/*! 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 83 You must push a crate across a f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You must push a crate across a floor to a docking bay. The crate weighs \(165 \mathrm{~N}\). The coefficient of static friction between crate and floor is \(0.510\), and the coefficient of kinetic friction is \(0.32\). Your force on the crate is directed horizontally. (a) What magnitude of your push puts the crate on the verge of sliding? (b) With what magnitude must you then push to keep the crate moving at a constant velocity? (c) If, instead, you then push with the same magnitude as the answer to (a), what is the magnitude of the crate's acceleration??

Short Answer

Expert verified
(a) \( 84.15\, \text{N} \), (b) \( 52.8\, \text{N} \), (c) \( 1.86\, \text{m/s}^2 \).

Step by step solution

01

Calculate the Force Needed to Start Sliding

To find the force needed to put the crate on the verge of sliding, use the static friction formula:\[ F_{\text{static}} = \mu_s \times N \]where \( \mu_s = 0.510 \) is the coefficient of static friction and \( N = 165\, \text{N} \) is the normal force, which is equal to the weight of the crate.\[ F_{\text{static}} = 0.510 \times 165 = 84.15\, \text{N} \]Thus, the force needed to start sliding is approximately \( 84.15\, \text{N} \).
02

Calculate the Force to Maintain Constant Velocity

Once the crate is sliding, the force required to keep it moving at constant velocity is determined by the kinetic friction:\[ F_{\text{kinetic}} = \mu_k \times N \]where \( \mu_k = 0.32 \) is the coefficient of kinetic friction.\[ F_{\text{kinetic}} = 0.32 \times 165 = 52.8\, \text{N} \]Therefore, the force required to maintain constant velocity is \( 52.8\, \text{N} \).
03

Calculate the Acceleration When Pushing With Initial Force

If the same force as in part (a), \( 84.15\, \text{N} \), is applied while the crate is moving, we calculate the acceleration by using Newton's second law, \( F = ma \), where \( F \) is the net force and \( m = \frac{165}{9.8}\, \text{kg} \) is the mass of the crate.First, determine the net force:\[ F_{\text{net}} = 84.15 - 52.8 = 31.35\, \text{N} \]Calculate the mass:\[ m = \frac{165}{9.8} \approx 16.84\, \text{kg} \]Now, find the acceleration:\[ a = \frac{F_{\text{net}}}{m} = \frac{31.35}{16.84} \approx 1.86 \, \text{m/s}^2 \]Thus, the crate's acceleration is approximately \( 1.86\, \text{m/s}^2 \).

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.

Static Friction
Static friction is the force that keeps an object at rest when a force is applied to it. It's like a stubborn friend who resists change. The key player here is the 'coefficient of static friction' which is denoted by \( \mu_s \). This is a ratio and tells us how strong the static friction is compared to the normal force, which is the force pressing the two surfaces together, like gravity in this case. For example, with a crate weighing 165 N, and a \( \mu_s \) of 0.510, the static friction force can be calculated using the formula:
  • \( F_{\text{static}} = \mu_s \times N \)
  • \( F_{\text{static}} = 0.510 \times 165 = 84.15 \, \text{N} \)
This means you need to apply at least 84.15 N of force to start moving the crate. If you apply less force, static friction will hold the crate in place.
Kinetic Friction
Once the object starts moving, it’s not static anymore but kinetic. Kinetic friction takes over and acts to slow down the moving object. This frictional force is usually less than static friction, which is why it's easier to keep the object moving after it starts than to get it moving in the first place.The coefficient of kinetic friction, \( \mu_k \), helps to determine this and is used to calculate the force needed to maintain motion:
  • \( F_{\text{kinetic}} = \mu_k \times N \)
  • \( F_{\text{kinetic}} = 0.32 \times 165 = 52.8 \, \text{N} \)
By applying a force of 52.8 N, you can keep the crate moving at a constant velocity. This balance occurs because the kinetic friction counteracts your applied force, showing why it isn't harder to maintain a constant velocity once achieved.
Newton's Second Law
Newton's Second Law of Motion tells us how forces affect the motion of an object. It's the principle stating that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. Mathematically, it's expressed as \( F = ma \) where:
  • \( F \) is the net force.
  • \( m \) is the mass of the object.
  • \( a \) is the acceleration.
In the context of the moving crate, if you initially push with the force needed to overcome static friction (84.15 N) while the crate is in motion, the net force was calculated as:
  • \( F_{\text{net}} = 84.15 - 52.8 = 31.35 \, \text{N} \)
Given that the weight of the crate gives its mass:
  • \( m = \frac{165}{9.8} \approx 16.84 \, \text{kg} \)
The acceleration \( a \) can then be found as:
  • \( a = \frac{31.35}{16.84} \approx 1.86 \, \text{m/s}^2 \)
This gives insight into how additional forces beyond kinetic friction affect an object's acceleration.
Force and Motion
Force and motion are interdependent concepts in physics. Forces are necessary to change the motion of an object. They can cause stationary objects to move or moving objects to stop, change direction, speed up, or slow down. When you understand this relationship:
  • Forces applied to a stationary object must overcome static friction first to start motion.
  • Once in motion, kinetic friction becomes relevant, often reducing the required force to maintain movement.
  • Consistent force applied to a moving object results in constant velocity, while unbalanced forces lead to acceleration as explained by Newton’s second law.
By applying these principles, you can predict and manipulate how objects will move under different forces, such as pushing a crate across a floor. Knowing how frictional forces act helps to achieve the desired motion efficiently and effectively, showing the vital role of forces in everyday physics.

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

ssM A car weighing \(10.7 \mathrm{kN}\) and traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) without negative lift attempts to round an unbanked curve with a radius of \(61.0 \mathrm{~m} .(\mathrm{a})\) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is \(0.350\), is the attempt at taking the curve successful?

Suppose the coefficient of static friction between the road and the tires on a car is \(0.60\) and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

A house is built on the top of a hill with a nearby slope at angle \(\theta=45^{\circ}\) (Fig. 6-55). An engineering study indicates that the slope angle should be reduced because the top layers of soil along the slope might slip past the lower layers. If the coefficient of static friction between two such layers is \(0.5\), what is the least angle \(\phi\) through which the present slope should be reduced to prevent slippage?

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?

ssM A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg}\). Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the net force on the bicycle from the road.

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.