/*! 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 8 A \(79-k g\) person sits on a \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 8

A \(79-k g\) person sits on a \(3.7-k g\) chair. Each leg of the chair makes contact with the floor in a circle that is \(1.3 \mathrm{cm}\) in diameter. Find the pressure exerted on the floor by each leg of the chair, assuming the weight is evenly distributed.

Short Answer

Expert verified
The pressure exerted on the floor by each leg of the chair is approximately 1,528,888.48 Pascals.

Step by step solution

01

Calculate Total Weight

First, add the weight of the person and the chair together. The person's mass is 79 kg and the chair's mass is 3.7 kg. The total weight in Newtons can be calculated by multiplying the total mass by the acceleration due to gravity, which is approximately 9.8 m/s虏. This gives us \[\text{Total weight} = (79 \text{ kg} + 3.7 \text{ kg}) \times 9.8 \text{ m/s}^2 = 811.26 \text{ N}.\]
02

Determine Weight per Leg

Assume that the total weight is evenly distributed among the four legs of the chair. The weight supported by each leg is\[\text{Weight per leg} = \frac{811.26 \text{ N}}{4} = 202.815 \text{ N}.\]
03

Find the Area of Contact per Leg

The contact area of each leg is a circle with a diameter of 1.3 cm. First, convert this diameter into meters: 1.3 cm = 0.013 m. The radius is half of the diameter, so the radius is 0.0065 m. Calculate the area using the formula for the area of a circle, \(A = \pi r^2\):\[A = \pi (0.0065 \text{ m})^2 = 1.327 x 10^{-4} \text{ m}^2.\]
04

Calculate the Pressure per Leg

The pressure exerted by each leg is calculated by dividing the weight supported by the leg by the area of contact. Use the formula for pressure: \(P = \frac{F}{A}\):\[P = \frac{202.815 \text{ N}}{1.327 \times 10^{-4} \text{ m}^2} \approx 1528888.48 \text{ Pa}.\]

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
Force is a push or pull exerted on an object. It is directly associated with motion and is an essential concept in physics. In our exercise, force is represented by the total weight of the person and chair combined. The formula for calculating force is given by Newton's second law:
  • The formula to determine force is \( F = ma \),where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
  • In this example, the acceleration is provided by gravity, which is approximately \( 9.8 \, \text{m/s}^2 \).
  • By adding the mass of the person and the chair, and multiplying by gravity, we obtain the total force acting through weight.
Force is pivotal to determine how much pressure is exerted on an object and it plays a crucial role in various physical calculations.
Area of a Circle
The area of a circle is crucial when determining how evenly distributed forces act on a surface, such as the ground in our exercise about the chair's legs. The formula for the area of a circle helps us know how much floor space is contacted by each leg:
  • The basic formula for the area of a circle is \( A = \pi r^2 \),where \( A \) is the area, \( r \) is the radius, and \( \pi \) is approximately 3.14159.
  • To find the radius, we divide the diameter by two. In our problem, the leg's diameter is 1.3 cm, or 0.013 meters, giving a radius of 0.0065 m.
  • Using the formula, the area of contact per leg translates into \( 1.327 \times 10^{-4} \, \text{m}^2 \).This shows how small each leg's footprint is on the floor surface.
Understanding the area helps determine the pressure, as pressure is the force applied over this specific area.
Gravity
Gravity is a natural phenomenon by which all things with mass are brought toward one another. It gives weight to physical objects and is responsible for the force that gravity applies to an object. In this problem:
  • We used the acceleration due to gravity, which is \( 9.8 \, \text{m/s}^2 \),to calculate the weight or force exerted by the person and the chair.
  • Weight is formulated as \( W = mg \),where \( W \) is weight (a force), \( m \) is mass, and \( g \) is the acceleration due to gravity.
  • This force due to gravity is what each chair leg distributes onto the floor. It鈥檚 a constant factor that influences the calculations and helps us comprehend the laws of motion and pressure.
Gravity is essential for providing the circumstances required to analyze pressure and the force on surfaces like the floor beneath a chair.

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 prospector finds a solid rock composed of granite \(\left(\rho=2650 \mathrm{kg} / \mathrm{m}^{3}\right)\) and gold. If the volume of the rock is \(3.55 \times 10^{-4} \mathrm{m}^{3}\), and its mass is \(3.81 \mathrm{kg}\) (a) what mass of gold is contained in the rock? What percentage of the rock is gold by (b) volume and \((c)\) mass?

Blo Bioluminescence Some species of dinoflagellate (a type of unicellular plankton) can produce light as the result of biochemical reactions within the cell. This light is an example of bioluminescence. It is found that bioluminescence in dinoflagellates can be triggered by deformation of the cell surface with a pressure as low as one dyne \(\left(10^{-5} \mathrm{N}\right)\) per square centimeter. What is this pressure in (a) pascals and (b) atmospheres?

BIO A patient is given an injection with a hypodermic needle \(3.3 \mathrm{cm}\) long and \(0.26 \mathrm{mm}\) in diameter. Assuming the solution being injected has the same density and viscosity as water at \(20^{\circ} \mathrm{C}\), find the pressure difference needed to inject the solution at the rate of \(1.5 \mathrm{g} / \mathrm{s}.\)

Water at \(20^{\circ} \mathrm{C}\) flows through a horizontal garden hose at the rate of \(5.0 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s} .\) The diameter of the garden hose is \(2.5 \mathrm{cm} .\) (a) What is the water speed in the hose? (b) What is the pressure drop across a 15 - \(\mathrm{m}\) length of hose? Suppose the crosssectional area of the hose is halved, but the length and pressure drop remain the same. (c) By what factor does the water speed change? (d) By what factor does the volume flow rate change? Explain.

A person floats in a boat in a small backyard swimming pool. Inside the boat with the person are several blocks of wood. Suppose the person now throws the blocks of wood into the pool, where they float. (a) Does the boat float higher, lower, or at the same level relative to the water? (b) Does the water level in the pool increase, decrease, or stay the same?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Famous Physicists

Read Explanation

Fluids

Read Explanation

Circular Motion and Gravitation

Read Explanation

Mechanics and Materials

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.