/*! 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 42 A \(2 \mathrm{~m} \times 3 \math... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2 \mathrm{~m} \times 3 \mathrm{~m}\) rectangular gate is located on the sloping side of a water reservoir such that the \(2-\mathrm{m}\) side of the gate is parallel to the water surface. The side of the reservoir (and the gate) slopes at an angle of \(60^{\circ}\) to the horizontal, and the top of the gate is \(2.5 \mathrm{~m}\) vertically below the water surface. Estimate the resultant hydrostatic force on the gate and the effective location of this resultant force, as measured vertically downward from the water surface.

Short Answer

Expert verified
The hydrostatic force is 206,010 N, located 3.5952 m below the water surface.

Step by step solution

01

Determine the Depth at the Center of Gravity

The center of gravity of the rectangular gate is the midpoint, located vertically below its top. Given the top of the gate is 2.5 m below the water surface and the height of the gate along the slope is 2 m, the center of gravity is at 3.5 m (2.5 m + \(\frac{2}{2}\) m) below the water surface.
02

Calculate the Hydrostatic Pressure at the Center of Gravity

The hydrostatic pressure at a depth is given by \( p = \rho g h \), where \( \rho \) is the density of water \(1000 \text{ kg/m}^3\), \( g \) is the acceleration due to gravity \(9.81 \text{ m/s}^2\), and \( h \) is the depth (3.5 m here). Thus, \( p = 1000 \times 9.81 \times 3.5 = 34335 \, \text{Pa} \).
03

Calculate the Hydrostatic Force on the Gate

The hydrostatic force \( F \) on the gate is the pressure at the center of gravity times the area of the gate. The area is \( 2 \text{ m} \times 3 \text{ m} = 6 \text{ m}^2 \). Thus, \( F = 34335 \times 6 = 206010 \, \text{N} \).
04

Calculate the Effective Depth of the Resultant Force

The effective depth \( y_R \) of the resultant force is given by \( y_R = h + \frac{I_g}{A y_c} \), where \( h \) is 3.5 m, \( I_g = \frac{b h^3}{12} = \frac{3 \times 2^3}{12} = 2 \,\text{m}^4 \), \( A = 6 \, \text{m}^2 \), and \( y_c = 3.5 \, \text{m} \). Thus, \( y_R = 3.5 + \frac{2}{6 \times 3.5} = 3.5 + 0.0952 = 3.5952 \, \text{m} \).

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Key Concepts

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

Hydrostatic Force
Hydrostatic force is essential in understanding how fluids exert pressure on submerged surfaces. In this context, we focus on how water exerts force on a submerged rectangular gate. Hydrostatic force is calculated by multiplying the hydrostatic pressure by the area of the surface. Here, the area is straightforward, as it is a rectangle: width times height. The pressure acts uniformly across the area at the depth of the center of gravity.
  • The hydrostatic pressure is contingent on the depth – the deeper the object, the higher the pressure.
  • The direction of the hydrostatic force is always perpendicular to the surface it acts upon.
For our rectangular gate, after determining the hydrostatic pressure at the center of gravity, the force on the gate was computed as 206,010 Newtons. Understanding this helps us design and analyze structures in contact with water.
Center of Gravity
The center of gravity is a pivotal concept, especially when dealing with submerged surfaces. For an object submerged in a fluid, its center of gravity is critical as it is the point where the weight of the body acts. It's crucial to accurately find this point to calculate the force and its effective location. In the case of the rectangular gate, the center of gravity is located at the midpoint, taking into consideration both the dimensions of the gate and its depth below the fluid surface.
  • The midpoint of the 2-meter side is at the depth of 3.5 meters from the water surface.
  • This depth influences the calculation of pressure acting at the center of gravity.
Locating the center of gravity accurately is vital to ensure that calculations related to forces and balance are correct. It simplifies complex systems and allows engineers to predict potential behaviors of structures.
Pressure Calculation
Pressure calculation in fluids is a cornerstone of hydrostatics. The formula used is derived from basic physics principles, and it ties the depth, density of the fluid, and acceleration due to gravity. This makes it a universally applicable concept.The pressure exerted by a fluid at a depth can be calculated using the equation: \( p = \rho g h \). In our scenario:
  • \( \rho \) is the density of water, typically around 1000 kg/m³.
  • \( g \), acceleration due to gravity, is approximately 9.81 m/s².
  • \( h \) is the effective depth of the surface from the fluid level, here 3.5 m to the center of gravity.
With these values, the pressure is calculated as 34,335 Pa (Pascal). Understanding this process empowers you to assess forces exerted by fluids in multiple contexts, not just with gates, but with any submerged surfaces.
Effective Depth
Effective depth is a critical concept in determining where the resultant hydrostatic force acts. While the center of gravity gives one part of this equation, the effective depth also takes into account the properties of the surface area and its moment of inertia.The formula for effective depth, \( y_R = h + \frac{I_g}{A y_c} \), reflects this:
  • \( h \) is the depth at the center of gravity.
  • \( A \) is the area of the gate.
  • \( I_g \) or moment of inertia helps describe how the area is distributed around the axis.
  • \( y_c \) stands for the center of gravity.
By these calculations, it was found as 3.5952 m. This parameter not only affects force magnitude but also impacts design considerations.

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Most popular questions from this chapter

Find the pressure head in millimeters of mercury (Hg) equivalent to \(80 \mathrm{~mm}\) of water plus \(60 \mathrm{~mm}\) of a fluid whose specific gravity is \(2.90 .\) The specific weight of mercury can be taken as \(133 \mathrm{kN} / \mathrm{m}^{3}\). Assume a temperature of \(20^{\circ} \mathrm{C}\).

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