Question

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

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

Step-by-step solution

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 (\(g\)): \(W_{water} = (1000 \ kg/m^3 * 10000h \ m^3) * 9.81 \ m/s^2 = 98100000h \ N\)

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: \(W_{walls} = (2500 * (101^2 - 100^2) * h) * 9.81 = 19902h \ N\)

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: \(F_{max} = \mu_{s} * N = 0.450 * 19902h = 8955.9h \ N\)

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 \(9.13 \ cm\).

Key concepts

Static Friction

Static friction is the force that prevents two surfaces from sliding past each other. It acts opposite to the direction of potential motion and needs to be overcome for motion to start. The static frictional force depends on two main factors: the normal force (the perpendicular force between the surfaces) and the coefficient of static friction, which is a measure of how "sticky" the surfaces are.

For this problem, the walls exert a certain normal force on the parking lot, determined by their weight. This weight produces a normal force, which when multiplied by the coefficient of static friction (0.450), gives the maximum static frictional force. This value indicates the highest force that the wall can withstand before it starts sliding due to the pool's water weight. If the frictional force is not exceeded, the walls remain in place and the pool is stable.

Density

Density is a measure of how much mass is contained in a given volume. It is typically expressed in units of grams per cubic centimeter ( ext{g/cm}^3) or kilograms per cubic meter ( ext{kg/m}^3). In this problem, the density of the concrete walls is crucial for calculating their weight, which influences the normal and static frictional forces.

The density of a substance is calculated as:

• Density = \( \frac{mass}{volume} \)

Given that the density of the pool walls is 2.50 ext{g/cm}^3, it informs us how heavy a given volume of concrete is. By converting this density to 2500 ext{kg/m}^3, it allows us to easily find the walls' mass in terms of cubic meters, which is necessary for the weight calculations. Understanding density helps in comprehending how much material is present in a given space, impacting stability in structures like the pool.

Gravitational Force

Gravitational force acts on all objects with mass, pulling them towards the center of the Earth. In physics problems, gravitational force is a crucial factor as it determines how much weight an object exerts. This weight is calculated by multiplying the object's mass by the gravitational acceleration ( \( g = 9.81 ext{ m/s}^2 \)).

In the problem involving the pool, the gravitational force affects both the water within the pool and the concrete walls. The weight of the water can be expressed as:- Water weight = Density of water × Volume of water × Gravitational acceleration- For the walls: Weight of walls = Density of walls × Volume of walls × Gravitational acceleration

Understanding gravitational force helps us perceive how heavy an object truly is in a given gravitational field, affecting the static friction calculations necessary for determining the stability of the pool structure.

Volume Calculation

Volume calculation involves determining the amount of three-dimensional space an object occupies. It's essential for figuring out both the water capacity of the pool and the volume of the walls, which subsequently affects their mass and weight calculations.

For the pool's water volume:

• The formula is \( V_{water} = ext{length}^2 \times ext{depth} \). Here, substituting known values gives \( V_{water} = 100^2 \times h \) leading to \( V_{water} = 10000h ext{ m}^3 \).
• The weight of this water contributes to the total gravitational force acting downward.

To calculate the volume of the pool walls:

• First find the outer dimension area \( (101^2) \) and subtract the inner area \( (100^2) \) to get the wall area.
• Then multiply by their thickness to find the volume.
• This contributes to the normal force acting on the ground.

These volumes are crucial for determining how much mass and resulting forces act on different components, influencing stability analysis in the given scenario.