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The coefficient of static friction between Teflon and scrambled eggs is about \(0.04 .\) What is the smallest angle from the horizontal that will cause the eggs to slide across the bottom of a Teflon-coated skillet?

Short Answer

Expert verified
The smallest angle is approximately 2.29 degrees.

Step by step solution

01

Identify the forces involved

To begin, identify the forces acting on the eggs at the point they start to slide. The main forces are gravity, which acts downward, and static friction, which opposes the motion.
02

Set up the equation for static friction

The static frictional force is given by the equation \( f_{s} = \mu_{s} \cdot N \), where \( \mu_{s} \) is the coefficient of static friction (0.04) and \( N \) is the normal force acting on the eggs.
03

Understating the normal force

The normal force \(N\) is equal to the component of the gravitational force perpendicular to the surface of the skillet. If \(m\) is the mass of the eggs, then \( N = m \cdot g \cdot \cos(\theta) \), where \(g\) is the acceleration due to gravity and \(\theta\) is the angle from the horizontal.
04

Balance the forces for motion

Eggs will start to slide when the parallel component of gravity equals the force of static friction. This means \( m \cdot g \cdot \sin(\theta) = \mu_{s} \cdot m \cdot g \cdot \cos(\theta) \).
05

Simplify the equation

Since mass \(m\) and gravity \(g\) appear on both sides of the equation, they cancel out. We are left with \( \tan(\theta) = \mu_{s} \).
06

Solve for the angle

Solve the equation \( \tan(\theta) = 0.04 \) to find \( \theta \). Thus, \( \theta = \arctan(0.04) \approx 2.29\) degrees.

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Key Concepts

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

Understanding the Coefficient of Static Friction
The coefficient of static friction, denoted as \(\mu_s\), is a crucial concept when dealing with the tendency of objects to move. It is a measure of how much resistance an object experiences before it begins to slide over a surface. This value is dimensionless and typically varies from 0 to 1, although it can exceed 1 in certain instances. A lower coefficient of static friction, such as 0.04 for Teflon and scrambled eggs, indicates that very little force is needed to initiate movement.
Key aspects of the static friction coefficient:
  • No units: It is a ratio and thus dimensionless.
  • Surface-dependent: It depends on the surfaces in contact. Different materials have different coefficients.
  • Force threshold: Represents the maximum force capable of being mobilized by static friction before movement occurs.
  • Independent of surface area: Unlike what might be assumed, it isn鈥檛 affected by the size of the surfaces in contact.
This value is pivotal in setting up your equations when calculating whether an object will stay still or start moving when a force is applied.
Examining the Normal Force
The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface, hence the term 鈥渘ormal.鈥 The normal force plays a pivotal role in determining the static frictional force, as it is a factor in the equation \( f_{s} = \mu_{s} \cdot N \).
In the context of a sloped surface, like a skillet, the normal force is calculated as:
  • \( N = m \cdot g \cdot \cos(\theta) \)
where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately 9.81 m/s虏 on Earth), and \(\theta\) represents the angle of the surface from the horizontal. This relationship shows how the normal force decreases as the angle increases, reducing the force available to resist sliding.
Understanding the normal force is crucial when analyzing the conditions that lead to movement on inclined planes.
Exploring Gravitational Force
Gravitational force is the attractive force that pulls objects towards each other, particularly towards Earth's center. For objects on a surface, this force contributes significantly to how they interact with the surface and if they will remain stationary or start sliding.
Key components of gravitational force analysis:
  • Formula: \( F_g = m \cdot g \), where \(F_g\) is the force due to gravity, \(m\) is mass, and \(g\) is gravitational acceleration.
  • Components on a slope: When on an incline, gravity can be divided into two components: parallel \(( m \cdot g \cdot \sin(\theta) )\) and perpendicular \(( m \cdot g \cdot \cos(\theta) )\) to the plane.
  • Critical transition: Movement begins when the parallel component of gravity outweighs the static frictional force.
Gravitational force is constantly present and influences how objects behave in various settings. On sloped surfaces, understanding it helps in predicting when an object will start to slide.

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Most popular questions from this chapter

\(A 1000 \mathrm{~kg}\) boat is traveling at \(90 \mathrm{~km} / \mathrm{h}\) when its engine is shut off. The magnitude of the frictional force \(\vec{f}_{k}\) between boat and water is proportional to the speed \(v\) of the boat: \(f_{k}=70 v,\) where \(v\) is in meters per second and \(f_{k}\) is in newtons. Find the time required for the boat to slow to \(45 \mathrm{~km} / \mathrm{h}\).

A worker pushes horizontally on a \(35 \mathrm{~kg}\) crate with a force of magnitude \(110 \mathrm{~N}\). The coefficient of static friction between the crate and the floor is 0.37 . (a) What is the value of \(f_{s, \max }\) under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker's \(110 \mathrm{~N}\) push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

A loaded penguin sled weighing \(80 \mathrm{~N}\) rests on plane inclined at angle \(\theta=20^{\circ}\) to the horizontal (Fig. \(6-23\) ). Between the sled and the plane, the coefficient of static friction is \(0.25,\) and the coefficient of kinetic friction is \(0.15 .\) (a) What is the least magnitude of the force \(\vec{F},\) parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude \(F\) that will start the sled moving up the plane? (c) What value of \(F\) is required to move the sled up the plane at constant velocity?

An \(8.00 \mathrm{~kg}\) block of steel is at rest on a horizontal table. The coef- ficient of static friction between the block and the table is \(0.450 .\) A force is to be applied to the block. To three significant figures, what is the magnitude of that applied force if it puts the block on the verge of sliding when the force is directed (a) horizontally, (b) upward at \(60.0^{\circ}\) from the horizontal, and (c) downward at \(60.0^{\circ}\) from the horizontal?

A certain string can withstand a maximum tension of \(40 \mathrm{~N}\) without breaking. A child ties a \(0.37 \mathrm{~kg}\) stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 \(\mathrm{m},\) slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? (b) What is the speed of the stone as the string breaks?

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