/*! 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 80 Dana loads luggage into an airpl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 80

Dana loads luggage into an airplane using a conveyor belt tilted at an angle of \(20^{\circ} .\) She places a few pieces of luggage on the belt before it starts to move, then she turns the belt on. It takes the belt \(0.70 \mathrm{s}\) to reach its top speed of \(1.2 \mathrm{m} / \mathrm{s}\). Does the luggage slip? Assume \(\mu_{\mathrm{s}}=0.50\) between the luggage and the belt.

Short Answer

Expert verified
Using the calculations, we will find if the conveyor belt's acceleration is greater than the maximum acceleration the luggage can have without slipping. Based on this comparison, we can determine whether the luggage will slip or not.

Step by step solution

01

Calculate Conveyor Belt's Acceleration

To find the acceleration of the conveyor belt, we can use the formula for acceleration, which is final speed minus initial speed divided by time. For our scenario, final speed is \(1.2 m/s\), the initial speed is \(0 m/s\) since the belt starts from rest, and the time it takes the belt to reach its top speed is \(0.70 s\). The calculation is as follows: \(a = (1.2 m/s - 0 m/s ) / 0.70 s\)
02

Calculate Maximum Acceleration without Slipping

Next, we calculate the maximum acceleration that the luggage can have without slipping by using the equation of motion under friction: \(f_{max} = \mu_{s} * m * g * cos\theta\), where \(g\) is the acceleration due to gravity (approximately \(9.8 m/s^2\)), \(\mu_{s}\) is the static friction coefficient and \(\theta\) is the angle of incline. Solve for maximum acceleration \(a_{max}\) by dividing \(f_{max}\) by mass \(m\): \(a_{max} = f_{max} / m = \mu_{s} * g * cos\theta / m\). Note that the mass \(m\) cancels out in both numerator and denominator, simplifying to: \(a_{max} = \mu_{s} * g * cos\theta\). In this problem, \(\mu_{s}\) is given as \(0.50\) and \(\theta\) as \(20^\circ\).
03

Compare Accelerations for Slipping

Finally, compare the acceleration \(a\) of the conveyor belt derived in Step 1 to the maximum acceleration \(a_{max}\) the luggage can have without slipping calculated in Step 2. If \(a > a_{max}\) then the luggage will slip, else it will not.

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 91影视!

Key Concepts

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

Static Friction
Static friction is the force that resists the initiation of sliding motion of two surfaces. Imagine you're trying to push a heavy box. Initially, it doesn't move because static friction holds it in place. This frictional force must be overcome for movement to begin. It depends on two factors:

  • The roughness of the surfaces in contact, described by the coefficient of static friction \(\mu_s\).
  • The normal force, which acts perpendicular to the contact surfaces.
The maximum static frictional force can be calculated with the formula \(f_{\text{max}} = \mu_s \cdot N\), where \(N\) is the normal force.

In our scenario, we determine the maximum acceleration the luggage can have without slipping. This ensures it remains stationary until the conveyor belt's acceleration surpasses this limit.
Inclined Planes
Inclined planes are surfaces tilted at an angle relative to a horizontal surface. They are common in physics problems because they introduce additional forces acting on objects. When dealing with an inclined plane, consider:

  • The angle of the incline, which impacts the gravitational force component acting along the plane.
  • The normal force, which acts perpendicular to the plane's surface.
In a problem like Dana's, the incline angle of \(20掳\) affects how the luggage interacts with the conveyor belt. The component of gravity down the plane is \(mg \sin \theta\), and the normal force is \(mg \cos \theta\). This normal force is crucial for calculating static friction.
Kinematics
Kinematics deals with the motion of objects without considering the forces causing them. In this problem, we focus on how the conveyor belt moves to determine if the luggage will slip. Key aspects include:

  • Initial and final velocity, which for the conveyor belt is \(0 \, \text{m/s}\) and \(1.2 \, \text{m/s}\) respectively.
  • Time, which is \(0.70 \, \text{s}\) for the belt to reach its top speed.
  • Acceleration, calculated as \(a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}}\).
Using these components, you can solve for the belt's acceleration and then compare it to the maximum allowed by static friction to determine if slipping occurs.
Newton's Laws of Motion
Newton's Laws describe the relationship between a body and the forces acting upon it. They are fundamental to understanding motion. Key law applicable here is Newton's Second Law:

  • Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass: \(F = ma\).
For the luggage, the force of static friction provides the net force required to prevent sliding. By solving for the maximum acceleration using \(f_{\text{max}} = \mu_s \cdot mg \cos \theta\), we apply Newton鈥檚 Second Law to find the limiting factor for luggage slipping on the moving belt.

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

Riders on the Tower of Doom, an amusement park ride, experience \(2.0 \mathrm{s}\) of free fall, after which they are slowed to a stop in 0.50 s. What is a 65 kg rider's apparent weight as the ride is coming to rest? By what factor does this exceed her actual weight?

A \(50 \mathrm{kg}\) box hangs from a rope. What is the tension in the rope if a. The box moves up at a steady \(5.0 \mathrm{m} / \mathrm{s} ?\) b. The box has \(v_{y}=5.0 \mathrm{m} / \mathrm{s}\) and is slowing down at \(5.0 \mathrm{m} / \mathrm{s}^{2} ?\)

A \(10 \mathrm{kg}\) crate is placed on a horizontal conveyor belt. The materials are such that \(\mu_{\mathrm{s}}=0.50\) and \(\mu_{\mathrm{k}}=0.30\) a. Draw a free-body diagram showing all the forces on the crate if the conveyer belt runs at constant speed. b. Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up. c. What is the maximum acceleration the belt can have without the crate slipping? If the acceleration of the belt exceeds the value determined in part c, what is the acceleration of the crate?

Josh starts his sled at the top of a \(3.0-\mathrm{m}\) -high hill that has a constant slope of \(25^{\circ} .\) After reaching the bottom, he slides across a horizontal patch of snow. Ignore friction on the hill, but assume that the coefficient of kinetic friction between his sled and the horizontal patch of snow is \(0.050 .\) How far from the base of the hill does he end up?

In a head-on collision, a car stops in 0.10 s from a speed of \(14 \mathrm{m} / \mathrm{s} .\) The driver has a mass of \(70 \mathrm{kg},\) and is, fortunately, tightly strapped into his seat. What force is applied to the driver by his seat belt during that fraction of a second?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Electricity

Read Explanation

Famous Physicists

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Rotational Dynamics

Read Explanation

Force

Read Explanation
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.