/*! 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 69 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 \(=m_{\mathrm{ca}} )\) of the washcloth rests on the table and part (mass \(=m_{\text { oft }} )\) does not. The coefficient of static friction between the table and the washcloth is 0.40 . Determine the maximum fraction \(\left[m_{\mathrm{oft}} /\left(m_{\mathrm{cn}}+m_{\mathrm{off}}\right)\right]\) 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

Understand the Problem

We have a washcloth with part of its mass resting on a table and part hanging over. The task is to find how much can hang over the edge without sliding off, given the table's friction coefficient.
02

Identify Forces Involved

The forces acting on the washcloth are the gravitational force and the frictional force. The frictional force prevents the washcloth from sliding and is calculated using the formula: \[ \text{Frictional Force} = \mu_s \cdot m_{\mathrm{ca}} \cdot g \] where \( \mu_s = 0.40 \) and \( g \) is the acceleration due to gravity.
03

Set Up the Equilibrium Condition

To prevent sliding, the frictional force must be equal to or greater than the gravitational force pulling on the overhanging part (\( m_{\mathrm{oft}} \cdot g \)). The equation is: \[ \mu_s \cdot m_{\mathrm{ca}} \cdot g \geq m_{\mathrm{oft}} \cdot g \] Simplifying this gives: \[ \mu_s \cdot m_{\mathrm{ca}} \geq m_{\mathrm{oft}} \]
04

Express in Terms of Total Mass

We need to find the ratio \( \frac{m_{\mathrm{oft}}}{m_{\mathrm{ca}} + m_{\mathrm{oft}}} \). Since \( m_{\mathrm{oft}} \leq \mu_s \cdot m_{\mathrm{ca}} \), replace \( m_{\mathrm{oft}} \) with \( \mu_s \cdot m_{\mathrm{ca}} \) in the numerator: \[ \frac{m_{\mathrm{oft}}}{m_{\mathrm{ca}} + m_{\mathrm{oft}}} = \frac{\mu_s \cdot m_{\mathrm{ca}}}{m_{\mathrm{ca}} + \mu_s \cdot m_{\mathrm{ca}}} \]
05

Simplify the Fraction

Simplify the fraction to: \[ \frac{\mu_s \cdot m_{\mathrm{ca}}}{m_{\mathrm{ca}} + \mu_s \cdot m_{\mathrm{ca}}} = \frac{\mu_s}{1 + \mu_s} \] Substituting \( \mu_s = 0.40 \) gives: \[ \frac{0.40}{1 + 0.40} = \frac{0.40}{1.40} = \frac{2}{7} \]
06

Conclude the Solution

The maximum fraction of the washcloth that can hang over the edge without sliding off is \( \frac{2}{7} \).

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Key Concepts

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

Equilibrium Condition
In physics, equilibrium means that all forces acting on an object are balanced, resulting in no movement. For the washcloth, our goal is for it to stay in place and not slide off the table. This means the forces pulling the washcloth towards the floor, primarily its overhanging part's weight, need to be countered by the frictional force on the table. These forces must reach a balance, which we refer to as the equilibrium condition.
  • The gravitational force pulling the overhanging part downwards is calculated as the mass times gravity, i.e., \( m_{\text{oft}} \cdot g \).
  • The frictional force that prevents it from sliding is determined by \( \mu_s \cdot m_{\text{ca}} \cdot g \), where \( \mu_s \) is the static friction coefficient.
To maintain equilibrium, the frictional force must equal or surpass the gravitational force of the overhang, leading us to our main inequality \( \mu_s \cdot m_{\text{ca}} \geq m_{\text{oft}} \). This defines the maximum portion that can hang without slipping.
Gravitational Force
Gravitational force is the pull that draws objects towards the center of the Earth. This force is a crucial factor in determining whether the washcloth can stay on the table or will fall completely off. Gravitational force can be calculated using the formula \( F_g = m \cdot g \), where \( m \) is mass and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth.
For our washcloth problem, two masses are involved:
  • \( m_{\text{ca}} \), the portion's mass on the table.
  • \( m_{\text{oft}} \), the portion's mass overhanging the edge.
The gravitational force acting on the overhanging part attempts to pull it down, threatening equilibrium. Hence, understanding and calculating gravitational force is essential to analyze how the cloth behaves under its influence.
Mass Distribution
Mass distribution in an object influences how it interacts with forces such as gravity and friction. With our washcloth, part is resting on the table while another hangs over, each part contributing differently to the system. It’s key to see how the mass is 'distributed' or spread out, as it affects equilibrium and the forces at play.

In our equation, we separate two different parts:
  • Mass on the table, \( m_{\text{ca}} \).
  • Mass over the edge, \( m_{\text{oft}} \).
This distinction is significant since a wide mass distribution can change the stability of our setup. If most of the mass is overhanging, friction may not suffice to prevent sliding. Therefore, understanding mass distribution is vital in calculating the maximum safe overhang.
Friction Coefficient
The friction coefficient, denoted as \( \mu_s \), quantifies how much friction acts between two surfaces. It varies between different material pairs. In our problem, it is given to be 0.40, which indicates a moderate level of friction.
The frictional force which stops the washcloth from sliding is calculated as \( \mu_s \cdot m_{\text{ca}} \cdot g \). This shows the role of \( \mu_s \) in providing the necessary force to resist the downward pull of gravity on the overhanging portion.
  • A higher coefficient implies stronger grip, supporting more mass without slipping.
  • A lower coefficient suggests weaker friction, allowing less mass to hang without sliding.
This coefficient allows us to define an inequality showing that the part’s weight should not exceed the frictional resistance — crucial for maintaining equilibrium and stability.

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Most popular questions from this chapter

A skater with an initial speed of 7.60 \(\mathrm{m}/\mathrm{s}\) stops propelling himself and begins to coast across the ice, eventually coming to rest. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is 0.100. Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest?

As part \(a\) of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of 412 \(\mathrm{N}\) , and the other has a weight of 908 \(\mathrm{N}\) . The rope and the pulleys are mass-less and there is no friction. (a) What is the acceleration of the lighter block? (b) Suppose that the heavier block is removed, and a downward force of 908 \(\mathrm{N}\) is provided by someone pulling on the rope, as part \(b\) of the drawing shows. Find the acceleration of the remaining block. (c) Explain why the answers in (a) and (b) are different.

ssm 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 292-kg motorcycle is accelerating up along a ramp that is inclined \(30.0^{\circ}\) above the horizontal. The propulsion force pushing the motorcycle up the ramp is \(3150 \mathrm{N},\) and air resistance produces a force of 250 \(\mathrm{N}\) that opposes the motion. Find the magnitude of the motorcycle's acceleration.

ssm A 95.0 -kg person stands on a scale in an elevator. What is the apparent weight when the elevator is \(\quad\) (a) accelerating upward with an acceleration of \(1.80 \mathrm{m} / \mathrm{s}^{2}, \quad\) (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.30 \(\mathrm{m} / \mathrm{s}^{2} ?\)

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