/*! 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 89 ssM A filing cabinet weighing \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

ssM A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68\), and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) 222 \(\mathrm{N}\), (b) \(334 \mathrm{~N},(\mathrm{c}) 445 \mathrm{~N}\), and \((\mathrm{d}) 556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

Short Answer

Expert verified
The cabinet moves in attempts (c) and (d).

Step by step solution

01

Calculate Maximum Static Friction

The maximum static friction is calculated by multiplying the weight of the cabinet by the coefficient of static friction: \( f_{s, \text{max}} = \mu_s \times \text{Weight} \). Here, \( \mu_s = 0.68 \) and the weight is \( 556 \mathrm{~N} \). So, \( f_{s, \text{max}} = 0.68 \times 556 = 378.08 \mathrm{~N} \).
02

Compare Applied Forces to Maximum Static Friction (Case a)

For a force of \( 222 \mathrm{~N} \): The applied force is less than \( 378.08 \mathrm{~N} \); hence, the frictional force is equal to the applied force, \( 222 \mathrm{~N} \). The cabinet doesn't move.
03

Compare Applied Forces to Maximum Static Friction (Case b)

For a force of \( 334 \mathrm{~N} \): Since \( 334 \mathrm{~N} < 378.08 \mathrm{~N} \), the frictional force matches the applied force, \( 334 \mathrm{~N} \). The cabinet remains stationary.
04

Compare Applied Forces to Maximum Static Friction (Case c)

For a force of \( 445 \mathrm{~N} \): This force exceeds \( 378.08 \mathrm{~N} \), so the cabinet starts moving. The frictional force switches to kinetic friction and is \( \mu_k \times 556 = 0.56 \times 556 = 311.36 \mathrm{~N} \).
05

Compare Applied Forces to Maximum Static Friction (Case d)

For a force of \( 556 \mathrm{~N} \): The cabinet moves as \( 556 \mathrm{~N} > 378.08 \mathrm{~N} \). The frictional force is given by kinetic friction, \( 311.36 \mathrm{~N} \).
06

Identify Attempts where Cabinet Moves

The cabinet starts moving when the applied force is greater than \( 378.08 \mathrm{~N} \); hence, it moves in attempts c and d.

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.

Frictional Force Calculation
When attempting to move an object like the filing cabinet in our example, understanding how to calculate frictional force is crucial. Frictional force is the force that resists the motion of two surfaces sliding against each other. There are two types of frictional forces, static and kinetic. Here, we focus on calculating the static frictional force that comes into play when the object is initially at rest.
To find the maximum static frictional force, use the formula:
  • \( f_{s, \text{max}} = \mu_s \times \text{Weight} \)
where \( f_{s, \text{max}} \) is the maximum static frictional force, \( \mu_s \) is the coefficient of static friction, and "Weight" is the weight of the object, here \( 556 \, \text{N} \).
Plugging the values in, we have:
  • \( f_{s, \text{max}} = 0.68 \times 556 = 378.08 \, \text{N} \)
This calculation tells us that the filing cabinet requires a force greater than \( 378.08 \, \text{N} \) to begin moving. If the applied force is less than \( 378.08 \, \text{N} \), the frictional force will equal the applied force, keeping the cabinet at rest.
Coefficient of Static Friction
The coefficient of static friction, denoted as \( \mu_s \), is a dimensionless value that represents the frictional resistance between two stationary objects. This coefficient varies depending on the materials in contact.
For our filing cabinet and floor system, \( \mu_s = 0.68 \). Higher coefficients indicate greater resistance to start moving.
To understand its role, consider how it affects the maximum static friction:
  • A greater \( \mu_s \) means the object requires more force to overcome the static friction and start moving.
  • It is crucial in determining whether an object will remain stationary or not when a force is applied.
In our scenario, when the applied force is less than \( 378.08 \, \text{N} \), \( \mu_s \) ensures the cabinet doesn't budge by providing sufficient static friction. Only when the force exceeds this threshold does the object begin to move, transitioning from static to kinetic friction.
Coefficient of Kinetic Friction
Once the object starts moving, the friction that acts upon it is known as kinetic friction. The coefficient of kinetic friction, \( \mu_k \), often has a smaller value compared to \( \mu_s \), indicating less resistance to continued motion once the object has started moving.
In our filing cabinet example, \( \mu_k = 0.56 \).
When the applied force surpasses the maximum static friction threshold, kinetic friction takes over. This is seen as:
  • The frictional force is calculated as \( f_k = \mu_k \times \text{Weight} \).
  • Plugging in the values gives us: \( f_k = 0.56 \times 556 = 311.36 \, \text{N} \).
Now, even though the cabinet moves, this frictional force counteracts to slow down or regulate its motion, making \( \mu_k \) vital in scenarios where constant force is applied to keep the object moving smoothly.

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

A roller-coaster car has a mass of \(1200 \mathrm{~kg}\) when fully loaded with passengers. As the car passes over the top of a circular hill of radius \(18 \mathrm{~m}\), its speed is not changing. At the top of the hill, what are the (a) magnitude \(F_{N}\) and (b) direction (up or down) of the normal force on the car from the track if the car's speed is \(v=11 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{N}\) and (d) the direction if \(v=\) \(14 \mathrm{~m} / \mathrm{s} ?\)

A cat dozes on a stationary merry-go-round, at a radius of \(5.4\) \(\mathrm{m}\) from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every \(6.0 \mathrm{~s}\). What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?

A baseball player with mass \(m=79 \mathrm{~kg}\), sliding into second base, is retarded by a frictional force of magnitude \(470 \mathrm{~N}\). What is the coefficient of kinetic friction \(\mu_{k}\) between the player and the ground?

A police officer in hot pursuit drives her car through a circular turn of radius \(300 \mathrm{~m}\) with a constant speed of \(80.0 \mathrm{~km} / \mathrm{h}\). Her mass is \(55.0 \mathrm{~kg}\). What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.)

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN}\), and the circle's radius is \(10 \mathrm{~m}\). At the top of the circle, what are the (a) magnitude \(F_{B}\) and (b) direction (up or down) of the force on the car from the boom if the car's speed is \(v=5.0 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{B}\) and (d) the direction if \(v=12 \mathrm{~m} / \mathrm{s} ?\)

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.