/*! 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 11 Tor the cellar of a new house, a... [FREE SOLUTION] | 91Ó°ÊÓ

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Tor the cellar of a new house, a hole is dug in the ground, -with vertical sides going down \(2.40 \mathrm{m}\). A concrete foundamall 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 strect fills up the spacc 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 \(2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times\) \(9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}\)

Short Answer

Expert verified
The force the water causes on the foundation wall is 10.32 kN.

Step by step solution

01

Identify Variables

First, identify all the variables from the problem: the height of the water \(h = 2.40 m\), the width of the excavation \(w = 9.60 m\), the distance to the wall \(d = 0.183 m\), the density of water \(\rho = 1000 kg/m^3\) and gravitational acceleration \(g = 9.80 m/s^2\).
02

Calculate Pressure

Next, calculate the hydrostatic pressure at a depth \(h\) in a fluid. The formula for pressure in a fluid at depth is \(P = \rho gh\). Substitute the given values into the formula to find the pressure: \(P = 1000 kg/m^3 * 9.81 m/s^2 * 2.40 m = 23544 Pa = 23.5 kPA\).
03

Calculate Force

Finally, calculate the total force exerted by the water on the wall. Force is pressure times area \(F = PA\). The area of the wall in contact with the water is given by \(A = hd = 2.40 m * 0.183 m = 0.4392 m^2\). Substitute the values for pressure and area into the force formula to find the force: \(F = 23.5 kPa * 0.4392 m^2 = 10.32 kN\).
04

Compare Forces

Compare the force the water exerts on the wall to the weight of the water. The weight of the water was given as \(41.3 kN\). Therefore, the force caused by the water on the wall is about a quarter of the weight of the water.

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Key Concepts

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

Fluid Mechanics
Fluid mechanics is an exciting branch of physics focusing on the behavior and properties of fluids, which are liquids and gases. The study of fluid mechanics helps us understand how these substances move and exert forces in different conditions.
A fluid is any substance that can flow, meaning it can change its shape depending on the container it's in or the forces applied to it.
An important aspect of fluid mechanics is understanding how pressure develops in a fluid and how it affects surfaces in contact with the fluid. When we talk about fluids, particularly liquids like water, there are two main types of pressure to consider:
  • Static pressure: This is the pressure a fluid exerts when it is not moving. An example is the pressure a puddle of water might exert against the sides of its container.
  • Dynamic pressure: This occurs when the fluid is in motion, and it's in addition to the static pressure.
In the exercise, the hydrostatic pressure, which is a type of static pressure, is a key idea. This is the pressure that a column of water or any fluid exerts due to the force of gravity acting on it. Knowing how to calculate and work with these pressures is essential in fields involving hydraulics, engineering, and even meteorology.
Hydrostatic Force
Hydrostatic force is the force exerted by a fluid at rest, due to the fluid's weight, on any surface in contact with it. In our exercise, this concept is crucial in determining how the water's weight affects the concrete wall.
Hydrostatic force is computed using the formula \( F = PA \), where \(P\) is the pressure at a certain depth, and \(A\) is the area over which this pressure acts. The pressure at a depth \( h \) in a liquid is determined by the equation \( P = \rho gh \), where \( \rho \) is the fluid density, \( g \) is the gravitational acceleration, and \( h \) is the height of the fluid column.
Let's break it down:
  • Density (\( \rho \)): For water, this is typically \( 1000 \, \text{kg/m}^3\).
  • Gravity (\( g \)): On Earth, this constant is approximately \( 9.8 \, \text{m/s}^2\).
  • Height (\( h \)): In a practical problem, it's simply the depth of the fluid above the surface.
These concepts converge to describe how fluids exert forces, and enable engineers to design structures capable of withstanding such forces, ensuring safety and stability.
Physics Problem Solving
When solving physics problems, especially in fluid mechanics, breaking the problem into systematic steps is essential. This exercise demonstrates how to approach such a problem efficiently and logically.
Here's a simplified approach to problem solving:
  • **Identify the Variables:** Start by recognizing all the known quantities given in the problem. This includes physical properties like height, distance, density, and gravity.
  • **Develop the Equations:** Use physics concepts and known formulas to develop equations involving the identified variables. Here, we calculated pressure and force using the hydrostatic equations.
  • **Calculation and Interpretation:** Perform the calculations to find the required information. Ensure the units make sense and the values are reasonable. Finally, interpret these values to check if the answers fit the problem's context.
By following a structured approach, you simplify complex problems and understand the results better. This helps in applying the learned concepts to diverse real-world scenarios, and reinforces a robust understanding of physics.

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Most popular questions from this chapter

(a) Calculate the absolute pressure at an ocean depth of \(1000 \mathrm{m} .\) Assume the density of seawater is \(1024 \mathrm{kg} / \mathrm{m}^{3}\) and that the air above exerts a pressure of \(101.3 \mathrm{kPa}\) (b) At this depth, what force must the frame around a circular submarine porthole having a diameter of \(30.0 \mathrm{cm}\) exert to counterbalance the force exerted by the water?

A large storage tank, open at the top and filled with water, develops a small hole in its side at a point \(16.0 \mathrm{m}\) below the water level. If the rate of flow from the leak is equal to \(2.50 \times 10^{-3} \mathrm{m}^{3} / \mathrm{min},\) determine (a) the specd at which the water leaves the hole and (b) the diameter of the hole.

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 United States possesses the eight largest warships in the world-aircraft carriers of the Nimilz class-and is building two more. Suppose one of the ships bobs up to float \(11.0 \mathrm{cm}\) higher in the water when 50 fighters take off from it in 25 min, at a location where the free-fall acceleration is \(9.78 \mathrm{m} / \mathrm{s}^{2} .\) Bristling with bombs and missiles, the planes have average mass 29 000 kg. Find the horizontal area enclosed by the waterline of the S1-billion ship. By comparison, its flight deck has area 18 000 \(\mathrm{m}^{2}\). Below decks are passageways hundreds of meters long, so narrow that two large men cannot pass each other.

In \(1983,\) the United States began coining the cent piece out of copper-clad rinc rather than pure copper. The mass of the old copper penny is \(3.083 \mathrm{g},\) while that of the new cent is 2.517 g. Calculate the percentage of sinc (by volume) in the new cent. The density of copper is \(8.960 \mathrm{g} / \mathrm{cm}^{3}\) and that of zinc is \(7.133 \mathrm{g} / \mathrm{cm}^{3} .\) The new and old coins have the same volume.

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