/*! 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 26 The sole of a man's size-10 shoe... [FREE SOLUTION] | 91Ó°ÊÓ

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The sole of a man's size-10 shoe is around \(11.0\) in. by \(4.00\) in. Determine the gauge pressure under the feet of a 200-lb man standing upright. Give your answer in both lb/in. \(^{2}\) and Pa. [Hint: \(1.00 \mathrm{lb} / \mathrm{in}^{2}=6895 \mathrm{~Pa}\). Check your work using \(1.00 \mathrm{in} .^{2}=6.45 \times\) \(10^{-4} \mathrm{~m}^{2}\) and \(\left.1.00 \mathrm{lb}=4.448 \mathrm{~N} .\right]\)

Short Answer

Expert verified
The gauge pressure is approximately 2.27 lb/in² or 15,659 Pa.

Step by step solution

01

Calculate the Total Area of Both Shoes

Each shoe has an area of \((11.0 \text{ in} \times 4.00 \text{ in})\). Since there are two shoes, the total area \(A\) is:\[A = 2 \times (11.0 \text{ in} \times 4.00 \text{ in}) = 88 \text{ in}^2\]
02

Calculate the Gauge Pressure in lb/in²

The gauge pressure \(P\) for the 200-lb man is given by dividing the weight \(W\) by the total area \(A\). Thus:\[P = \frac{W}{A} = \frac{200 \text{ lb}}{88 \text{ in}^2} \approx 2.27 \text{ lb/in}^2\]
03

Convert Pressure to Pascals (Pa)

Using the conversion \(1 \text{ lb/in}^2 = 6895 \text{ Pa}\), convert the pressure to Pascal:\[P = 2.27 \text{ lb/in}^2 \times 6895 \text{ Pa/lb/in}^2 \approx 15,659 \text{ Pa}\]
04

Verify Conversion Using Alternate Units

First, convert inches squared to meters squared and pounds to Newtons:- \(1 \text{ in}^2 = 6.45 \times 10^{-4} \text{ m}^2\)- \(1 \text{ lb} = 4.448 \text{ N}\)The area in meters squared:\[88 \text{ in}^2 \times 6.45 \times 10^{-4} \text{ m}^2/ ext{in}^2 = 0.05676 \text{ m}^2\]The force in Newtons:\[200 \text{ lb} \times 4.448 \text{ N/lb} = 889.6 \text{ N}\]Pressure in Pascal:\[P = \frac{889.6 \text{ N}}{0.05676 \text{ m}^2} \approx 15,667.4 \text{ Pa}\]This verifies our previous conversion result.

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Key Concepts

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

Pressure Conversion
Converting pressure units is an essential skill in understanding physical phenomena because it enables us to express pressure values in different systems of measurement. In the context of this problem, we need to convert pressure given in pounds per square inch ( ext{lb/in}^2) to Pascals (Pa), which is the SI unit for pressure.

The exercise provides the hint that 1 ext{lb/in}^2 is equivalent to 6895 Pa. This conversion factor is crucial here. When you calculate the pressure exerted by the man on the ground as 2.27 ext{lb/in}^2, it must be translated into a form that's more widely used in scientific contexts.

To convert the pressure to Pascals, multiply the value in ext{lb/in}^2 by 6895. This is because each pound-force per square inch exerts 6895 Pascals of pressure onto a surface. It is also good practice to verify the result through different conversion routes, as shown in the exercise, to ensure accuracy.
SI Units
SI Units, short for the International System of Units, serve as the standardized metric system used globally. They ensure uniformity of measurements in science, industry, and commerce. In this exercise, understanding SI Units helped verify the gauge pressure calculation.

When converting pressure from ext{lb/in}^2 to Pascals, we use conversions not just for pressure but also for area and force:
  • 1 inch squared is equivalent to 6.45 x 10^{-4} square meters.
  • 1 pound-force equals 4.448 Newtons in the SI system.

The problem involves these conversions to ultimately express the pressure in Pascals. Let's break down why each conversion is essential:

  • Area conversion ensures that dimensions are expressed in meters squared, which is the SI unit for area. This step is necessary when calculating pressure in the SI system since pressure is force per unit area.
  • Force conversion is necessary since pressure is defined as force per unit area, and we need the force in Newtons to use it in the Pascals formula correctly.

By converting all relevant measurements into SI units, we make the pressure calculation compatible with international standards.
Footwear Analysis
Analyzing the pressure exerted by footwear can provide insights into design considerations for comfort and safety. In the given exercise, the analysis determines how much force a person’s footwear applies to the surface area they are standing on.

The exercise focuses on a specific shoe size, and the calculated pressure shows how each square inch of the shoe’s sole bears the man's weight. Here are some aspects analyzed:
  • Total Contact Area: By calculating the total contact area under the shoes, we assess the distribution of weight. Larger soles generally mean a greater area and less pressure, assuming weight remains constant. This is why some athletic and work shoes opt for broader soles.
  • Pressure Distribution: The pressure calculated here is a uniform distribution, which means that ideally, pressure should spread evenly across the entire shoe sole. Uneven pressure can lead to discomfort and even injury over time.

Ideally, understanding and analyzing these factors can help shoemakers design footwear that minimizes excessive pressure points, contributes to better wear comfort, and enhances the overall stability of footwear.

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Most popular questions from this chapter

A wooden cylinder has a mass \(m\) and a base area \(A\). It floats in water with its axis vertical. Show that the cylinder undergoes SHM if given a small vertical displacement. Find the frequency of its motion. When the cylinder is pushed down a distance \(y\), it displaces an additional volume Ay of water. Because this additional displaced volume has mass \(A y_{\rho w}\), an additional buoyant force \(A y_{\rho w g}\) acts on the cylinder, where \(\rho_{w}\) is the density of water. This is an unbalanced force on the cylinder and is a restoring force. In addition, the force is proportional to the displacement and so is a Hooke's Law force. Therefore, the cylinder will undergo SHM, as described in Chapter 11 . Comparing \(F_{B}=A \rho_{w} g y\) with Hooke's Law in the form \(F=k y\), we see that the elastic constant for the motion is \(k=A \rho_{w} g .\) This, acting on the cylinder of mass \(m\), causes it to have a vibrational frequency of $$ f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{A \rho_{w} g}{m}} $$

The diameter of the large piston of a hydraulic press is \(20 \mathrm{~cm}\), and the area of the small piston is \(0.50 \mathrm{~cm}^{2}\). If a force of \(400 \mathrm{~N}\) is applied to the small piston, (a) what is the resulting force exerted on the large piston? (b) What is the increase in pressure underneath the small piston? ( \(c\) ) Underneath the large piston?

A 60 -kg performer balances on a cane. The end of the cane in contact with the floor has an area of \(0.92 \mathrm{~cm}^{2}\). Find the pressure exerted on the floor by the cane. (Neglect the weight of the cane.)

A barrel will rupture when the gauge pressure within it reaches 350 \(\mathrm{kPa}\). It is attached to the lower end of a vertical pipe, with the pipe and barrel filled with oil \(\left(\rho=890 \mathrm{~kg} / \mathrm{m}^{3}\right)\). How long can the pipe be if the barrel is not to rupture? From \(P=\rho g h\) we have $$ h=\frac{P}{\rho g}=\frac{350 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}}{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(890 \mathrm{~kg} / \mathrm{m}^{3}\right)}=40.1 \mathrm{~m} $$

A 60 -kg rectangular box, open at the top, has base dimensions of \(1.0 \mathrm{~m}\) by \(0.80 \mathrm{~m}\) and a depth of \(0.50 \mathrm{~m}\). (a) How deep will it sink in fresh water? (b) What weight \(F_{W b}\) of ballast will cause it to sink to a depth of \(30 \mathrm{~cm}\) ? (a) Assuming that the box floats, \(F_{B}=\) Weight of displaced water \(=\) Weight of box \(\left.1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times y)=(60 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) where \(y\) is the depth the box sinks. Solving yields \(y=0.075 \mathrm{~m}\). Because this is smaller than \(0.50 \mathrm{~m}\), our assumption is shown to be correct. (b) \(F_{B}=\) weight of box \(+\) weight of ballast But the \(F_{B}\) is equal to the weight of the displaced water. Therefore, the above equation becomes \(\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times 0.30 \mathrm{~m})=(60)(9.81) \mathrm{N}+F_{W b}\) from which \(F_{W b}=1760 \mathrm{~N}=1.8 \mathrm{kN}\). The ballast must have a mass of \((1760 / 9.81) \mathrm{kg}=180 \mathrm{~kg}\).

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