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(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?

Short Answer

Expert verified
The minimum coefficient of static friction is 0.20.

Step by step solution

01

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 \( g = 9.8 \, \text{m/s}^2 \).
02

Identify Known Quantities

The acceleration of the train \( a \) is given as \( 0.20 \times g \). Let \( a = 0.20g = 0.20 \times 9.8 \, \text{m/s}^2 = 1.96 \, \text{m/s}^2 \).
03

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.
04

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, \( N = mg \).
05

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., \( \mu_s mg = ma \).
06

Solve for the Coefficient of Static Friction

Rearranging the equation, we have \( \mu_s = \frac{a}{g} = \frac{0.20g}{g} = 0.20 \).

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Key Concepts

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

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.
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.
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 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.
By understanding gravitational acceleration and its relationship with other forces, you can determine the coefficient of static friction necessary to prevent sliding.

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Most popular questions from this chapter

(II) A small block of mass \(m\) is given an initial speed \(v_{0}\) up a ramp inclined at angle \(\theta\) to the horizontal. It travels a distance \(d\) up the ramp and comes to rest. ( \(a\) ) Determine a formula for the coefficient of kinetic friction between block and ramp. (b) What can you say about the value of the coefficient of static friction?

(II) How many revolutions per minute would a \(22-\mathrm{m}-\) diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

(II) The design of a new road includes a straight stretch that is horizontal and flat but that suddenly dips down a steep hill at \(22^{\circ} .\) The transition should be rounded with what minimum radius so that cars traveling \(95 \mathrm{~km} / \mathrm{h}\) will not leave the road (Fig. \(5-45) ?\)

A car drives at a constant speed around a banked circular track with a diameter of 127 \(\mathrm{m}\) . The motion of the car can be described in a coordinate system with its origin at the center of the circle. At a particular instant the car's accel- eration in the horizontal plane is given by $$\vec{\mathbf{a}}=(-15.7 \hat{\mathbf{i}}-23.2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}$$ (a) What is the car's speed? (b) Where \((x\) and \(y)\) is the car at this instant?

(III) The coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between two surfaces is not strictly independent of the velocity of the object. A possible expression for \(\mu_{\mathrm{k}}\) for wood on wood is $$\mu_{\mathrm{k}}=\frac{0.20}{\left(1+0.0020 v^{2}\right)^{2}},$$ where \(v\) is in \(\mathrm{m} / \mathrm{s} .\) A wooden block of mass 8.0 \(\mathrm{kg}\) is at rest on a wooden floor, and a constant horizontal force of 41 \(\mathrm{N}\) acts on the block. Use numerical integration to determine and graph \((a)\) the speed of the block, and \((b)\) its position, as a function of time from 0 to 5.0 \(\mathrm{s}\) (c) Determine the percent difference for the speed and position at 5.0 \(\mathrm{s}\) if \(\mu_{\mathrm{k}}\) is constant and equal to \(0.20 .\)

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