/*! 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 90 Friction in an Elevator. You are... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Friction in an Elevator. You are riding in an elevator on the way to the 18th floor of your dormitory. The elevator is accelerating upward with \(a=190 \mathrm{m} / \mathrm{s}^{2}\) . Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg}\) . While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

Short Answer

Expert verified
The applied force must be approximately 105.05 N.

Step by step solution

01

Identify known quantities

We know that the mass of the box, \(m\), is 28.0 kg. The acceleration of the elevator, \(a\), is 1.90 m/s². The coefficient of kinetic friction, \(\mu_k\), is 0.32. The force we need to find is the horizontal force \(F\) to slide the box at constant speed.
02

Calculate the apparent weight

When the elevator accelerates upward, the apparent weight of the box changes. The apparent weight is the normal force \(N\), which can be calculated as: \( N = m(g + a) \). Here, \(g\) is the acceleration due to gravity, approximately 9.81 m/s². So, substitute the values: \( N = 28.0 (9.81 + 1.90) \).
03

Calculate the frictional force

The frictional force \(F_f\) can be found using the formula: \( F_f = \mu_k \cdot N \). Substitute the normal force from the previous step: \( F_f = 0.32 \cdot N \).
04

Set frictional force equal to applied force

To slide the box at a constant speed, the horizontal applied force \(F\) must counteract the frictional force (Newton's 1st law for constant velocity). Thus, \( F = F_f \). Use the result from the frictional force to find \( F \).
05

Solve for the applied force

Using the expression for \(F_f\), calculate \(F\). First, solve for \(N\): \( N = 28.0 (9.81 + 1.90) = 328.28 \) N. Then substitute into the frictional force: \( F_f = 0.32 \cdot 328.28 \approx 105.05 \) N. Thus, the required applied force \(F\) is 105.05 N.

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.

Kinetic Friction
When you slide any object, such as a box, across a surface, you face a physical phenomenon called friction. Friction is the force that tries to stop the motion between two surfaces that are in contact. There are two main types of friction: static and kinetic. Static friction exists when the object is not moving, while kinetic friction occurs when the object slides across a surface. In the context of our elevator scenario, kinetic friction is what you experience as you push the box across the elevator floor.

Kinetic friction depends on two main factors:
  • The nature of the surfaces in contact (characterized by the coefficient of kinetic friction, \( \mu_k \)). Each pair of surfaces has its own unique coefficient. Here, between the box and the elevator floor, it's 0.32.
  • The normal force exerted perpendicular to the contact surfaces.
The formula for calculating kinetic frictional force, \( F_f \), is:\[ F_f = \mu_k \cdot N \]Where \( N \) is the normal force. In our case, the frictional force must be offset by your applied force to move the box at a constant speed.
Normal Force
The normal force is a crucial concept in understanding problems involving friction. It is the force exerted by a surface, like the elevator floor, perpendicular to the object resting on it — in our case, the box. In physics, it acts as a counterbalance to the object's weight and any additional forces acting on it.

In the scenario where you are in a moving elevator, the normal force does not simply equal the weight of the object. Due to the upward acceleration of the elevator, the normal force increases. The formula for the normal force in this context becomes:\[ N = m(g + a) \]Here:
  • \( m \) is the mass of the box (28.0 kg).
  • \( g \) is the standard acceleration due to gravity (approximately 9.81 m/s²).
  • \( a \) is the upward acceleration of the elevator (1.90 m/s²).
Using these values, you can calculate the normal force, \( N \), as approximately 328.28 N. This adjusted normal force becomes the critical factor in determining the frictional force experienced by the box.
Applied Force
The applied force is what you exert when pushing the box toward the elevator door. For the box to maintain a constant speed as it slides, the applied force must exactly match the kinetic frictional force. This balance ensures that the net force acting on the box is zero (Newton's First Law of Motion).

When you apply just enough force to match the frictional resistance:\[ F = F_f \]Where:
  • \( F \) is the force you apply.
  • \( F_f \) is the kinetic frictional force.
In our situation, after computing the normal force, the frictional force calculated will be matched by your applied force. By determining that the frictional force \( F_f \) is roughly 105.05 N, you understand that you need to exert an applied force of 105.05 N to slide the box steadily.
Elevator Physics
Elevator physics involves understanding how motion and force interact within an accelerating system. Here, you are in an elevator accelerating upwards, impacting how forces are perceived. As the elevator speeds up, the apparent weight of everything inside changes, because the entire system feels extra force due to the upward acceleration.

This apparent weight change is critical in calculating forces acting on objects inside, like the box containing your new computer. When calculating the apparent weight, or normal force, you take into account both gravity and the elevator's acceleration:\[ N = m(g + a) \]This means that while the box’s real weight under gravity alone (without any elevator motion) would just be its mass times the gravitational force, the added upward push makes it seem heavier inside the elevator.

Grasping these changes allows you to accurately determine the necessary forces to apply — like the force needed to slide the box — making elevator physics a fascinating, real-world application of dynamics.

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

Rolling Friction. Two bicycle tires are set rolling with the same initial speed of 3.50 \(\mathrm{m} / \mathrm{s}\) on a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes 18.1 \(\mathrm{m}\) ; the other is at 105 psi and goes 92.9 \(\mathrm{m}\) . What is the coefficient of rolling friction \(\mu_{r}\) for each? Assume that the net horizontal force is due to rolling friction only.

You find it takes 200 \(\mathrm{N}\) of horizontal force to move an empty pickup truck along a level road at a speed of 2.4 \(\mathrm{m} / \mathrm{s}\) . You then load the pickup and pump up its tires so that its total weight increases by 42\(\%\) while the coefficient of rolling friction decreases by 19\(\%\) . Now what horizontal force will you need to move the pickup along the same road at the same speed? The speed is low enough that you can ignore air resistance.

Angle for Minimum Force. A box with weight w is pulled at constant speed along a level floor by a force \(\vec{F}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{k}\) (a) In terms of \(\theta, \mu_{k},\) and \(w,\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{k}=0.25\) , calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) . (c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F,\) required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta .\) ) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{x}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part \((\mathrm{b})\) .

A 75.0 -kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.0 \(\mathrm{kg}\) . (a) Find the maximum and minimum tension in the chain. \((b)\) What is the tension at a point three-fourths of the way up from the bottom of the chain?

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

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.