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Chapter 11: Problem 2

A petite young woman distributes her \(500 \mathrm{~N}\) weight equally over the heels of her high-heeled shoes. Each heel has an area of \(0.750 \mathrm{~cm}^{2}\). (a) What pressure is exerted on the floor by each heel? (b) With the same pressure, how much weight could be supported by two flat- bottomed sandals, each of area \(200 \mathrm{~cm}^{2} ?\)

Short Answer

Expert verified
(a) 3,333,333.33 Pa, (b) 133,333.33 N.

Step by step solution

01

Understand the Known Values

We know the weight of the woman is 500 N, and it's equally distributed over two heels. The area of each heel is given as 0.750 cm squared.
02

Convert Units for Area to m虏

First, convert the area from cm虏 to m虏, because the standard unit of pressure is Pascal (Pa), which is N/m虏. 1 cm虏 = 0.0001 m虏, so 0.750 cm虏 is 0.750 * 0.0001 = 0.000075 m虏.
03

Calculate Force on Each Heel

Since the weight is equally distributed over two heels, each heel supports half the total weight. Thus, each heel supports 500 N / 2 = 250 N.
04

Calculate Pressure Exerted by Each Heel

Pressure is calculated as force divided by the area over which the force is applied. Therefore, the pressure exerted by each heel is \( \frac{250 \text{ N}}{0.000075 \text{ m}^2} \).Calculate this to find the pressure: \(\frac{250}{0.000075} \approx 3,333,333.33 \text{ Pa}\).
05

Understand Pressure for the Two Sandals

We will use the same pressure exerted by each heel to find out how much weight can be supported by the two sandals, each of area 200 cm虏.
06

Convert Units for Sandal Area to m虏

Convert the area of each sandal from cm虏 to m虏. 200 cm虏 = 200 * 0.0001 = 0.02 m虏.
07

Calculate Total Area for Two Sandals

The total area for the two sandals is 2 * 0.02 m虏 = 0.04 m虏.
08

Calculate Maximum Weight Supported by Sandals

Using the pressure of 3,333,333.33 Pa (from the heel calculation), apply the formula\( \text{Weight} = \text{Pressure} \times \text{Total Area} \).Thus, \( \text{Weight} = 3,333,333.33 \times 0.04 \) N.Calculate this to find the maximum weight:133,333.33 N.

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Key Concepts

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

Pressure Calculation
To understand pressure calculation, let's explore how pressure is defined. Pressure is the amount of force applied over a particular area, and it's calculated using the formula:
  • Pressure = Force / Area
The unit of pressure is the Pascal, symbolized as Pa. One Pascal corresponds to one Newton per square meter. In scenarios like the one with the high heels, we are curious about how much force is spread out over the small area of the heel. The smaller the area with the same force, the higher the pressure applied. In our exercise, each heel exerts a pressure of about 3,333,333.33 Pa, illustrating how concentrated force can amplify pressure significantly when distributed over a tiny area.
Unit Conversion
Unit conversion is a fundamental skill, especially in physics, where different systems of measurement are used. In our exercise, the area was initially given in square centimeters (cm虏). However, pressure calculations require the area in square meters (m虏) because pressure is expressed in Pascals (N/m虏). This necessitates a conversion:
  • 1 cm虏 = 0.0001 m虏
Hence, converting 0.750 cm虏 to square meters becomes:
  • 0.750 cm虏 = 0.000075 m虏
Performing these conversions ensures accurate calculations, adhering to standard scientific units.
Force Distribution
Force distribution describes how a force is spread across different areas or points. In the given problem, the woman鈥檚 weight of 500 N is distributed evenly over two heels. This means each heel supports half of the total weight, calculated as:
  • Weight per heel = Total Weight / Number of heels
  • 250 N = 500 N / 2
Understanding this distribution is crucial because it affects the pressure exerted on the floor. With each heel bearing 250 N, this distributed force becomes the starting point for calculating pressure.
Area and Pressure Relationship
The relationship between area and pressure is inverse. That means when you apply the same force over an area, the smaller the area, the higher the pressure - and vice versa. This principle becomes clear through the heel and sandal scenario:
  • Small heel area = High pressure
  • Larger sandal area = Lower pressure for the same force
For instance, the sandals, covering more area, could support a substantial weight when maintaining the same pressure calculated for the heels. With each sandal having a larger area (200 cm虏), converted into 0.02 m虏, and a total area for two sandals as 0.04 m虏, we assess that these sandals could support a heavier load. This illustrates how varying the area impacts the weight a surface can support under constant pressure.

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Most popular questions from this chapter

Achilles tendon. The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is \(78.1 \mathrm{~mm}^{2}\), its average length is \(25 \mathrm{~cm},\) and its average Young's modulus is 1474 MPa. (a) How much tensile stress is required to stretch this muscle by \(5.0 \%\) of its length? (b) If we model the tendon as a spring, what is its force constant? (c) If a \(75 \mathrm{~kg}\) sprinter exerts a force of 13 times his weight on his Achilles tendon, by how much will it stretch?

A thin, light wire \(75.0 \mathrm{~cm}\) long having a circular cross section \(0.550 \mathrm{~mm}\) in diameter has a \(25.0 \mathrm{~kg}\) weight attached to it, causing it to stretch by \(1.10 \mathrm{~mm}\). (a) What is the stress in this wire? (b) What is the strain of the wire? (c) Find Young's modulus for the material of the wire.

An object suspended from a spring vibrates with simple harmonic motion. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is kinetic and what fraction is potential?

A steel wire has the following properties: $$ \begin{array}{l} \text { Length }=5.00 \mathrm{~m} \\ \text { Cross-sectional area }=0.040 \mathrm{~cm}^{2} \end{array} $$ Young's modulus \(=2.0 \times 10^{11} \mathrm{~Pa}\) Shear modulus \(=0.84 \times 10^{11} \mathrm{~Pa}\) Proportional limit \(=3.60 \times 10^{8} \mathrm{~Pa}\) Breaking stress \(=11.0 \times 10^{8} \mathrm{~Pa}\) The proportional limit is the maximum stress for which the wire still obeys Hooke's law (see point \(\mathrm{B}\) in Figure 11.12 ). The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

"Seeing" surfaces at the nanoscale. One technique for making images of surfaces at the nanometer scale, including membranes and biomolecules, is dynamic atomic force microscopy. In this technique, a small tip is attached to a cantilever, which is a flexible, rectangular slab supported at one end, like a diving board (Figure 11.40 ). The cantilever vibrates, so the tip moves up and down in simple harmonic motion. In one mode of operation, the resonant frequency for a cantilever with force constant \(k=1000 \mathrm{~N} / \mathrm{m}\) is \(100 \mathrm{kHz}\). As the oscillating tip is brought within a few nanometers of the surface of a sample (as shown in the figure), it experiences an attractive force from the surface. For an oscillation with a small amplitude (typically \(0.050 \mathrm{nm}),\) the force \(F\) that the sample surface exerts on the tip varies linearly with the displacement \(x\) of the tip, \(|F|=k_{\text {surf }} x,\) where \(k_{\text {surf }}\) is the effective force constant for this force. The net force on the tip is therefore \(\left(k+k_{\text {surf }}\right) x,\) and the frequency of the oscillation changes slightly due to the interaction with the surface. Measurements of the frequency as the tip moves over different parts of the sample's surface can provide information about the sample. If we model the vibrating system as a mass on a spring, what is the mass necessary to achieve the desired resonant frequency when the tip is not interacting with the surface? A. \(25 \mathrm{ng}\) B. \(100 \mathrm{ng}\) C. \(25 \mu \mathrm{g}\) D. \(100 \mu \mathrm{g}\)
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