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Chapter 6: Problem 5

Hopping into your Porsche, you floor it and accelerate at \(12 \mathrm{m} / \mathrm{s}^{2}\) without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible.

Short Answer

Expert verified
The minimum coefficient of static friction needed is approximately 1.22.

Step by step solution

01

Understand the Problem

The problem asks us to find the minimum coefficient of static friction necessary for a car to accelerate at a rate of \(12 \text{ m/s}^2\) without slipping. Static friction is what prevents the tires from slipping on the road during acceleration.
02

Identify Relevant Equations

We need to use Newton's second law \( F = ma \), where \( F \) is the force of static friction, \( m \) is the mass of the car, and \( a \) is the acceleration. The force of static friction can also be expressed as \( F = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force, which equals \( mg \) (the weight of the car) when the car is on a flat surface.
03

Set Up the Equation

Set the two expressions for force equal to one another since frictional force provides the necessary force for acceleration: \( ma = \mu_s \cdot mg \). Here, \( m \) is the mass of the car, \( a \) is 12 m/s虏, \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏), and \( \mu_s \) is what we're solving for.
04

Solve for the Coefficient of Static Friction

From \( ma = \mu_s \cdot mg \), we can simplify the equation to \( a = \mu_s \cdot g \) after canceling \( m \) from both sides. Solve for \( \mu_s \) by rearranging to get \( \mu_s = \frac{a}{g} \). Substitute \( a = 12 \text{ m/s}^2 \) and \( g = 9.8 \text{ m/s}^2 \) to find \( \mu_s = \frac{12}{9.8} \approx 1.22 \).
05

Conclusion

The minimum coefficient of static friction required to accelerate at \(12 \text{ m/s}^2\) without slipping is approximately 1.22. This means the tires need to have a high level of grip on the road surface.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a cornerstone of classical mechanics, bridging the gap between an object's motion and the forces acting on it. This law can be expressed with the equation \( F = ma \), which signifies that the force \( F \) applied to an object is equal to the mass \( m \) of the object multiplied by the acceleration \( a \).
Understanding this relationship helps explain how an object will respond when subjected to external forces. For instance, in our exercise with the Porsche, the force required to accelerate the car was determined through this very law.
  • The mass \( m \) acts as a measure of an object's resistance to changes in motion
  • The acceleration \( a \) is the rate of change of velocity
  • Force \( F \) is what causes the acceleration

This ensures that the force of static friction will be strong enough to maintain grip, preventing the tires from slipping as the car accelerates.
Acceleration
Acceleration refers to the rate at which an object changes its velocity. It is a vector quantity, meaning it also has a direction, not just a magnitude. The acceleration in our exercise is given as \( 12 \text{ m/s}^2 \), which indicates how quickly the car's speed is increasing.
It is essential to understand:
  • Acceleration results from the application of force
  • The greater the force, the greater the acceleration, assuming mass remains constant

In circumstances like driving a car, the tires provide the necessary friction to translate forces into motion efficiently. Thus, achieving a prescribed acceleration depends on factors like the coefficient of static friction.
Coefficient of Static Friction
The coefficient of static friction, denoted by \( \mu_s \), is a dimensionless quantity that represents the ratio of the maximum static frictional force to the normal force. In simpler terms, it tells us how much grip the tires can have on the road before slipping occurs.
For the Porsche to accelerate at \( 12 \text{ m/s}^2 \), the minimum coefficient of static friction required, calculated as \( \mu_s = \frac{a}{g} \), turns out to be approximately \( 1.22 \).
  • Static friction is necessary to prevent the tires from sliding.
  • The higher the coefficient, the stronger the grip.
  • Typical values for \( \mu_s \) can vary based on the surfaces in contact and their conditions.

Understanding \( \mu_s \) is critical not only for the physics of motion but also for real-world applications like road safety and vehicle performance.

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Most popular questions from this chapter

At the airport, you pull a 18 -kg suitcase across the floor with a strap that is at an angle of \(45^{\circ}\) above the horizontal. Find (a) the normal force and (b) the tension in the strap, given that the suitcase moves with constant speed and that the coefficient of kinetic friction between the suitcase and the floor is 0.38 .

Force Times Time At the local hockey rink, a puck with a mass of \(0.12 \mathrm{kg}\) is given an initial speed of \(v_{0}=6.7 \mathrm{m} / \mathrm{s}\). (a) If the coefficient of kinetic friction between the ice and the puck is 0.13 , how much time \(t\) does it take for the puck to come to rest? (b) If the mass of the puck is doubled, does the frictional force \(F\) exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping time of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that \(F t=m v_{0}\). (The significance of this result will be discussed in Chapter \(9,\) where we will see that \(m v\) is the momentum of an object.)

Find the tension in each of the two ropes supporting a hammock if one is at an angle of \(18^{\circ}\) above the horizontal and the other is at an angle of \(35^{\circ}\) above the horizontal. The person sleeping in the hammock (unconcerned about tensions and ropes) has a mass of \(68 \mathrm{kg}\).

A physics textbook weighing \(22 \mathrm{N}\) rests on a table. The coefficient of static friction between the book and the table is \(\mu_{\mathrm{s}}=0.60 ;\) the coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.40 .\) You push horizontally on the book with a force that gradually increases from 0 to \(15 \mathrm{N}\), and then slowly decreases to \(5.0 \mathrm{N}\), as in dicated in the following table. For each value of the applied force given in the table, give the magnitude of the force of friction and state whether the book is accelerating, decelerating, at rest, or moving with constant speed.

A 48-kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to \(26^{\circ}\), the crate begins to slide downward. (a) What is the coefficient of static friction between the crate and the ramp? (b) At what angle does the crate begin to slide if its mass is doubled?
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