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Chapter 2: Problem 13

A square aperture in the vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are \(2 \mathrm{~m}\) long and the tank contains a liquid of relative density 1.15. The centre of the aperture is \(1.5 \mathrm{~m}\) below the free surface. Calculate the net hydrostatic thrust on the plate, the moment of this thrust about the hinge and the position of the centre of pressure.

Short Answer

Expert verified
Thrust is 33855 N. Moment of thrust is 54536.1 Nm. Center of pressure is 1.611 m below the surface.

Step by step solution

01

- Calculate the Area of the Aperture

The aperture is a square with a diagonal of length 2 meters. We need to find the side length of the square. Using the relationship for the diagonal of a square, which is side length times \sqrt{2}, we have: \[ \text{Diagonal} = \text{side} \times \sqrt{2} = 2 \Rightarrow \text{side} = \frac{2}{\sqrt{2}} = \sqrt{2} \mathrm{m} \] Therefore, the area (A) of the square is: \[ A = (\sqrt{2})^2 = 2 \mathrm{m}^2 \]
02

- Determine the Hydrostatic Pressure at the Center

The center of the aperture is 1.5 meters below the free surface. Hydrostatic pressure (P) is given by: \[ P = \rho g h \] where \(\rho\) is the density of the liquid, \(\rho = 1.15 \times 1000 \mathrm{kg/m}^3\), \(g\) is the acceleration due to gravity, and \(h\) is the depth. Thus, \[ P = 1.15 \times 1000 \mathrm{kg/m}^3 \times 9.81 \mathrm{m/s}^2 \times 1.5 \mathrm{m} = 16927.5 \mathrm{Pa} \]
03

- Calculate the Hydrostatic Thrust on the Plate

The net hydrostatic thrust (F) on the plate is the product of the pressure at the center and the area of the aperture: \[ F = P \times A = 16927.5 \mathrm{Pa} \times 2 \mathrm{m}^2 = 33855 \mathrm{N} \]
04

- Find the Moment of Thrust about the Hinge

The center of the aperture is 1.5 meters below the surface, and the center of pressure for a vertically submerged plane surface is given by: \[ y_{cp} = \bar{y} + \frac{I_{G}}{A\bar{y}} \] where \(\bar{y}\) is the distance from the surface to the center of the plate (1.5 meters), \(I_{G}\) is the second moment of area about the centroid: \[ I_{G} = \frac{(\text{side})^4}{12} = \frac{(\sqrt{2})^4}{12} = \frac{4}{12} = \frac{1}{3} \mathrm{m}^4 \] Therefore, \[ y_{cp} = 1.5 + \frac{\frac{1}{3}}{2 \times 1.5} = 1.5 + \frac{1}{9} = 1.611 \mathrm{m} \] The moment of thrust (M) about the hinge is: \[ M = F \times y_{cp} = 33855 \mathrm{N} \times 1.611 \mathrm{m} = 54536.1 \mathrm{Nm} \]

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Key Concepts

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

fluid mechanics
Fluid mechanics deals with the behavior of fluids (liquids and gases) at rest and in motion. This field includes the study of how forces interact with fluids, which is crucial for applications in engineering and physical sciences. Understanding fluid mechanics helps us calculate properties like pressure, velocity, and flow of fluids.
In this exercise, we are dealing with a liquid in a tank, examining hydrostatic pressure, thrust, and center of pressure. Although the liquid is at rest, the field of fluid mechanics still applies, as it provides the principles to determine forces and moments in static fluids.
hydrostatic thrust
Hydrostatic thrust is the force exerted by a fluid at rest on a submerged surface. This force depends on the fluid's density, the gravitational acceleration, the depth of the fluid, and the area of the surface.
In the given problem, we calculate the hydrostatic thrust on a square aperture submerged in a tank. The thrust (\text{F}) is computed by multiplying the hydrostatic pressure at the center of the aperture with the area of the aperture:
\[ F = P \times A \]
This yields the units of force (Newtons, N) and quantifies the pushing force the fluid exerts on the submerged surface.
center of pressure
The center of pressure is the point on a submerged surface where the total hydrostatic force acts. It is essential for understanding how to balance and support structures submerged in fluids.
In our problem, the calculation of the center of pressure (\(y_{cp}\)) includes finding the second moment of area (\(I_{G}\)) about the centroid of the submerged surface and its distance from the surface of the fluid.
The formula is:
\[ y_{cp} = \bar{y} + \frac{I_{G}}{A\bar{y}} \]
Here, \(\bar{y}\) is the centroid depth, \(I_{G}\) is the second moment of area, and \(A\) is the area. The calculated center of pressure helps determine where the hydrostatic force effectively acts upon the plate.
hydrostatic equations
Hydrostatic equations describe the relationship between fluid properties (like pressure) and their depth, density, and gravitational pull when at rest. These equations are vital for understanding pressure variations in fluids.
In this exercise, the hydrostatic pressure (\(P\)) is computed using the equation:
\[ P = \rho g h \]
Here, \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(h\) is the depth below the fluid's surface. This equation helps determine the pressure at any given point in the fluid, allowing us to compute hydrostatic forces and thrusts on submerged surfaces.
The application of hydrostatic equations ensures accurate calculations of forces in fluid mechanics, crucial for designing and analyzing submerged structures.

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Most popular questions from this chapter

A solid uniform cylinder of length \(150 \mathrm{~mm}\) and diameter \(75 \mathrm{~mm}\) is to float upright in water. Between what limits must its mass be?

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