/*! 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 65 If you weigh yourself on a bathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If you weigh yourself on a bathroom scale at the equator, is the reading you get greater than, less than, or equal to the reading you get if you weigh yourself at the North Pole?

Short Answer

Expert verified
You weigh less at the equator than at the North Pole.

Step by step solution

01

Understanding the Problem

We need to determine if a person's weight is different at the equator compared to the North Pole. It's important to remember that weight is the force exerted by gravity on a mass, which can vary slightly depending on location due to Earth's shape and rotation.
02

Consider Earth's Shape

Earth is not a perfect sphere; it is an oblate spheroid. This means it is slightly flattened at the poles and bulges at the equator. Therefore, the distance from the center of the Earth to its surface is greater at the equator than at the poles.
03

Analyze the Effects of Earth's Rotation

At the equator, the rotation of the Earth causes centrifugal force, which acts outward and decreases the effective gravitational force. This is because the Earth rotates fastest at the equator, reducing the apparent weight of objects there.
04

Compare Gravitational Force

The gravitational force decreases with distance from the center of the Earth. Since the surface is farther from the Earth's center at the equator than at the poles, the gravitational pull is weaker at the equator.
05

Conclusion

Considering the effects of Earth's shape and rotation, a person would weigh less at the equator than at the North Pole due to the reduced gravitational pull and the effect of centrifugal 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 91Ó°ÊÓ!

Key Concepts

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

Earth's Shape
The Earth may seem like a perfect sphere from space, but it's actually an oblate spheroid. Imagine a slightly squashed ball leading to a flattened top and bottom. This flattening is due to the Earth's rotation. As a result, the equator "bulges" out more compared to the poles.

This shape has an effect on gravity. The distance from the Earth's center to its surface is not uniform. At the equator, this distance is longer than it is at the poles. The increased distance at the equator means that the gravitational force experienced there is a little less than at the poles. Understanding this concept helps explain why there are variations in weight at different points on Earth.
  • The Earth is an oblate spheroid, not a perfect sphere.
  • There's more distance from the center to the surface at the equator than at the poles.
  • This affects the gravitational pull experienced by objects.
Gravitational Force
Gravitational force is the force that pulls objects towards the center of the Earth. It's what gives objects weight. The formula for gravitational force is \[ F_g = \, \frac{G \cdot m_1 \cdot m_2}{r^2} \]where \( F_g \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses involved, and \( r \) is the distance between the centers of the two masses.

As we discussed earlier, because the Earth isn't a perfect sphere, this distance, \( r \), changes depending on where you are - it's greater at the equator than at the poles. The larger the \( r \), the smaller the gravitational force. So, a person will experience a slightly weaker gravitational pull at the equator compared to at the poles. This directly influences the weight they would measure on a scale.
  • Gravitational force is weaker with greater distance from Earth's center.
  • This changing force results in weight variations between different Earth locations.
  • The equator experiences weaker gravitational force due to the Earth's shape.
Centrifugal Force
The Earth's rotation introduces another force that affects weight - centrifugal force. It's the force that acts outward on a body moving in a circle, away from the center around which it rotates. Imagine swinging a bucket of water in a circle; the water pushes outward.

The faster rotation at the equator means more centrifugal force. This force "pulls" objects outward and slightly reduces their apparent weight. This effect is not present at the poles because they are the axis of rotation and do not move.
  • Centrifugal force acts outward away from the center of rotation.
  • This force is strongest at the equator due to faster rotational speed.
  • The effect of centrifugal force reduces the effective weight of objects.
  • There's no significant centrifugal force at the poles because the rotation effect is minimal.

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

Force Times Time At the local hockey rink, a puck with a mass of \(0.12 \mathrm{kg}\) is given an initial speed of \(v_{0}=6.7 \mathrm{m} / \mathrm{s}\). (a) If the coefficient of kinetic friction between the ice and the puck is 0.13 , how much time \(t\) does it take for the puck to come to rest? (b) If the mass of the puck is doubled, does the frictional force \(F\) exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping time of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that \(F t=m v_{0}\). (The significance of this result will be discussed in Chapter \(9,\) where we will see that \(m v\) is the momentum of an object.)

A car is driven with constant speed around a circular Answer each of the following questions with "Yes" or "No." (a) Is the car's velocity constant? (b) Is its speed constant? (c) Is the magnitude of its acceleration constant? (d) Is the direction of its acceleration constant?

Find the coefficient of kinetic friction between a \(4.7-\mathrm{kg}\) block and the horizontal surface on which it rests if an \(89-\mathrm{N} / \mathrm{m}\) spring must be stretched by \(2.2 \mathrm{cm}\) to pull the block with constant speed. Assume the spring pulls in a direction \(13^{\circ}\) above the horizontal.

Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver \(2 ?\) (b) Choose the best explanation from among the following: I. Locking up the brakes gives the greatest possible braking force. II. The same tires on the same road result in the same force of friction. III. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.

A child goes down a playground slide that is inclined at an angle of \(26.5^{\circ}\) below the horizontal. Find the acceleration of the child given that the coefficient of kinetic friction between the child and the slide is 0.315.

See all solutions

Recommended explanations on Physics Textbooks

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.