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Chapter 13: Problem 37

A square pool with \(100 .-\mathrm{m}\) -long sides is created in a concrete parking lot. The walls are concrete \(50.0 \mathrm{~cm}\) thick and have a density of \(2.50 \mathrm{~g} / \mathrm{cm}^{3}\). The coefficient of static friction between the walls and the parking lot is \(0.450 .\) What is the maximum possible depth of the pool?

Short Answer

Expert verified
Answer: The maximum possible depth of the pool before sliding is 9.13 cm.

Step by step solution

01

Find Volume and Weight of Water in the Pool

Inorder to find the volume of the water, multiply the side length (\(100 m\)) squared by the depth (\(h\)): \(V_{water} = (100 * 100 * h) = 10000h \ m^3\) Assuming the density of water is \(1000 \ kg/m^3\), we can find the weight of the water by multiplying the mass of water by gravitational acceleration (\(g\)): \(W_{water} = (1000 \ kg/m^3 * 10000h \ m^3) * 9.81 \ m/s^2 = 98100000h \ N\)
02

Find Volume and Weight of Pool Walls

First, find the volume of the pool walls: - Outer area of the pool: \((100 + 2 * 0.5)^2 = 101^2\) - Inner area of the pool: \(100^2\) - Area of the pool walls: \(101^2 - 100^2\) - Volume of the pool walls: \((101^2 - 100^2) * h\) The density of the pool walls is given as \(2500 \ kg/m^3\). Then, let's find the mass and weight of the pool walls: \(W_{walls} = (2500 * (101^2 - 100^2) * h) * 9.81 = 19902h \ N\)
03

Calculate Normal Force and Maximum Static Frictional Force

The normal force exerted by the pool walls on the parking lot is equal to the weight of the walls: \(N = W_{walls} = 19902h \ N\) The maximum static frictional force between the walls and parking lot is given by: \(F_{max} = \mu_{s} * N = 0.450 * 19902h = 8955.9h \ N\)
04

Balance Forces and Solve for Depth

To find the maximum depth, equate the weight of the water and the maximum static frictional force: \(98100000h = 8955.9h\) Divide both sides by \(h\) and solve for the maximum depth: \(h = \frac{8955.9}{98100000 - 8955.9} = 0.00009129 \ m = 9.13 \ cm\) Thus, the maximum possible depth of the pool before sliding is \(9.13 \ cm\).

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Most popular questions from this chapter

A sealed vertical cylinder of radius \(R\) and height \(h=0.60 \mathrm{~m}\) is initially filled halfway with water, and the upper half is filled with air. The air is initially at standard atmospheric pressure, \(p_{0}=1.01 \cdot 10^{5} \mathrm{~Pa}\). A small valve at the bottom of the cylinder is opened, and water flows out of the cylinder until the reduced pressure of the air in the upper part of the cylinder prevents any further water from escaping. By what distance is the depth of the water lowered? (Assume that the temperature of water and air do not change and that no air leaks into the cylinder.)

Calculate the ratio of the lifting powers of helium (He) gas and hydrogen (H \(_{2}\) ) gas under identical circumstances. Assume that the molar mass of air is \(29.5 \mathrm{~g} / \mathrm{mol}\).

Which of the following assumptions is not made in the derivation of Bernoulli's Equation? a) Streamlines do not cross. c) There is negligible friction. b) There is negligible d) There is no turbulence. viscosity. e) There is negligible gravity.

In many problems involving application of Newton's Second Law to the motion of solid objects, friction is neglected for the sake of making the solution easier. The counterpart of friction between solids is viscosity of liquids. Do problems involving fluid flow become simpler if viscosity is neglected? Explain.

A very large balloon with mass \(M=10.0 \mathrm{~kg}\) is inflated to a volume of \(20.0 \mathrm{~m}^{3}\) using a gas of density \(\rho_{\text {eas }}=\) \(0.20 \mathrm{~kg} / \mathrm{m}^{3}\). What is the maximum mass \(m\) that can be tied to the balloon using a \(2.00 \mathrm{~kg}\) piece of rope without the balloon falling to the ground? (Assume that the density of air is \(1.30 \mathrm{~kg} / \mathrm{m}^{3}\) and that the volume of the gas is equal to the volume of the inflated balloon).
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