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Chapter 11: Problem 14

Interactive Solution \(\underline{11.14}\) at presents a model for solving this problem. A solid concrete block weighs \(169 \mathrm{~N}\) and is resting on the ground. Its dimensions are \(0.400 \mathrm{~m} \times 0.200 \mathrm{~m} \times 0.100 \mathrm{~m}\). A number of identical blocks are stacked on top of this one. What is the smallest number of whole blocks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block?

Short Answer

Expert verified
The minimum number of blocks needed is 96.

Step by step solution

01

Understand the Problem

A concrete block is resting on the ground, and additional blocks are stacked on top. We need to find the smallest number of blocks required so that the weight causes a pressure of at least two atmospheres on the ground. One atmosphere is equivalent to a pressure of 101,325 Pa.
02

Calculate Surface Area of Base Block

The base dimensions of the block are given as 0.400 m x 0.200 m. The surface area in contact with the ground, which is the base, is calculated as: \[ A = 0.400 \, \text{m} \times 0.200 \, \text{m} = 0.080 \, \text{m}^2 \]
03

Determine Required Pressure

We want the pressure to be at least two atmospheres. First, convert 2 atm to Pascals:\[ 2 \, \text{atm} = 2 \times 101,325 \, \text{Pa} = 202,650 \, \text{Pa} \]
04

Use Pressure Formula to Calculate Total Required Weight

Pressure is defined as force per unit area: \[ P = \frac{F}{A} \] Solving for total force weight, \( F \):\[ F = P \times A \] Substitute the known values:\[ F = 202,650 \, \text{Pa} \times 0.080 \, \text{m}^2 = 16,212 \, \text{N} \]
05

Calculate Number of Blocks Needed

Each block weighs 169 N. To find the minimum number of blocks needed to have a total weight of at least 16,212 N: \[ n \times 169 \, \text{N} \geq 16,212 \, \text{N} \] Solving for \( n \): \[ n \geq \frac{16,212}{169} \approx 95.96 \] Since \( n \) must be a whole number, round up to the nearest whole number: \( n = 96 \).

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Key Concepts

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

Pressure Calculation
Pressure is the force exerted uniformly over a surface. It's crucial in understanding how forces distribute over areas. The formula for pressure is:
  • \( P = \frac{F}{A} \)
where \( P \) is pressure, \( F \) is the force applied, and \( A \) is the area. In this problem, we're given the task of stacking concrete blocks to create a minimum pressure of two atmospheres. Pressure helps us understand how the weight from the blocks will spread on the ground beneath the bottom block.
To solve, we need to know:
  • The area over which the weight is applied.
  • The weight needed to achieve the desired pressure.
Force and Weight
Force and weight are closely connected in physics. Weight is the force exerted by gravity on an object. It's calculated using:
  • \( W = m \times g \)
where \( W \) is weight, \( m \) is mass, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). In the exercise, each concrete block has a weight of 169 N. When calculating pressure, the total force (weight from all stacked blocks) is essential. Calculating the total force helps ensure the pressure condition is met.
Unit Conversion
Unit conversion involves changing one unit of measurement to another, which is necessary for consistent and correct calculations. In the given problem, pressure is initially provided in atmospheres. To use it in our calculations, we convert it to Pascals (Pa), the SI unit for pressure.
  • 1 atmosphere (atm) = 101,325 Pa
  • 2 atm = 2 脳 101,325 Pa = 202,650 Pa
This conversion allows us to compute the force required using the pressure formula in SI units, ensuring accurate results.
Surface Area Measurement
Surface area measurement is vital for pressure calculations, as it determines where the force is applied. Surface area is calculated from the dimensions given for a solid.
  • For the base of our block: \( 0.400 \, \text{m} \times 0.200 \, \text{m} = 0.080 \, \text{m}^2 \)
The base surface area is where all pressure from the blocks accumulates. Knowing this area allows us to calculate how much force is needed to achieve a specific pressure. In this exercise, understanding the surface area helps us determine how many blocks are required to meet the pressure goal.

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Most popular questions from this chapter

In the human body, blood vessels can dilate, or increase their radii, in response to various stimuli, so that the volume flow rate of the blood increases. Assume that the pressure at either end of a blood vessel, the length of the vessel, and the viscosity of the blood remain the same, and determine the factor \(R_{\text {dilated }} / R_{\text {normal }}\) by which the radius of a vessel must change in order to double the volume flow rate of the blood through the vessel.

Prairie dogs are burrowing rodents. They do not suffocate in their burrows, because the effect of air speed on pressure creates sufficient air circulation. The animals maintain a difference in the shapes of two entrances to the burrow, and because of this difference, the air \(\left(\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)\) blows past the openings at different speeds, as the drawing indicates. Assuming that the openings are at the same vertical level, find the difference in air pressure between the openings and indicate which way the air circulates.

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open at the top. Friction is absent. The spring constant of the spring is \(3600 \mathrm{~N} /\) \(\mathrm{m}\). The piston has a negligible mass and a radius of \(0.025 \mathrm{~m}\). (a) When air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress? (b) How much work does the atmospheric pressure do in compressing the spring?

A solid cylinder (radius \(=0.150 \mathrm{~m},\) height \(=0.120 \mathrm{~m}\) ) has a mass of \(7.00 \mathrm{~kg}\). This cylinder is floating in water. Then oil \(\left(\rho=725 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is poured on top of the water until the situation shown in the drawing results. How much of the height of the cylinder is in the oil?

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. It is contained inside a large "medicine dropper," into which the liquid is drawn using a squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). (a) Compared to the weight \(W\) of the tube, how much buoyant force is needed to make the tube float in either battery acid or antifreeze? (b) Is a greater volume of battery acid or a greater volume of antifreeze displaced by the hydrometer to provide the necessary buoyant force? (c) Which mark is farther up from the bottom of the tube? Justify your answers.
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