/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A damp washcloth is hung over th... [FREE SOLUTION] | 91Ó°ÊÓ

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A damp washcloth is hung over the edge of a table to dry. Thus, part (mass \(=m_{\text {on }}\) ) of the washcloth rests on the table and part (mass \(=m_{\text {off }}\) ) does not. The coefficient of static friction between the table and the washcloth is \(0.40 .\) Determine the maximum fraction \(\left[m_{\text {off }} /\left(m_{\text {on }}+m_{\text {off }}\right)\right]\) that can hang over the edge without causing the whole washcloth to slide off the table

Short Answer

Expert verified
The maximum fraction is \( \frac{2}{7} \).

Step by step solution

01

Understanding Forces at Play

The frictional force preventing the cloth from sliding is static friction, which equals the coefficient of static friction multiplied by the normal force. The gravitational force on the portion hanging off tries to slide it off the table.
02

Express Static Friction

The static frictional force is given by \( f_{s} = \mu_{s} \cdot m_{\text{on}} \cdot g \), where \( \mu_{s} = 0.40 \) and \( g \) is the acceleration due to gravity.
03

Express Gravitational Force

The gravitational force acting on the hanging part is \( m_{\text{off}} \cdot g \). This is the force trying to pull the cloth off the table.
04

Set Up the Equation for Equilibrium

For the washcloth not to slide, the static frictional force must be equal to or greater than the gravitational force: \[ \mu_{s} \cdot m_{\text{on}} \cdot g \geq m_{\text{off}} \cdot g \]
05

Simplify the Equation

Cancel \( g \) from both sides of the equation: \[ \mu_{s} \cdot m_{\text{on}} \geq m_{\text{off}} \]
06

Solve for Maximum Fraction

We want the maximum fraction \( \frac{m_{\text{off}}}{m_{\text{on}} + m_{\text{off}}} \). Substituting \( m_{\text{off}} = \mu_{s} \cdot m_{\text{on}} \) gives us: \[ \frac{m_{\text{off}}}{m_{\text{on}} + m_{\text{off}}} = \frac{\mu_{s} \cdot m_{\text{on}}}{m_{\text{on}} + \mu_{s} \cdot m_{\text{on}}} \] \[ = \frac{0.40 \cdot m_{\text{on}}}{m_{\text{on}} + 0.40 \cdot m_{\text{on}}} \]
07

Simplify Solution for Maximum Fraction

Simplify the expression: \[ \frac{0.40 \cdot m_{\text{on}}}{1.40 \cdot m_{\text{on}}} = \frac{0.40}{1.40} = \frac{2}{7} \] This is the maximum fraction that can hang over without sliding.

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Key Concepts

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

Coefficient of Friction
The coefficient of friction, denoted as \( \mu \), is a dimensionless number that represents the frictional properties between two surfaces. In this context, we're focusing on the coefficient of static friction, \( \mu_s \), which is the force resisting the initiation of sliding motion. It is crucial in determining how much force is needed to start moving an object at rest.

For instance, when a washcloth is placed on a table, static friction prevents it from sliding off. The coefficient of static friction depends on both the surfaces involved— in this case, the damp washcloth and the table's surface.

In practical scenarios, knowing the coefficient of friction helps in calculating how steep an incline can be before objects start slipping, or how much force needs to be applied to move objects. In our exercise, it specifically helps us calculate the maximum portion of the washcloth that can hang over the edge without sliding off.
  • Static Friction is directly proportional to the normal force—the greater the weight pressing the washcloth down, the greater the static friction.
  • In our exercise, we see this as \( f_s = \mu_s \cdot m_{\text{on}} \cdot g \), where \( f_s \) is the static frictional force ensuring the washcloth doesn't slide.
Gravitational Force
Gravitational force is a natural phenomenon by which all things with mass are brought toward one another. In physics, this force is described as \( F_g = m \cdot g \), where \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.

In our exercise, the gravitational force is what pulls the hanging part of the washcloth downward, trying to slide it off the table. It is an opposing force to static friction.

Understanding gravitational force is essential in physics as it provides a basis for predicting motion and equilibrium. It's what makes this simple exercise of the washcloth significant—the challenge is to balance gravitational pull with static friction.

When dealing with situations like this, it's important to consider:
  • Gravitational force acts directly downward, aligning perpendicular to the surface.
  • In our case, it ensures that forces in the problem setup always have a gravitational component, influencing motion and equilibrium.
Equilibrium in Physics
Equilibrium in physics refers to the state where all forces acting on an object are balanced, resulting in no net movement. Specifically, we deal with static equilibrium when dealing with non-moving objects or systems at rest.

In this exercise, equilibrium is achieved when the static frictional force counterbalances the gravitational force acting on the washcloth's overhanging part. When these forces are balanced, the system is in equilibrium, meaning the washcloth remains stationary.

This scenario highlights two main aspects of equilibrium:
  • Translational Equilibrium: No net force exists that would otherwise cause the washcloth to slide off the table. The frictional force counters exactly the portion of the gravitational force trying to slide it off.
  • Rotational Equilibrium: No net torque is acting that might tilt or rotate the system. However, in this simplified scenario, rotational factors are less of concern given the flat nature.
This concept demonstrates why equilibrium is vital in engineering and design, ensuring stability in structures and devices. Understanding and applying forces in equilibrium allows us to create systems that are stable and functional under varied conditions.

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Most popular questions from this chapter

The central ideas in this problem are reviewed in Multiple-Concept Example \(9 .\) One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches \(47.0 \mathrm{~N}\), the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?

Refer to Concept Simulation \(4.4\) at for background relating to this problem. The drawing shows a large cube (mass \(=25 \mathrm{~kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\vec{P}\). A small cube \((\mathrm{mass}=4.0 \mathrm{~kg})\) is in contact with the front surface of the large cube and will slide downward unless \(\vec{P}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71\). What is the smallest magnitude that \(\overrightarrow{\mathrm{P}}\) can have in order to keep the small cube from sliding downward?

A \(5.00-\mathrm{kg}\) block is placed on top of a \(12.0-\mathrm{kg}\) block that rests on a frictionless table. The coefficient of static friction between the two blocks is \(0.600 .\) What is the maximum horizontal force that can be applied before the \(5.00-\mathrm{kg}\) block begins to slip relative to the \(12.0-\mathrm{kg}\) block, if the force is applied to (a) the more massive block and (b) the less massive block?

The earth exerts a gravitational force on a falling raindrop, pulling it down. Does the raindrop exert a gravitational force on the earth, pulling it up? If so, is this force greater than, less than, or equal to the force that the earth exerts on the raindrop? Provide a reason for your answer. Problem A raindrop has a mass of \(5.2 \times 10^{-7} \mathrm{~kg}\), and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.

A neutron star has a mass of \(2.0 \times 10^{30} \mathrm{~kg}\) (about the mass of our sun) and a radius of \(5.0 \times 10^{3} \mathrm{~m}\) (about the height of a goodsized mountain). Suppose an object falls from rest near the surface of such a star. How fast would it be moving after it had fallen a distance of \(0.010 \mathrm{~m} ?\) (Assume that the gravitational force is constant over the distance of the fall, and that the star is not rotating.)

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