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Chapter 6: Problem 29

(II) What will a spring scale read for the weight of a \(53-\mathrm{kg}\) woman in an elevator that moves \((a)\) upward with constant speed \(5.0 \mathrm{~m} / \mathrm{s},(b)\) downward with constant speed \(5.0 \mathrm{~m} / \mathrm{s}\) (c) upward with acceleration \(0.33 g,(d)\) downward with acceleration \(0.33 g\), and \((e)\) in free fall?

Short Answer

Expert verified
(a) and (b) 519.4 N, (c) 696.1 N, (d) 345.9 N, (e) 0 N.

Step by step solution

01

Understanding the Problem

We need to find what the spring scale will read (the apparent weight) of a woman inside an elevator under different motion conditions.
02

Analyze the Forces

The forces acting on the woman are the gravitational force (weight) and the normal force exerted by the scale. The scale reading will be equal to the magnitude of the normal force.
03

Calculate Weight with Constant Speed

For parts (a) and (b), since the elevator moves with constant speed, the acceleration is zero. The apparent weight is equal to the real weight. \[W = mg = 53 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 519.4 \, \text{N}\]
04

Calculate Weight with Upward Acceleration

For part (c), when the elevator accelerates upward, the apparent weight is greater due to the additional upward force:\[W' = m(g + a) = 53 \, \text{kg} \times (9.8 + 0.33 \times 9.8) \, \text{m/s}^2\]Calculate this as:\[W' = 53 \, \text{kg} \times 13.134 \, \text{m/s}^2 = 696.102 \, \text{N}\]
05

Calculate Weight with Downward Acceleration

For part (d), when the elevator accelerates downward, the apparent weight is reduced:\[W' = m(g - a) = 53 \, \text{kg} \times (9.8 - 0.33 \times 9.8) \, \text{m/s}^2\]Calculate this as:\[W' = 53 \, \text{kg} \times 6.526 \, \text{m/s}^2 = 345.878 \, \text{N}\]
06

Calculate Weight in Free Fall

For part (e), in free fall, the acceleration is equal to gravity, thus the apparent weight is zero because the woman and the scale are both accelerating downward at the same rate:\[W' = m(g - g) = 0 \, \text{N}\]
07

Conclude with Results

Summarize the results: (a) and (b) = 519.4 N (c) = 696.102 N (d) = 345.878 N (e) = 0 N

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Key Concepts

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

Apparent Weight
The concept of apparent weight is one that we often experience, especially in scenarios involving elevators or amusement park rides.
Apparent weight is how heavy an object feels under different conditions of acceleration.
It stems from the normal force, which is the supportive force exerted by a surface, like a floor or a scale.
  • If you are in an elevator moving at a constant speed or stationary, your apparent weight feels like your actual weight.
  • However, if the elevator speeds up or slows down, your apparent weight may feel different.
In calm or constant speed scenarios, the forces are balanced, and the apparent weight is equal to the gravitational force acting on your body.
When there's acceleration, your perceived weight changes due to the additional forces involved.
Elevator Physics
Elevator physics is an intriguing subject that merges everyday experiences with fundamental physics concepts. Elevators often serve as a prime example of moving reference frames in physics.
The main forces at play include gravitational force and the normal force from the elevator floor.
When the elevator is ascending or descending at constant speeds, these forces balance out, creating a scenario of zero net force, meaning no change in the apparent weight.
  • Upward Acceleration: Causes an increase in apparent weight as the additional force required to move upwards is aligned with gravity.
  • Downward Acceleration: Leads to a decrease in apparent weight because the required force opposes gravity.
  • Free Fall: Your apparent weight drops to zero because both you and the elevator are accelerating downwards at the same rate.
In essence, the behavior of elevators is an excellent touchpoint for understanding motion and forces in a straightforward, relatable manner.
Gravitational Force
Gravitational force is a ubiquitous and constant force that acts on all objects with mass.
It is the pull that attracts objects towards each other, with its strongest example being Earth's pull on objects towards its center.
Mathematically, gravitational force is described as:
  • Weight ( W ) = mass ( m ) 脳 gravitational acceleration ( g ).
  • On Earth, g is approximately 9.8 ext{m/s}^2.
This force is what gives objects weight and informs how much force you need to apply to lift something.
The gravitational force remains consistent as it depends solely on your mass and the gravity constant, making it unaffected by the elevator's speed or movement direction.
Normal Force
Normal force is the supportive force that surfaces exert perpendicularly against an object resting on them.
This force is crucial as it balances the gravitational force when you are standing on a surface, preventing you from falling through.
  • In elevators moving at constant speeds, the normal force equals gravitational force, resulting in no change to your actual weight.
  • During acceleration, the normal force shifts, either increasing or decreasing, to account for the additional forces.
  • In free-fall scenarios, the normal force equates to zero because there is no support from the elevator floor, signifying zero apparent weight.
The normal force adapts based on the environment and the conditions, playing a versatile role that ensures balance against gravitational pull.

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Most popular questions from this chapter

(III) It can be shown (Appendix D) that for a uniform sphere the force of gravity at a point inside the sphere depends only on the mass closer to the center than that point. The net force of gravity due to points outside the radius of the point cancels. How far would you have to drill into the Earth, to reach a point where your weight is reduced by \(5.0 \% ?\) Approximate the Earth as a uniform sphere.

You are an astronaut in the space shuttle pursuing a satellite in need of repair. You are in a circular orbit of the same radius as the satellite \((400 \mathrm{km}\) above the Earth), but 25 \(\mathrm{km}\) behind it. (a) How long will it take to overtake the satellite if you reduce your orbital radius by 1.0 \(\mathrm{km}\) ? (b) By how much must you reduce your orbital radius to catch up in 7.0 \(\mathrm{h}\) ?

Jupiter is about 320 times as massive as the Earth. Thus, it has been claimed that a person would be crushed by the force of gravity on a planet the size of Jupiter since people can't survive more than a few \(g\) 's. Calculate the number of \(g\) 's a person would experience at the equator of such a planet. Use the following data for Jupiter: mass \(=1.9 \times 10^{27} \mathrm{~kg}, \quad\) equatorial radius \(=7.1 \times 10^{4} \mathrm{~km}\) rotation period \(=9 \mathrm{hr} 55 \mathrm{~min} .\) Take the centripetal acceleration into account.

What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is \(\frac{1}{10}\) of its value at the Earth's surface?

How long would a day be if the Earth were rotating so fast that objects at the equator were apparently weightless?
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