/*! 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 51 Suppose you stand on a bathroom ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you stand on a bathroom scale and get a reading of \(700 \mathrm{N}\). In principle, would the scale read more, less, or the same if the Earth did not rotate?

Short Answer

Expert verified
The scale would read more if the Earth did not rotate.

Step by step solution

01

Understand the Problem

You are standing on a bathroom scale, which measures your weight. The weight is given as the force exerted by the scale on you, and currently reads 700 N. We are asked to determine how this reading would change if the Earth did not rotate.
02

Identify the Forces acting on You

Normally, the gravitational force acts downward on you, and the scale provides an equal upward normal force of 700 N. Due to the Earth's rotation, there is also a small centrifugal force acting outward which slightly reduces the normal force read by the scale.
03

Consider Changes due to Earth's Non-Rotation

If the Earth did not rotate, the centrifugal force would be absent. As a result, the normal force (and hence the scale reading) would increase to equal the full gravitational force.
04

Relate Changes in Force to Scale Reading

Without the centrifugal force, the force exerted by the Earth on you (true weight) does not change, but the reading by the scale (apparent weight) would be higher because it no longer has to counteract any outward force due to rotation.
05

Conclusion of Effects on Scale Reading

Therefore, without the Earth's rotation reducing the scale reading through centrifugal force effects, the scale would read a higher number than 700 N.

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Key Concepts

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

Gravitational Force
Gravitational force is the attractive force that the Earth exerts on objects. It pulls objects towards its center and is given by the formula:
  • \( F_g = mg \), where \( F_g \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity.
This force is what we commonly refer to as "weight." Imagine a magnet attracting a metal piece—gravity does something similar by pulling everything towards Earth.
When you stand on a scale, the scale measures this gravitational force acting on your body. That's why your scale reading reflects your weight. In our scenario, the gravitational force is the same whether Earth rotates or not. However, the scale reading might not always match the gravitational force exactly if other forces, like centrifugal force, are involved.Understanding gravitational force helps us unravel why scales show a different reading when Earth's rotation comes into play. The gravitational force remains constant, but how it's balanced or appears to be balanced (apparent weight) by other forces varies.
Normal Force
Normal force is the support force exerted by surfaces perpendicular to the surface. When you stand on a scale, the scale provides an upward force equal to your gravitational force, which we call the normal force.
It counteracts the force of gravity, ensuring that you remain in equilibrium and do not fall through the scale. Since the scale shows the force it exerts (the normal force), in the absence of other active forces, it reflects the gravitational force directly. However, if the Earth's rotation is considered, a small reduction occurs in the normal force due to the centrifugal force acting outward, which we will explore in the next section. This slight reduction is what makes the scale read less than the gravitational force under normal circumstances. But if Earth did not rotate, this centrifugal effect wouldn't exist, and the normal force would be unrestricted by any reducing factors, resulting in a higher reading on the scale.
Earth's Rotation
The Earth's rotation introduces an interesting, albeit slightly complex, force called centrifugal force. This force acts outward from the axis of rotation. Due to Earth's fast spin, centrifugal force affects objects on its surface, including you, by slightly reducing the apparent weight measured by a scale.
This happens because while the gravitational force pulls you down, the centrifugal force tries to push you away from the axis of rotation. It slightly counteracts gravity, diminishing the normal force exerted by the Earth. Thus, the scale displays a lower weight than your true gravitational weight. If Earth stopped rotating, the centrifugal effect would vanish. Without anything opposing the gravitational pull, the normal force would fully balance the gravitational force without any reduction. This makes scales show the complete gravitational weight, producing a higher reading compared to a rotating Earth scenario.

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Most popular questions from this chapter

Find the centripetal acceleration at the top of a test tube in a centrifuge, given that the top is \(4.2 \mathrm{cm}\) from the axis of rotation and that its linear speed is \(77 \mathrm{m} / \mathrm{s}\).

Your friend's 13.6-g graduation tassel hangs on a string from his rearview mirror. (a) When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car. Explain. (b) If the tassel hangs at an angle of \(6.44^{\circ}\) relative to the vertical, what is the acceleration of the car?

Playing a Violin The tension in a violin string is \(2.7 \mathrm{N}\). When pushed down against the neck of the violin, the string makes an angle of \(4.1^{\circ}\) with the horizontal. (a) With what force must you push down on the string to bring it into contact with the neck? (b) If the angle were less than \(4.1^{\circ}\), would the required force be greater than, less than, or the same as in part (a)? Explain.

A 0.16-g spider hangs from the middle of the first thread of its future web. The thread makes an angle of \(7.2^{\circ}\) with the horizontal on both sides of the spider. (a) What is the tension in the thread? (b) If the angle made by the thread had been less than \(7.2^{\circ}\), would its tension have been greater than, less than, or the same as in part (a)? Explain.

Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 52,000 times the acceleration of gravity. The distance from the axis of rotation to the bottom of the test tube is \(7.5 \mathrm{cm}\).

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