/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 A vertical rectangular gate of h... [FREE SOLUTION] | 91Ó°ÊÓ

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A vertical rectangular gate of height \(h\) and width \(w\) is installed such that the top of the gate is a distance \(d\) below the water surface in a reservoir. Calculate the hydrostatic force on the gate and show that the location of the center of pressure is given by $$ y_{\mathrm{cp}}=d+\frac{L(3 d+2 h)}{6 d+3 h} $$

Short Answer

Expert verified
The hydrostatic force is \(F = \rho g w h (d + \frac{h}{2})\) and the center of pressure is \(y_{cp} = d + \frac{h(3d + 2h)}{6d + 3h}\).

Step by step solution

01

Understand the Problem

We need to calculate the hydrostatic force on a vertical rectangular gate submerged in a fluid, and also determine the center of pressure of this force. The gate has height \(h\), width \(w\), and its top edge is situated \(d\) below the water surface.
02

Define Hydrostatic Force Formula

The hydrostatic force \(F\) on a submerged vertical surface is given by \(F = \rho g A h_c\), where \(\rho\) is the fluid density, \(g\) is gravitational acceleration, \(A\) is the area, and \(h_c\) is the depth of the centroid.
03

Calculate the Area and Centroid Depth

The area \(A\) of the rectangle is \(A = h \times w\). The centroid \(h_c\) of a vertically submerged rectangle is at depth \(d + \frac{h}{2}\) from the surface.
04

Compute the Hydrostatic Force

Substitute the area and centroid depth into the force formula: \( F = \rho g (h \times w) \left(d + \frac{h}{2}\right) \).
05

Define the Center of Pressure Formula

The center of pressure \(y_{cp}\) for a vertical surface is given by \(y_{cp} = d + I_g/(A \cdot h_c)\), where \(I_g = \frac{w h^3}{12}\) is the second moment of area about the horizontal axis through the centroid.
06

Substitute into Center of Pressure Equation

Substitute \(I_g = \frac{w h^3}{12}\), \(A = h \times w\), and \(h_c = d + \frac{h}{2}\) into the center of pressure equation: \[y_{cp} = d + \frac{\frac{w h^3}{12}}{h \times w \cdot (d + \frac{h}{2})}\].
07

Simplify the Center of Pressure Equation

Simplify to find \(y_{cp}\): \[y_{cp} = d + \frac{h^3}{12h(d + \frac{h}{2})} = d + \frac{h^2}{12(d + \frac{h}{2})}\].Simplifying further, we achieve: \[y_{cp} = d + \frac{h(3d + 2h)}{6d + 3h}\].

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Key Concepts

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

Center of Pressure
The center of pressure is a crucial concept in fluid mechanics that helps us understand where the resultant force of a fluid acts on a submerged object. Unlike the centroid, which is the geometric center, the center of pressure is influenced by the pressure distribution over the submerged surface.
In a vertical rectangular gate submerged in fluid, the pressure increases linearly with depth, which shifts the center of pressure below the centroid. To find the specific location of this center, we use the formula:
\[ y_{cp} = d + \frac{I_g}{A \cdot h_c} \]
where:
  • \( y_{cp} \) is the center of pressure
  • \( I_g \) is the second moment of area about the horizontal axis through the centroid, given as \( \frac{wh^3}{12} \)
  • \( A \) is the area of the gate
  • \( h_c \) is the depth of the centroid
This calculation reveals that the center of pressure depends on both the geometry of the gate and the depth at which it is submerged, allowing engineers to design structures that can withstand fluid forces effectively.
Fluid Mechanics
Fluid mechanics is the study of how fluids behave both at rest and in motion. It covers both liquids, like water, and gases, like air. This branch of physics deals with the forces and the mechanics involved, providing the foundation for understanding hydrostatic forces.
In the context of the vertical rectangular gate submerged in water, fluid mechanics principles help us understand how water exerts force. As the gate is submerged deeper, the force exerted by the fluid on the gate increases due to the increasing pressure, making the understanding of fluid statics essential.
Fluid mechanics can be divided into fluid statics, which concerns fluids at rest, and fluid dynamics, which concerns fluids in motion. For problems like calculating the force on a submerged gate, fluid statics is the key area of study, as it deals with pressure in free-standing fluids.
Hydrostatic Pressure
Hydrostatic pressure is a fundamental concept in fluid mechanics, defining the pressure exerted by a fluid at equilibrium due to the force of gravity. This pressure increases with depth due to the weight of the fluid above.
The formula for hydrostatic pressure \( p \) at a depth \( h \) is:
\[ p = \rho g h \]
where:
  • \( \rho \) is the fluid's density
  • \( g \) is the acceleration due to gravity
  • \( h \) is the depth below the free surface
This concept is crucial when determining the force on submerged surfaces, like our vertical rectangular gate. The hydrostatic force \( F \) on the surface can be computed using:
\[ F = \rho g A h_c \]
where \( A \) is the area of the surface and \( h_c \) is the depth to the centroid. This shows that as depth increases, so does the hydrostatic force, affecting how structures withstand underwater pressures.
Vertical Rectangular Gate
The vertical rectangular gate is a common structure in fluid mechanics problems and applications. These gates are often used to control the flow of water in reservoirs, dams, and other water management systems.
Key characteristics include:
  • The width \( w \) and the height \( h \) of the gate, which together define its area \( A = h \times w \).
  • The positioning of the gate with its top edge \( d \) below the water surface, which is critical for calculating the hydrostatic force impact.
Configuring vertical gates correctly ensures the stability of water control systems. The position and dimensions of these gates directly influence the pressure distribution from water, necessitating careful calculation of the forces involved. Understanding the dynamics of these forces with the center of pressure offers assurance in the design process, making the practical considerations of fluid mechanics more predictable and manageable.

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Most popular questions from this chapter

(a) A large tank contains water at \(20^{\circ} \mathrm{C},\) and the absolute pressure \(12 \mathrm{~m}\) below the surface of the water is measured as \(200 \mathrm{kPa}\). Estimate the atmospheric pressure above the tank. (b) If the water in the tank was replaced by a liquid with a specific gravity of \(0.85,\) what absolute pressure and what gauge pressure would be measured \(6 \mathrm{~m}\) below the surface of the liquid?

A container that is open to the atmosphere stores two immiscible liquids. The liquid on top has a thickness of \(7 \mathrm{~m}\) and specific weight of \(9 \mathrm{kN} / \mathrm{m}^{3},\) and the liquid on the bottom has a thickness of \(2.3 \mathrm{~m}\). If a transducer measures a (gauge) pressure of \(92 \mathrm{kPa}\) on the bottom of the tank, estimate the specific gravity of the liquid on the bottom. Must the liquid on the bottom necessarily be denser than the liquid on the top? Explain.

Water pressure at a pipeline junction measures \(450 \mathrm{kPa}\). What is the corresponding pressure head (expressed as a height of water)?

Two piston diameters are being considered for use in a hydraulic system: a 25-mmdiameter piston and a 100 -mm-diameter piston. If an applied force of \(500 \mathrm{~N}\) to the 25 -mm piston is found to be satisfactory, what force on the 100 -mm piston at the same location would be required so as not to compromise the performance of the hydraulic system?

A water truck is mounted with a cylindrical tank that has a diameter of \(2 \mathrm{~m}\) and a length of \(10 \mathrm{~m}\). The long axis of the tank is oriented with the direction of truck motion. If the tank is filled with water at \(20^{\circ} \mathrm{C}\) and the truck accelerates at a rate of \(2 \mathrm{~m} / \mathrm{s}^{2}\), estimate the difference in magnitude between the resultant hydrostatic force on the front and back of the tank.

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