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Chapter 5: Problem 33

The coefficient of kinetic friction between a \(40 \mathrm{~kg}\) crate and the warehouse floor is \(70 \%\) of the corresponding coefficient of static friction. The crate falls off a forklift that is moving at \(3 \mathrm{~m} / \mathrm{s}\) and then slides along the warehouse floor for a distance of \(2.5 \mathrm{~m}\) before coming to rest. What is the coefficient of static friction between the crate and the floor?

Short Answer

Expert verified
The coefficient of static friction is approximately 0.2624.

Step by step solution

01

Understanding the Problem

We need to calculate the coefficient of static friction (\( \mu_s \)) based on the information given about the kinetic friction and the motion of the crate.
02

Calculate Deceleration from Slide

Given the crate slides a distance of \( 2.5 \text{ m} \) after falling off with an initial velocity of \( 3 \text{ m/s} \) and comes to rest, we can use the formula \( v^2 = u^2 + 2as \) to find the deceleration \( a \). Here, \( v = 0 \), \( u = 3 \text{ m/s} \), and \( s = 2.5 \text{ m} \). Substituting these values gives:\[ 0 = (3)^2 + 2a(2.5) \]\[ 0 = 9 + 5a \]\[ a = -\frac{9}{5} = -1.8 \text{ m/s}^2 \]
03

Relating Deceleration to Kinetic Friction

The force causing the deceleration is due to kinetic friction. We use the formula \( F = ma = \mu_k mg \), where \( \mu_k \) is the coefficient of kinetic friction and \( g = 9.8 \text{ m/s}^2 \). Substituting \( a = -1.8 \text{ m/s}^2 \):\[ 40(-1.8) = \mu_k (40)(9.8) \]\[ \mu_k = \frac{1.8}{9.8} \approx 0.1837 \]
04

Determine Static Friction Coefficient

We are given that the kinetic friction coefficient is \( 70\% \) of the static friction coefficient, i.e., \( \mu_k = 0.7\mu_s \). We can now express this as:\[ \mu_s = \frac{\mu_k}{0.7} \]Substituting the kinetic friction value:\[ \mu_s = \frac{0.1837}{0.7} \approx 0.2624 \]
05

Final Answer

The coefficient of static friction between the crate and the floor is approximately \( 0.2624 \).

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Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the motion of two surfaces sliding past each other. In this lesson, imagine pushing a crate across a floor. As you push, kinetic friction acts in the opposite direction of your push. This force is crucial because it determines how quickly an object slows down once it's set into motion.
  • The magnitude of kinetic friction depends on two factors: the nature of the surfaces in contact and the normal force, which is the force perpendicular to the contact surface.
  • The coefficient of kinetic friction (\(\mu_k\)) is a dimensionless value that represents how much frictional force will be generated between different materials.
  • This coefficient is generally less than the coefficient of static friction, meaning it's easier to keep an object moving than to start moving it from rest.
Understanding kinetic friction helps in predicting the motion of objects and is essential for solving many physics problems involving moving objects.
Deceleration Calculation
Calculating deceleration is all about understanding how quickly an object slows down. In our exercise, a crate falls off a forklift at an initial speed and then stops after sliding a certain distance.
  • The formula used for calculating deceleration when an object comes to rest is given by \(v^2 = u^2 + 2as\).
  • Here, \(v\) is the final velocity (0, because the crate stops), \(u\) is the initial velocity, \(a\) is the acceleration (or deceleration, which is negative in slowing down), and \(s\) is the distance over which the object slows down.
Using this formula helps to find out how forces like friction affect the movement of objects and enables the prediction of how different materials interact when moving.
Physics Problem Solving
Physics problem solving involves a structured approach to understanding the principles at play in given situations, like our crate falling off a forklift.
  • First, clearly understand all given information and what the problem is asking for.
  • Identify the underlying physics concepts, such as friction, motion, or force, involved in the scenario.
  • Apply relevant equations and mathematical formulas, such as the equations of motion or frictional force equations, to solve for the unknown variables.
  • Finally, reason through your solution to ensure it makes sense in the context of the problem and check your calculations for accuracy.
Developing these skills is key to solving a wide range of physics problems effectively and efficiently.
Force and Motion
Force and motion are fundamental concepts in physics that describe how objects move and interact. When discussing the crate and the warehouse floor, these concepts explain how and why the crate decelerates after falling.
  • Force is an interaction that changes the motion of an object - it can start, stop, or change the direction or speed of an object.
  • Newton's Second Law of Motion, \(F = ma\), tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
  • In our scenario, the force of friction is acting opposite to the motion, causing the crate to decelerate.
  • Understanding the relationship between force and motion helps us predict how objects will move under various conditions.
These principles not only apply to crates in a warehouse but are also essential in many real-world applications, from designing vehicles to building infrastructure.

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Most popular questions from this chapter

An average person can reach a maximum height of about \(60 \mathrm{~cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around \(50 \mathrm{~cm}\). To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of \(60 \mathrm{~cm} ?\) (b) Make a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(\underline{W}\). what force does the ground exert on him or her during the jump?

A \(2 \mathrm{~kg}\) book sits at rest on a horizontal table. The coefficient of static friction between the book and the surface is \(0.40,\) and the coefficient of kinetic friction is 0.20 . (a) What is the normal force acting on the book? (b) Is there a friction force on the book? (c) What minimum horizontal force would be required to cause the book to slide on the table? (d) If you give the book a strong horizontal push so that it begins sliding, what kind of force will cause it to come to rest? (e) What is the magnitude of this force?

A \(750.0 \mathrm{~kg}\) boulder is raised from a quarry \(125 \mathrm{~m}\) deep by a long chain having a mass of \(575 \mathrm{~kg} .\) This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking. (a) What is the maximum acceleration the boulder can have and still get out of the quarry, and (b) how long does it take for the boulder to be lifted out at maximum acceleration if it started from rest?

An \(80 \mathrm{~N}\) box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and the table are \(\frac{1}{4}\) and \(\frac{1}{2},\) respectively. What is the friction (b) \(25 \mathrm{~N}\), force on this box if the tension in the rope is (a) \(0 \mathrm{~N},]\) (c) \(39 \mathrm{~N},(\mathrm{~d}) 41 \mathrm{~N},(\mathrm{e}) 150 \mathrm{~N} ?\)

One straightforward way to measure the coefficients of friction between a box and a wooden surface is illustrated in Figure \(5.59 .\) The sheet of wood can be raised by pivoting it about one edge. It is first raised to an angle \(\theta_{1}\) (which is measured) for which the box just begins to slide downward. The sheet is then immediately lowered to an angle \(\theta_{2}\) (which is also measured) for which the box slides with constant speed down the sheet. Apply Newton's second law to the box in both cases to find the coefficients of kinetic and static friction between it and the wooden sheet in terms of the measured angles \(\theta_{1}\) and \(\theta_{2}\).
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