/*! 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 56 \(\mathrm{A}\) 75 \(\mathrm{kg}\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 56

\(\mathrm{A}\) 75 \(\mathrm{kg}\) man weighs himself at the north pole and at the equator. Which scale reading is higher? By how much? Assume the earth is a perfect sphere. Explain why the readings differ.

Short Answer

Expert verified
The scale reading is higher at the North Pole, where the man weighs 737.25 N compared to his weight of 733.5 N at the equator. The difference is 3.75 N due to the centrifugal force at the equator and the shape of the earth.

Step by step solution

01

Calculate the man's weight at the North Pole

The weight of an object is the force of gravity on it. We can find it by using the equation \(W = m \cdot g\), where \(m\) is the mass and \(g\) is acceleration due to gravity. At the North Pole, the acceleration due to gravity is approximately \(9.83 \, m/s^2\). So the weight of the man at the North Pole \(W_{np}\) is \(75 \, kg \cdot 9.83 \, m/s^2 = 737.25 \, N\)
02

Calculate the man's weight at the Equator

At the equator, the acceleration due to gravity is slightly less, approximately \(9.78 \, m/s^2\), since the centrifugal force reduces the weight of the man. So the weight \(W_{eq}\) of the man at the equator is \(75 \, kg \cdot 9.78 \, m/s^2 = 733.5 \, N\)
03

Compare the two weights

Comparing the two weights calculated, it's clear that the man's weight is greater at the North Pole than at the equator by a difference of \(737.25 \, N - 733.5 \, N = 3.75 \, N\)
04

Explain the difference

The slightly greater weight at the North Pole compared to the equator can be explained by the Earth's rotation. The centrifugal force due to Earth's rotation is stronger at the equator and works in the opposite direction of gravity, slightly decreasing the force of gravity and hence the weight of objects there. Also, the earth isn't a perfect sphere but is slightly flatter at the poles and bulges at the equator, which means objects at the equator are further from the center of Earth, and therefore experience less gravitational force.

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.

Centrifugal Force
When you think about how the Earth spins, this spinning introduces a force known as the centrifugal force. Imagine the Earth as a spinning top. As it spins, any object on its surface, like you or me, experiences a tendency to be "pushed" outward, away from the center.
Centrifugal force is not a real force in the way gravity is, but it's a useful concept when considering how things behave on a rotating system. This "force" increases as you move toward the equator because the Earth's rotation speed is higher there.
In our example, the man experiences more centrifugal force at the equator.
  • The centrifugal effect works against gravity, making objects feel lighter.
  • At the North Pole, this force is minimal as there's less outward push, so weight there feels heavier compared to the equator.
Acceleration due to Gravity
Gravity is a force of attraction that pulls objects toward one another. The acceleration due to gravity is the rate at which an object speeds up as it falls. On Earth, this acceleration is generally around 9.8 m/s虏.
The exact value can differ depending on where you are, such as at the North Pole or equator.
At the North Pole, gravity pulls you with an acceleration of about 9.83 m/s虏. But at the equator, it's a bit less at 9.78 m/s虏.
  • The reason is the Earth's shape and rotation.
  • Less gravitational pull at the equator means slightly less weight as experienced by the man in the problem.
Earth's Rotation
The Earth rotates on its axis, which is an imaginary line that runs from the North Pole to the South Pole. This rotation causes days and nights to happen and greatly influences forces experienced on Earth.
One major effect of Earth's rotation is the centrifugal force, which operates most strongly at the equator.
This rotation also causes a bulging effect at the equator:
  • Objects at the equator are further from Earth's center compared to those at the poles.
  • This distance can affect the gravitational pull, slightly reducing it at the equator.
These aspects make Earth's rotation important in understanding weight variations across different latitudes.
Gravitational Force
Gravitational force is the attraction between two masses, and for us on Earth, it often refers to the force Earth exerts on objects to pull them towards its center.
It is this force that gives us our weight, defined as the pull Earth has on our mass.
On Earth's surface, this force isn't uniform everywhere due to various factors like Earth's shape and rotation:
  • At the North Pole, the gravitational pull is stronger because you're closer to Earth's center.
  • At the equator, because of Earth's bulging shape and rotational speed, you're slightly farther from the center, which reduces gravitational force.
This variation in gravitational force explains why the man's weight differs between the North Pole and the equator.

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

You are driving your car through a roundabout that has a radius of \(9.0 \mathrm{m} .\) Your physics textbook is lying on the seat next to you. What is the fastest speed at which you can go around the curve without the book sliding? The coefficient of static friction between the book and the seat is 0.30 .

What is the free-fall acceleration at the surface of (a) Mars and (b) Jupiter?

The ultracentrifuge is an important tool for separating and analyzing proteins in biological research. Because of the enor mous centripetal accelerations that can be achieved, the apparatus (see Figure 6.18 ) must be carefully balanced so that each sample is matched by another on the opposite side of the rotor shaft. Any difference in mass of the opposing samples will cause a net force in the horizontal plane on the shaft of the rotor; this force can actually be large enough to destroy the centrifuge. Suppose that a scientist makes a slight error in sample preparation, and one sample has a mass 10 mg greater than the opposing sample. If the samples are \(10 \mathrm{cm}\) from the axis of the rotor and the ultracentrifuge spins at 70,000 rpm, what is the magnitude of the net force on the rotor due to the unbalanced samples?

The material that comprises the side of the moon facing the earth is actually slightly more dense than the material on the far side. When the spacecraft is above a more dense area of the surface, the moon's gravitational force on the craft is a bit stronger. In order to stay in a circular orbit of constant height and speed, the spacecraft could fire its rockets while passing over the denser area. The rockets should be fired so as to generate a force on the craft A. Away from the surface of the moon. B. In the direction of motion. C. Toward the center of the moon. D. Opposite the direction of motion.

Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface. In planning a future mission to Europa, what is the fastest that an astronaut with legs of length \(0.70 \mathrm{m}\) could walk on the surface of Europa? Europa is \(3100 \mathrm{km}\) in diameter and has a mass of \(4.8 \times 10^{22} \mathrm{kg}\).
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Energy Physics

Read Explanation

Rotational Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Thermodynamics

Read Explanation

Geometrical and Physical Optics

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.