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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

An airplane is moving through the air at a velocity \(v=200 . \mathrm{m} / \mathrm{s} .\) Streamlines just over the top of the wing are compressed to \(80.0 \%\) of their original area, and those under the wing are not compressed at all. a) Determine the velocity of the air just over the wing. b) Find the difference in the pressure between the air just over the wing, \(P\), and that under the wing, \(P\). c) Find the net upward force on both wings due to the pressure difference, if the area of the wing is \(40.0 \mathrm{~m}^{2}\) and the density of the air is \(1.30 \mathrm{~kg} / \mathrm{m}^{3}\).

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