find the total force exerted on the side and circular base of the tank, we need to integrate this expression over the entire volume of the pool. We can do this by dividing the pool into small elements and summing up the forces acting on each element.

Using integration, we can develop a formula for calculating the mass of an object based on its density function. This formula is:

M = ∫'a to b' ρ(x) dx

where ρ(x) is the density function, and 'a to b' represents the interval over which we are integrating.

Applying this formula to our problem, we can find the mass of the water in the pool by integrating the density function over the volume of the pool. This gives us:

M = ∫'0 to 2.5' ρ(y) dy

Substituting the values given in the problem, we get:

M = ∫'0 to 2.5' 1000 kg/m^3 dy

= 12500 kg

To find the force exerted on the side and circular base of the tank, we need to multiply the mass of the water by the acceleration due to gravity.

F = M g

= 12500 kg x 9.8 m/s^2

= 122500 N

Therefore, the force exerted on the side and circular base of the tank is approximately 122500 Newtons.

In conclusion, this problem demonstrates how integration can be used to solve real-world problems involving fluid mechanics. By applying the concept of hydrostatic pressure and integrating the density function, we were able to find the force exerted on the side and circular base of a tank. This technique has numerous applications in various fields, including engineering, physics, and mathematics.

Force exrted on the side and circular base of the tank