Problem 3
Question
(I) Suppose you are standing on a train accelerating at \(0.20 \mathrm{~g}\). What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?
Step-by-Step Solution
Verified Answer
The minimum coefficient of static friction is 0.20.
1Step 1: Understanding the Problem
The problem asks for the minimum coefficient of static friction required to prevent sliding while standing on an accelerating train. The given acceleration of the train is in terms of gravitational acceleration, denoted as \( g \), where \( g = 9.8 \, \text{m/s}^2 \).
2Step 2: Identify Known Quantities
The acceleration of the train \( a \) is given as \( 0.20 \times g \). Let \( a = 0.20g = 0.20 \times 9.8 \, \text{m/s}^2 = 1.96 \, \text{m/s}^2 \).
3Step 3: Apply Newton's Laws
According to Newton's first law, static friction must counteract the horizontal force caused by the train's acceleration to prevent sliding. This force is given by \( F = ma \), where \( m \) is mass and \( a \) is acceleration.
4Step 4: Relate Static Friction to Force and Coefficient
The force of static friction \( F_{\text{friction}} \) is determined by \( F_{\text{friction}} = \mu_s N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. Here, \( N = mg \).
5Step 5: Set Up Equation for Minimum Friction
To prevent sliding, static friction must be at least equal to the force due to horizontal acceleration, i.e., \( \mu_s mg = ma \).
6Step 6: Solve for the Coefficient of Static Friction
Rearranging the equation, we have \( \mu_s = \frac{a}{g} = \frac{0.20g}{g} = 0.20 \).
Key Concepts
Newton's LawsStatic FrictionGravitational Acceleration
Newton's Laws
Newton's Laws of motion are essential for understanding the behavior of objects in motion or at rest. In the context of this problem, Newton's first law, also known as the law of inertia, is particularly relevant. This law states that an object will remain at rest or in uniform motion unless acted upon by a net external force.
When you're standing on an accelerating train, the key challenge is to remain stationary relative to the train. The acceleration of the train creates a horizontal force pushing against your body. According to Newton's first law, this requires a counteracting force to prevent sliding.
Newton's second law, given by the formula \( F = ma \), helps us calculate the amount of force acting on your body due to the train's acceleration. Here, \( F \) is the force, \( m \) is your mass, and \( a \) is the train's acceleration. By calculating this force, we can determine the amount of static friction necessary to prevent sliding.
When you're standing on an accelerating train, the key challenge is to remain stationary relative to the train. The acceleration of the train creates a horizontal force pushing against your body. According to Newton's first law, this requires a counteracting force to prevent sliding.
Newton's second law, given by the formula \( F = ma \), helps us calculate the amount of force acting on your body due to the train's acceleration. Here, \( F \) is the force, \( m \) is your mass, and \( a \) is the train's acceleration. By calculating this force, we can determine the amount of static friction necessary to prevent sliding.
Static Friction
Static friction is the force that keeps objects stationary relative to each other, preventing them from sliding across surfaces. It acts between your feet and the train's floor when you stand. The static friction force must be strong enough to counteract the force exerted by the train's acceleration.
The static friction force \( F_{\text{friction}} \) can be expressed as \( F_{\text{friction}} = \mu_s N \), where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force. The normal force in this scenario equals the gravitational force on your body, which is \( mg \), where \( m \) is your mass and \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \)).
To prevent sliding, the static friction force must be at least equal to the force generated by the train's acceleration. Thus, the coefficient of static friction \( \mu_s \) plays a crucial role in ensuring you remain standing without moving.
The static friction force \( F_{\text{friction}} \) can be expressed as \( F_{\text{friction}} = \mu_s N \), where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force. The normal force in this scenario equals the gravitational force on your body, which is \( mg \), where \( m \) is your mass and \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \)).
To prevent sliding, the static friction force must be at least equal to the force generated by the train's acceleration. Thus, the coefficient of static friction \( \mu_s \) plays a crucial role in ensuring you remain standing without moving.
Gravitational Acceleration
Gravitational acceleration is the acceleration due to Earth's gravity, pulling objects toward the ground. On Earth, this is approximately \( 9.8 \, \text{m/s}^2 \). In our problem, gravitational acceleration is used as a reference to express the train's acceleration.
The train's acceleration is given as a fraction of gravitational acceleration—\( 0.20g \). Thus, it is crucial in calculating the force you experience while standing on the accelerating train.
The train's acceleration is given as a fraction of gravitational acceleration—\( 0.20g \). Thus, it is crucial in calculating the force you experience while standing on the accelerating train.
- The gravitational force acting on you is calculated as \( mg \), where \( m \) is your mass.
- The horizontal force due to the train’s acceleration is \( ma \).
- The static frictional force must balance this horizontal force to keep you steady.
Other exercises in this chapter
Problem 1
The Problems in this Section are ranked \(1,\) II, or III according to estimated difficulty, with \((1)\) Problems being easiest. Level (III) Problems are meant
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(I) A force of \(35.0 \mathrm{~N}\) is required to start a 6.0 -kg box moving across a horizontal concrete floor. ( \(a\) ) What is the coefficient of static fr
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(I) The coefficient of static friction between hard rubber and normal street pavement is about \(0.90 .\) On how steep a hill (maximum angle) can you leave a ca
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(I) What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is \(0.90 ?\)
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