/*! 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 53 Force on a window A diving pool ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 53

Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows. The window is a square, 0.5 m on a side, with the lower edge of the window 1 m from the bottom of the pool.

Short Answer

Expert verified
Answer: The force acting on the square window is approximately 1531.25 N.

Step by step solution

01

Understand the given problem.

We have a square window with side 0.5 m that is located 1m above the bottom of a 4 m deep diving pool. We need to determine the force exerted on the window by the water due to its depth in the pool.
02

Calculate water pressure at a given depth.

To calculate the water pressure at a given depth, we can use the formula P = 蟻gh, where P is the pressure, 蟻 is the density of water (1000 kg/m^3), g is the acceleration due to gravity (9.81 m/s^2), and h is the depth of the water.
03

Integrate the water pressure across the area of the window.

Since the window is a square, the pressure won't be uniform across it because pressure varies with depth. We need to integrate the pressure variation across the entire area to get the force acting on the window. Let x be the horizontal position from the right edge of the window, and y be the vertical position from the bottom edge of the window. Then the differential force exerted on a small section of the window with area dA = dy dx will be: dF = P dA = 蟻g(y + 1) dy dx Now, we'll integrate the force for x values ranging from 0 to 0.5 m and y values ranging from 0 to 0.5 m. Force F = int(蟻g(y + 1) dy dx, x=0..0.5, y=0..0.5)
04

Perform the integration to find the force.

Start by integrating in the y-direction first. int(蟻g(y + 1) dy, y=0..0.5) = [蟻g((y^2 / 2) + y)] from 0 to 0.5 Substituting the values, we have: = [蟻g((0.25^2 / 2) + 0.25)] - [蟻g((0^2 / 2) + 0)] = 蟻g(0.3125) Now, integrate in the x-direction: int(蟻g(0.3125) dx, x=0..0.5) = [蟻g(0.3125)x] from 0 to 0.5 = 蟻g(0.3125)(0.5) - 蟻g(0.3125)(0) = 蟻g(0.3125)(0.5)
05

Calculate the force.

Now, we can compute the force acting on the window using the given values for density and gravity: Force F = 蟻g(0.3125)(0.5) = 1000 kg/m^3 脳 9.81 m/s^2 脳 0.3125 脳 0.5 = 1531.25 N The force acting on the square window is approximately 1531.25 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, which deals with the pressure exerted by a fluid at rest. It is particularly important when analyzing the forces on submerged surfaces, such as the window of a diving pool in our exercise. The formula for hydrostatic pressure (\( P \) is given by \( P = \rho gh \), where \( \rho \) represents the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid above the point in question.

Understanding hydrostatic pressure allows us to calculate the force applied by the water on the submerged window. Since pressure increases with depth, the bottom of the window experiences a higher pressure than the top, leading to a non-uniform distribution of force across the window's surface. This variation in pressure must be taken into account to accurately determine the total force exerted on the window.
Integration in Physics
Integration in physics is a powerful mathematical tool used to sum up small quantities over a continuous range, which is essential in dealing with problems involving variable forces. In our exercise, the pressure on the pool window varies with depth, making it necessary to integrate over the entire window area to determine the total force.

To achieve this, we divide the window into small differential sections, with an infinitesimally small force \( dF \) acting on each section. The force on each section is the product of the pressure at that section and the small area \( dA \) (\( dF = P dA \). We then perform a double integration over both dimensions of the window (\( x \) and \( y \) axes) to sum the forces over the entire window area. The result of this integration is the total force exerted by the water on the window.
Fluid Statics
Fluid statics, or hydrostatics, is the branch of physics that studies fluids at rest. A central theme in fluid statics is understanding how forces like pressure are distributed within a fluid. As we've seen in the problem, the distribution of pressure due to the weight of the fluid creates a force on submerged surfaces.

In the context of a submerged window in a pool, fluid statics tells us that the pressure increases with the depth of the fluid. This is vital to calculating the force on the submerged window because the pressure is not the same at all points. It's higher at greater depths, which means the lower part of the window bears a larger part of the force. Thus, fluid statics principles are key to determining the overall force on any submerged object by taking into account the depth-dependent hydrostatic pressure.

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Most popular questions from this chapter

Consider the upper half of the astroid described by \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3},\) where \(a>0\) and \(|x| \leq a\) Find the area of the surface generated when this curve is revolved about the \(x\) -axis. Note that the function describing the curve is not differentiable at \(0 .\) However, the surface area integral can be evaluated using symmetry and methods you know.

Mass of one-dimensional objects Find the mass of the following thin bars with the given density function. $$\rho(x)=1+\sin x, \text { for } 0 \leq x \leq \pi$$

Shell method Use the shell method to find the volume of the following solids. The solid formed when a hole of radius 3 is drilled symmetrically along the axis of a right circular cone of radius 6 and height 9

Solids from integrals Sketch a solid of revolution whose volume by the disk method is given by the following integrals. Indicate the function that generates the solid. Solutions are not unique. $$\text { a. } \int_{0}^{\pi} \pi \sin ^{2} x d x$$ $$\text { b. } \int_{0}^{2} \pi\left(x^{2}+2 x+1\right) d x$$

Work in a gravitational field For large distances from the surface of Earth, the gravitational force is given by \(F(x)=\frac{G M m}{(x+R)^{2}}\) where \(G=6.7 \times 10^{-11} \mathrm{N} \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant. \(M=6 \times 10^{24} \mathrm{kg}\) is the mass of Earth, \(m\) is the mass of the object in the gravitational field, \(R=6.378 \times 10^{6} \mathrm{m}\) is the radius of Farth, and \(x \geq 0\) is the distance above the surface of Earth (in meters). a. How much work is required to launch a rocket with a mass of \(500 \mathrm{kg}\) in a vertical flight path to a height of 2500 km (from Earth's surface)? b. Find the work required to launch the rocket to a height of \(x\) kilometers, for \(x>0\) c. How much work is required to reach outer space \((x \rightarrow \infty) ?\) d. Equate the work in part (c) to the initial kinetic energy of the rocket. \(\frac{1}{2} m v^{2},\) to compute the escape velocity of the rocket.
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