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Chapter 4: Problem 46

A cup of coffee is on a table in an airplane flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the plane accelerates forward, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?

Short Answer

Expert verified
Maximum acceleration is 2.94 m/s虏.

Step by step solution

01

Understand the Forces Involved

The cup of coffee is affected by both gravitational force and friction when the plane accelerates. The static friction between the cup and the table is the key force that prevents the cup from sliding backward. This frictional force must be greater than or equal to the force that the acceleration of the plane exerts on the cup to prevent sliding.
02

Identifying Frictional Force

The maximum static frictional force (\( f_{s,max} \)) can be computed using the formula: \( f_{s,max} = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction (0.30) and N is the normal force. Since the plane is flying at constant altitude, the normal force equals the weight of the cup (\( N = mg \)). Thus, \( f_{s,max} = \mu_s \cdot mg \).
03

Setting Up Newton's Second Law

As the plane accelerates, a force (\( F = ma \)) acts on the cup due to the plane's acceleration. The maximum acceleration before sliding can be found by setting this equal to the maximum frictional force: \( ma = \mu_s \cdot mg \).
04

Solving for Maximum Acceleration

We can solve for the maximum acceleration \( a \) by canceling the mass \( m \) from both sides of the equation: \( a = \mu_s \cdot g \). With \( \mu_s = 0.30 \) and \( g = 9.8 \text{ m/s}^2 \), the calculation is \( a = 0.30 \cdot 9.8 \text{ m/s}^2 = 2.94 \text{ m/s}^2 \).

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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 of Motion is fundamental in understanding how objects move when subjected to forces. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. This can be represented by the equation: \( F = ma \). Here,
  • \( F \) is the net force applied,
  • \( m \) is the mass of the object, and
  • \( a \) is the acceleration produced.
When the plane in our scenario starts to accelerate, it applies a force on the cup of coffee sitting on the table. This force tries to move the cup backward. However, the object will only move if the force is greater than the static friction holding it in place. In simpler words, the plane's acceleration creates a forward force against which the static friction must act to keep the cup stationary.
This principle helps in determining how the forces need to balance out to avoid any sliding motion.
Maximum Acceleration
In any scenario with moving objects and surfaces, like our airplane and cup, it鈥檚 crucial to understand the concept of maximum acceleration. Maximum acceleration is basically the greatest rate the plane can change its velocity without the external forces overcoming the static friction.The problem tells us that the plane undergoes acceleration. We then determine the plane's acceleration at which the cup just begins to slide鈥攅ssentially the point where it can't stay at rest. This is the crux of the problem.
To find this, we use the equation derived from Newton's Second Law: \( ma = \mu_s \cdot mg \), which simplifies to \( a = \mu_s \cdot g \). This equation stems from balancing the frictional force with the force due to the plane鈥檚 acceleration. Using the static friction coefficient \( \mu_s = 0.30 \) and gravity \( g = 9.8 \text{ m/s}^2 \), we calculate that the plane鈥檚 maximum acceleration is \( a = 2.94 \text{ m/s}^2 \). This value shows the threshold at which the cup will start sliding backward due to the plane's forward acceleration.
Coefficient of Friction
The coefficient of friction is a crucial factor in any problem involving surfaces in contact, and there are two types: static and kinetic. In this exercise, we focus on static friction, which is the force preventing relative motion between the cup and the table while the plane accelerates.
The coefficient of static friction \( \mu_s \) is a measure of how much force is necessary to start moving an object that is at rest. Every pair of surfaces has a unique friction coefficient based on their material properties.
The given coefficient in this problem is \( 0.30 \), meaning that the frictional force is 30% of the gravitational force acting on the cup. This allows us to determine the threshold force, or the maximum static friction before sliding occurs. The higher this coefficient, the more resistance to motion, making it essential in calculating the maximum acceleration before the cup starts its unwanted journey across the table.
  • Helps predict when sliding begins,
  • Depends on materials in contact, and
  • A key factor in real-world applications involving motion.
Understanding the role of \( \mu_s \) is vital to accurately predicting motion or ensuring stability.

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Most popular questions from this chapter

A 6.00-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is \(1.20 \mathrm{m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) accelerating downward with an acceleration whose magnitude is \(1.20 \mathrm{m} / \mathrm{s}^{2}\).

A man seeking to set a world record wants to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to the airplane. The mass of the man is 85 kg, and the coefficient of static friction between his shoes and the runway is 0.77. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

The helicopter in the drawing is moving horizontally to the right at a constant velocity \(\overrightarrow{\mathbf{v}}\). The weight of the helicopter is \(W=53800 \mathrm{N}\). The lift force \(\overrightarrow{\mathbf{L}}\) generated by the rotating blade makes an angle of \(21.0^{\circ}\) with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance \(\overrightarrow{\mathbf{R}}\) that opposes the motion.

A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg. The ball approaches the cushion with a velocity of \(+2.1 \mathrm{m} / \mathrm{s}\) and rebounds with a velocity of \(-2.0 \mathrm{m} / \mathrm{s} .\) The ball remains in contact with the cushion for a time of \(3.3 \times 10^{-3}\) s. What is the average net force (magnitude and direction) exerted on the ball by the cushion?

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m},\) and the mass of the earth is 81.4 times as great as that of the moon.
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