/*! 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 48 An elephant that has a mass of \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 48

An elephant that has a mass of \(6000 \mathrm{~kg}\) evenly distributes her weight on all four feet. (a) If her feet are approximately circular and each has a diameter of \(50 \mathrm{~cm}\), estimate the pressure on each foot. (b) Compare the answer in part (a) with the pressure on each of your feet when you are standing up. Make some rough but reasonable assumptions about the area of your feet.

Short Answer

Expert verified
The pressure exerted by each foot of the elephant is estimated to be \(75\,kPa\) and the pressure exerted by each foot of a human is estimated to be \(6.86\,kPa\). Therefore, although the elephant is much heavier, the pressure it exerts is less due to the larger contact area of its feet.

Step by step solution

01

Calculate force exerted by the elephant

First, we need to determine the weight of the elephant, which is the force it exerts due to its mass. Force can be calculated using the formula \(F = m \cdot g\), where \(m = 6000 \,kg\) is the mass of the elephant and \(g = 9.8 \,m/s^2\) is the acceleration due to gravity. Thus, \(F = 6000\,kg \cdot 9.8\,m/s^2 = 58800\,N\). Since this force is distributed evenly across all four feet, the force per foot is \(F/4 = 58800/4\,N = 14700\,N\).
02

Calculate the area of each foot

The area of the foot can be calculated using the formula for the area of a circle, \(A = pi \cdot (d/2)^2\), with \(d = 0.5\,m\) being the diameter. Hence, \(A = 3.1415 \cdot (0.5/2)^2 = 0.196\, m^2\).
03

Calculate the pressure exerted by each foot

The pressure can be calculated using the formula \(P = F/A\), where \(F = 14700\,N\) is the force per foot and \(A = 0.196\,m^2\) is the area of each foot. Thus, \(P = 14700/0.196\,Pa = 75000\,Pa\) or \(75\,kPa\).
04

Calculation of pressure on human foot

Assume a person with a weight of \(70\,kg\), which equals a force of \(F=70\,kg \cdot 9.8\,m/s^2 = 686\,N\). Assume also the bottom of a human foot to measure \(25\,cm\) x \(10\,cm\). The contact area of both feet is therefore \(2 \cdot 25\,cm/100\,m \cdot 10\,cm/100\,m = 0.05\,m^2\). The pressure exerted by each foot of the person is \(P=686/2/0.05\,Pa = 6860\,Pa\) or \(6.86\,kPa\).
05

Compare the pressures

Comparing the pressure exerted by each foot of the elephant and the human, the elephant exerts a pressure of \(75\,kPa\) whereas the human exerts a pressure of \(6.86\,kPa\). Even though the elephant is much heavier, the pressure it exerts is less due to the larger contact area of its feet.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Force Distribution
When an elephant walks, the force from its weight is distributed evenly across all four of its feet. This means that instead of the entire weight being concentrated on one point, it's spread out evenly. Let's break this down:
- The elephant weighs 6000 kg, but we need to calculate the force. This is done by multiplying the mass by gravity: \( F = m \times g \). Here, \( g = 9.8 \, m/s^2 \), so the total force is \( 58800 \, N \).
- Since this force is evenly distributed over four feet, each foot bears a quarter of the total weight, giving us \( 14700 \, N \) per foot.
Understanding how force is distributed helps when estimating pressures on surfaces, which we'll dive into next.
Area Calculation
To find the pressure exerted by the elephant on the ground, you need to know the area over which the force is applied. The foot of an elephant is approximately circular, and we can calculate the area using the formula for the area of a circle:- The diameter of each foot is 50 cm, or 0.5 meters.
- The radius, therefore, is \( 0.25 \, m \) (since the radius is half of the diameter).
- Using the formula \( A = \pi \times (d/2)^2 \), the area comes out to be \( 0.196 \, m^2 \).
With the area calculated, it becomes easier to move onto determining the pressure by combining this with the force on each foot.
Comparison of Pressures
Once we know both the force and the area, calculating the pressure is just a matter of division. Pressure is the force applied per unit area. For the elephant:- Pressure \( P \) is calculated as \( P = F/A \).
- With a force of \( 14700 \, N \) and an area of \( 0.196 \, m^2 \), the pressure is \( 75000 \, Pa \, (75 \, kPa) \).
To see how this compares to a human, let's consider someone weighing 70 kg:- The force exerted by this person is \( 686 \, N \).
- Assuming a combined foot area of \( 0.05 \, m^2 \), the pressure is \( 6860 \, Pa \, (6.86 \, kPa) \).
Although the elephant is much heavier, the larger area of its feet distributes the force, resulting in lower pressure compared to human feet.
Physics Problem Solving
Solving physics problems like this one involves systematic steps: 1. **Understand the Problem:** What do you need to find out? In this case, the pressure on the feet of an elephant and a human.
2. **Gather the Known Data:** Here, this includes the mass of the elephant, gravity, the diameter of the elephant's foot, and assumptions about a human's weight and foot area.
3. **Calculate Step-by-Step:** Start with determining the force, then the area, and finally the pressure by applying appropriate formulas at each step.
4. **Analyze the Results:** Compare the calculated pressures to understand the effect of force distribution over different areas.
This methodical approach helps simplify complex physics problems and can be applied to various scenarios in mechanics and beyond.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A diver is \(10.0 \mathrm{~m}\) below the surface of the ocean. The surface pressure is \(1 \mathrm{~atm}\). What is the absolute pressure and gauge pressure he experiences? The density of seawater is \(1.025 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).

Two identically shaped containers in the shape of a truncated cone are placed on a table, but one is inverted such that the small end is resting on the table. The containers are filled with the same height of water. The pressure at the bottom of each container is the same. However, the weight of the water in each container is different. Explain why this statement is correct.

A set of density rods is designed to illustrate the concept of density. The idea is to create cylinders of equal diameters and masses, but varying lengths, to show which have the largest and smallest densities. (a) Derive a formula that will predict the ratio of the length of one rod to the length of another rod, as a function of the densities of the two rods. Assume the radius and mass of each cylinder is the same and express your answer in terms of the specific gravity of each rod. (b) If you want to design a set of five density rods (made of aluminum, iron, copper, brass, and lead), determine the ratios of the lengths of each rod to the length of the lead rod, the densest material in the group. The specific gravities of the elements are approximately \(\mathrm{SG}_{\text {alum }}=2.7, \mathrm{SG}_{\text {iron }}=7.8\), \(\mathrm{SG}_{\text {copper }}=8.9, \mathrm{SG}_{\text {brass }}=8.5, \mathrm{SG}_{\text {lead }}=11.3\).

Astronomy Evidence from the Mars rover Discovery suggests that oceans as deep as \(0.50 \mathrm{~km}\) once may have existed on Mars. The acceleration due to gravity on Mars is \(0.379 \mathrm{~g}\) and the current atmospheric pressure is \(650 \mathrm{~N} / \mathrm{m}^{2}\). (a) If there were any organisms in the Martian ocean in the distant past, what pressure (absolute and gauge) would they have experienced at the bottom, assuming the surface pressure was the same as today? Assume that the salinity of Martian oceans was the same as oceans on Earth. (b) If the bottom-dwelling organisms in part (a) were brought from Mars to Earth, how deep could they go in our ocean without exceeding the maximum pressure they experienced on Mars?

Medical The aorta is approximately \(25 \mathrm{~mm}\) in diameter. The mean pressure there is about \(100 \mathrm{mmHg}\) and the blood flows through the aorta at approximately \(60 \mathrm{~cm} / \mathrm{s}\). Suppose that at a certain point a portion of the aorta is blocked so that the cross-sectional area is reduced to \(3 / 4\) of its original area. The density of blood is \(1060 \mathrm{~kg} / \mathrm{m}^{3}\). (a) How fast is the blood moving just as it enters the blocked portion of the aorta? (b) What is the gauge pressure (in \(\mathrm{mmHg}\) ) of the blood just as it has entered the blocked portion of the aorta?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Dynamics

Read Explanation

Waves Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fundamentals of Physics

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.