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Chapter 5: Problem 84

If the coefficient of static friction between a table and a uniform, massive rope is \(\mu_{\mathrm{s}},\) what fraction of the rope can hang over the edge of the table without the rope sliding?

Short Answer

Expert verified
The fraction of the rope that can hang over the table without the rope sliding is \(\frac{\mu_{\mathrm{s}}}{1 + \mu_{\mathrm{s}}}\).

Step by step solution

01

Identify known values

We know that the coefficient of static friction is \(\mu_{\mathrm{s}}\). The length of the rope is not specified, so we can theoretically say it is of length L. We need to divide it into two pieces, one part being on the table (length x) and the other one hanging off the table (length L-x).
02

Calculate gravitational force

The force of gravity on the hanging part of the rope is \(F_{\mathrm{gravity}} = (L - x) \cdot g \cdot \rho\), where \(\rho\) is the linear mass density of the rope, which is mass/length.
03

Calculate frictional force

The frictional force on the rope is \(F_{\mathrm{friction}} = \mu_{\mathrm{s}} \cdot F_{\mathrm{normal}}\), where \(F_{\mathrm{normal}} = x \cdot g \cdot \rho\) is the normal force, equivalent to the weight of the piece of the rope lying on the table.
04

Equate frictional force and gravitational force

When the rope is just on the verge of sliding, the frictional force is equal to the gravitational force. Thus we have \(\mu_{\mathrm{s}} \cdot x \cdot g \cdot \rho = (L - x) \cdot g \cdot \rho\). We can cancel out \(g\) and \(\rho\) from both sides to find \(x = \frac{L}{1 + \mu_{\mathrm{s}}}\).
05

Determine the fraction of the rope that can hang

The asked fraction of the rope that can hang over the table is \(\frac{L - x}{L} = \frac{\mu_{\mathrm{s}}}{1 + \mu_{\mathrm{s}}}\)

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Key Concepts

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

Coefficient of Static Friction
Imagine trying to push a heavy box across the floor. Initially, it resists your push and refuses to budge. This resistance is due to static friction, a force that counteracts the initiation of motion. The coefficient of static friction, often represented as \( \mu_s \), is a dimensionless value that quantifies the friction between two static surfaces. A higher \( \mu_s \) indicates greater resistance to the start of motion.

In physics problems, \( \mu_s \) is crucial for calculating the maximum force that can be applied before two objects start sliding against each other. It is not a universal constant, but rather depends on the pairing of materials, such as rubber on concrete or wood on metal. In our textbook exercise, knowing the \( \mu_s \) of the rope and table interaction allowed us to determine what fraction of the rope can dangle off the edge before it slips.
Gravitational Force Calculation
Every object with mass exerts a gravitational pull, and this pull can be calculated using the universal law of gravitation. In most classroom physics problems, we simplify the calculation by considering the force of gravity near the Earth's surface. This force, known as weight, is found using the equation \( F_{gravity} = m \cdot g \), 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 rope problem, we applied this principle to find the gravitational force on the section of the rope hanging off the table. By multiplying the length of this section by the gravitational acceleration and the linear mass density \( \rho \) of the rope, we could calculate the downward force exerted by gravity.
Normal Force in Physics
When an object rests on a surface, the surface provides an upward force that balances the object's weight. This force is known as the normal force, symbolized as \( F_{normal} \). The term 'normal' refers to the force being perpendicular to the contact surface. The magnitude of the normal force is typically equal to the object's weight when it lies flat on a horizontal surface and there are no other vertical forces at play.

In our scenario with the rope on the table, we calculated the normal force (\( F_{normal} \) ) by considering the weight of the portion of the rope on the table. This force is essential in determining the maximum static friction, as it is a key variable in the equation \( F_{friction} = \mu_s \cdot F_{normal} \). Understanding the normal force is pivotal for analyzing situations of equilibrium, motion, and friction in physics.

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Most popular questions from this chapter

Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's) assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?

A flat (unbanked) curve on a highway has a radius of \(170.0 \mathrm{~m}\). A car rounds the curve at a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). (a) What is the minimum coefficient of static friction that will prevent sliding? (b) Suppose that the highway is icy and the coefficient of static friction between the tires and pavement is only one-third of what you found in part (a). What should be the maximum speed of the car so that it can round the curve safely?

A block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a hanging block with mass \(m_{2}\) by a cord passing over a small, friction less pulley (Fig. P5.74). The coefficient of static friction is \(\mu_{\mathrm{s}}\), and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}\). (a) Find the value of \(m_{2}\) for which the block of mass \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the value of \(m_{2}\) for which the block of mass \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

A small rock with mass \(m\) is attached to a light string of length \(L\) and whirled in a vertical circle of radius \(R\). (a) What is the minimum speed \(v\) at the rock's highest point for which it stays in a circular path? (b) If the speed at the rock's lowest point in its circular path is twice the value found in part (a), what is the tension in the string when the rock is at this point?

DATA In your physics lab, a block of mass m is at rest on a horizontal surface. You attach a light cord to the block and apply a horizontal force to the free end of the cord. You find that the block remains at rest until the tension \(T\) in the cord exceeds \(20.0 \mathrm{~N}\). For \(T>20.0 \mathrm{~N},\) you measure the acceleration of the block when \(T\) is maintained at a constant value, and you plot the results (Fig. \(\mathrm{P} 5.109)\). The equation for the straight line that best fits your data is \(a=\left[0.182 \mathrm{~m} /\left(\mathrm{N} \cdot \mathrm{s}^{2}\right)\right] T-2.842 \mathrm{~m} / \mathrm{s}^{2}\) For this block and surface, what are (a) the coefficient of static friction and (b) the coefficient of kinetic friction? (c) If the experiment were done on the earth's moon, where \(g\) is much smaller than on the earth, would the graph of \(a\) versus \(T\) still be fit well by a straight line? If so, how would the slope and intercept of the line differ from the values in Fig. \(\mathrm{P} 5.109 ?\) Or, would each of them be the same?
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