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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) 3.33 × 10^6 Pa per heel. (b) 133,200 N can be supported by the sandals.

Step by step solution

01

Calculate Pressure on Each Heel

To find the pressure exerted by each heel, we use the pressure formula. Pressure \( P \) is given by the formula \( P = \frac{F}{A} \), where \( F \) is the force (or weight) and \( A \) is the area. For each heel, \( F = \frac{500 \text{ N}}{2} = 250 \text{ N} \). The area \( A = 0.750 \text{ cm}^2 \), which needs to be converted to meters squared: \( 0.750 \text{ cm}^2 = 0.750 \times 10^{-4} \text{ m}^2 \). Substitute these into the pressure formula to find pressure on one heel: \[ P = \frac{250 \text{ N}}{0.750 \times 10^{-4} \text{ m}^2} \].
02

Convert Units for Pressure

Substitute the known values into the formula to calculate pressure, remembering to convert \( cm^2 \) to \( m^2 \): \( P = \frac{250}{0.750 \times 10^{-4}} \). Calculating yields \( P = 3.33 \times 10^{6} \text{ N/m}^2 \). This means each heel exerts a pressure of \( 3.33 \times 10^{6} \text{ Pa} \) on the floor.
03

Calculate Weight Supported by Sandals

Using the pressure we found, determine how much weight could be supported by the sandals. The pressure \( P \) remains the same, and you use the pressure formula rearranged to find the force \( F = PA \). Here, \( A = 200 \text{ cm}^2 = 0.02 \text{ m}^2 \) per sandal, so for two sandals \( A_{total} = 2 \times 0.02 \text{ m}^2 = 0.04 \text{ m}^2 \). Substitute these values into the formula. \[ F = (3.33 \times 10^{6} \text{ N/m}^2) \times 0.04 \text{ m}^2 \].
04

Compute Final Weight Supported

Perform the multiplication from the previous step: \( F = 3.33 \times 10^{6} \times 0.04 = 133,200 \text{ N} \). This means that, at the same pressure, the sandals could support a total weight of 133,200 N.

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Key Concepts

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

Pressure Formula
Pressure is a concept that describes how a force is distributed over an area. To calculate it, we use the pressure formula: \[ P = \frac{F}{A} \] - **P** is the pressure.- **F** represents the force applied.- **A** is the area over which the force is applied.
This formula helps us understand how concentrated a force is on a specific area. For example, in the case of high-heeled shoes, the small area of the heel means that the pressure exerted on the floor is quite high. The smaller the area over which a force acts, the larger the pressure. This is why sharp objects can easily puncture surfaces compared to blunt ones, even if the force is the same.
Unit Conversion
Before doing any calculations in physics, it’s important to ensure all units are consistent. In the exercise, area was initially given in square centimeters \( \text{cm}^2 \), but pressure often needs to be calculated using area in square meters \( \text{m}^2 \). Therefore, converting units is crucial: - To convert from \( \text{cm}^2 \) to \( \text{m}^2 \), multiply by \( 10^{-4} \).- For example, \( 0.750 \text{ cm}^2 \) becomes \( 0.750 \times 10^{-4} \text{ m}^2 \).
By converting units properly, you make sure your calculations are correct, allowing for accurate results. This step is crucial to getting correct calculations when working with pressure and forces in real-world applications.
Area Calculation
Calculating the area is a key step when it comes to determining pressure. In the example, the area of each heel and the area of the sandals were measured. Here's why area is so important: - The smaller the area the force is applied to, the greater the pressure exerted (assuming force is constant). - The heels applied pressure on a tiny area compared to the sandals which distribute the force over a larger area.
By understanding how to calculate and use areas, you can correctly assess how pressure will change when the area changes, which has practical applications in engineering and everyday life. For instance, wide-bottomed shoes exert less pressure than high-heeled shoes, making them less likely to damage delicate floors.
Force and Pressure Relationship
Force and pressure share a direct relationship that explains many phenomena we observe. Pressure increases as force increases if the area stays the same. Similarly, reducing the area increases the pressure if the force remains unchanged: - The formula \( P = \frac{F}{A} \) encapsulates this relationship.- In the example, doubling the area over which the same force acts reduces the pressure by half.
Understanding this relationship is crucial for controlling pressure in practical situations. Engineers must often manage this to ensure safety and effectiveness, such as in designing footwear that can distribute a person’s weight evenly to prevent floor damage while maintaining comfort.

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Most popular questions from this chapter

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

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A steel cable with cross-sectional area 3.00 \(\mathrm{cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8}\) Pa. Find the maximum upward acceleration that can be given a \(1200-\mathrm{kg}\) elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

A bar with cross-sectional area \(A\) is subjected to equal and opposite tensile forces \(\vec{F}\) at its ends. Consider a plane through the bar making an angle through the bar making an angle \(\theta\) with a plane at right angles to the bar (Fig. 11.64\()\) . (a) What is the tensile (normal) stress at this plane in terms of \(F, A,\) and \(\theta ?(b)\) What is the shear (tangential) stress at the plane in terms of \(F, A,\) and \(\theta ?\) (c) For what value of \(\theta\) is the tensile stress a maximum? (d) For what value of \(\theta\) is the shear stress a maximum?

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