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Chapter 14: Problem 12

A swimming pool has dimensions \(30.0 \mathrm{m} \times 10.0 \mathrm{m}\) and a flat bottom. When the pool is filled to a depth of \(2.00 \mathrm{m}\) with fresh water, what is the force caused by the water on the bottom? On each end? On each side?

Short Answer

Expert verified
The force exerted by water on the bottom of the pool is \(5880000 \mathrm{N}\), on the ends it's \(784000 \mathrm{N}\), and on the sides it's \(2352000 \mathrm{N}\).

Step by step solution

01

Calculate Force on Bottom

Firstly, calculate the force exerted by the water on the bottom of the pool. The area of the bottom is \(30.0 \mathrm{m} \times 10.0 \mathrm{m} = 300.0 \mathrm{m^2}\) and the height is \(2.00 \mathrm{m}\). Use the formula for pressure \(P = \rho \cdot g \cdot h\) where \(\rho\) is the density of water equal to \(1000 \mathrm{kg/m^3}\), \(g\) is the gravitational constant equal to \(9.8 \mathrm{m/s^2}\), and \(h\) is the height of the water. This gives \(P = 1000 \mathrm{kg/m^3} \cdot 9.8 \mathrm{m/s^2} \cdot 2 \mathrm{m} = 19600 \mathrm{Pa}\). Then calculate the force on the bottom using \(F = P \cdot A\) where \(A\) is the area of the bottom. This gives \(F = 19600 \mathrm{Pa} \cdot 300.0 \mathrm{m^2} = 5880000 \mathrm{N}\)
02

Calculate Force on Ends

Now calculate the force on one end of the pool. The area of each end is \(2.0 \mathrm{m} \times 10.0 \mathrm{m} = 20.0 \mathrm{m^2}\). From Step 1, the pressure at a depth of 2.0 m is \(19600 \mathrm{Pa}\). Consequently, the force on one end can be given as \(F = P \cdot A = 19600 \mathrm{Pa} \cdot 20.0 \mathrm{m^2} = 392000 \mathrm{N}\). This is for one end, so multiply by 2 to get the total force on both ends which comes out as \(784000 \mathrm{N}\).
03

Calculate Force on Sides

Finally, calculate the force on one side of the pool. The area of each side is \(2.0 \mathrm{m} \times 30.0 \mathrm{m} = 60.0 \mathrm{m^2}\). Using the pressure from Step 1, the force on one side is \(F = P \cdot A = 19600 \mathrm{Pa} \cdot 60.0 \mathrm{m^2} = 1176000 \mathrm{N}\). Again, this is for one side, so multiply by 2 to get the total force on both sides which equals to \(2352000 \mathrm{N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrostatic Pressure
Hydrostatic pressure is a fundamental concept in fluid mechanics that describes the pressure exerted by a stationary liquid due to its weight. It increases with depth because as you go deeper into the liquid, there is more weight pushing down due to the liquid above. This concept is crucial when considering forces in a fluid environment, such as a swimming pool.

The formula to calculate hydrostatic pressure is given by: \[ P = \rho \cdot g \cdot h \] where \( P \) is the pressure, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the height or depth of the fluid column above the point where the pressure is being calculated.

In our swimming pool example, the hydrostatic pressure at the bottom of the pool is calculated using the density of fresh water (approximately \(1000 \mathrm{kg/m^3}\)) and gravitational acceleration (\(9.8 \mathrm{m/s^2}\)). The depth of the water (\(2.00 \mathrm{m}\)) directly correlates to the height in the formula, providing us with the specific pressure at the bottom of the pool.
Force Calculation
Once we understand hydrostatic pressure, we can use this information to calculate the force being exerted on surfaces submerged in a fluid, such as the bottom and walls of a swimming pool. Force calculation is an integral part of understanding fluid mechanics and engineering design.

The force exerted by a fluid on a surface is calculated by: \[ F = P \cdot A \] where \( F \) is the force, \( P \) is the pressure exerted by the fluid, and \( A \) is the area of the surface. Hence, the force is the product of the pressure at a given depth and the surface area of the submerged part of the structure.

In the context of our example, once the hydrostatic pressure at the pool鈥檚 depth is known, it can be multiplied by the area of the pool's bottom to find the total force exerted on it. Likewise, we calculate the force on the pool's ends and sides using the same pressure but adjusting the area to match those specific surfaces.
Buoyancy and Fluid Statics
Buoyancy is a key principle in fluid statics, which is the branch of fluid mechanics that studies fluids at rest and the forces in them. It refers to the upward force that a fluid exerts on an object that is submerged, which is equal to the weight of the fluid displaced by the object. This is commonly known as Archimedes' principle.

The formula for buoyant force (\( F_b \)) is: \[ F_b = \rho \cdot g \cdot V \] where \( \rho \) is the density of the fluid, \( g \) is the gravitational acceleration, and \( V \) is the volume of the fluid displaced by the object.

In our swimming pool scenario, the concept of buoyancy would apply if we were to consider an object floating in the pool. The object would experience a buoyant force, and this force would be analyzed using the same principles that govern the hydrostatic pressure and force calculation on the pool鈥檚 surfaces. Understanding buoyancy is essential for applications ranging from swimming to shipbuilding, where the buoyancy must balance the weight to keep objects afloat.

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Most popular questions from this chapter

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. (a) Suppose that a horizontal wind blows in streamline flow with a speed of \(11.2 \mathrm{m} / \mathrm{s}\) outside a large pane of plate glass with dimensions \(4.00 \mathrm{m} \times 1.50 \mathrm{m}\) Assume the density of the air to be uniform at \(1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? (b) What If? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is \(22.4 \mathrm{m} / \mathrm{s}\), twice as high.

For the cellar of a new house, a hole is dug in the ground, with vertical sides going down \(2.40 \mathrm{m} .\) A concrete foundation wall is built all the way across the \(9.60-\mathrm{m}\) width of the excavation. This foundation wall is \(0.183 \mathrm{m}\) away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by \(2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times\) \(9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}\).

The small piston of a hydraulic lift has a cross-scrional area of \(3.00 \mathrm{cm}^{2},\) and its large piston has a cross-sectional area of \(\left.200 \mathrm{cm}^{2} \text { (Figure } 14.4\right) .\) What force must be applied to the small piston for the lift to raise a load of \(15.0 \mathrm{kN} ?\) (In service stations, this force is usually exerted by compressed air.)

The weight of a rectangular block of low-density material is 15.0 N. With a thin string, the center of the horizontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When \(25.0 \%\) of the block's volume is submerged, the tension in the string is \(10.0 \mathrm{N}\) (a) Sketch a free-body diagram for the block, showing all forces acting on it. (b) Find the buoyant force on the block. (c) Oil of density \(800 \mathrm{kg} / \mathrm{m}^{3}\) is now steadily added to the beaker, forming a layer above the water and surrounding the block. The oil exerts forces on each of the four side walls of the block that the oil touches. What are the directions of these forces? (d) What happens to the string tension as the oil is added? Explain how the oil has this Effect on the string tension. (c) The string breaks when its tension reaches \(60.0 \mathrm{N}\). At this moment, \(25.0 \%\) of the block's volume is still below the water line; what additional fraction of the block's volume is below the top surface of the oil? (f) After the string breaks, the block comes to a new equilibrium position in the beaker. It is now in contact only with the oil. What fraction of the block's volume is submerged?

A hypodermic syringe contains a medicine with the density of water (Figure \(\mathbf{P} 1 \mathbf{1} . \mathbf{5 3}\) ). The barrel of the syringe has a cross sectional area \(A=2.50 \times 10^{-5} \mathrm{m}^{2},\) and the needle has a cross-sectional area \(a=1.00 \times 10^{-8} \mathrm{m}^{2} .\) In the absence of a force on the plunger, the pressure everywhere is 1 atm. \(A\) force \(\mathbf{F}\) of magnitude \(2.00 \mathrm{N}\) acts on the plunger, making medicine squirt horizontally from the needle. Determine the speed of the medicine as it leaves the needle's tip.
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