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If the coefficient of static friction between a table and a uniform massive rope is \(\mu_{s}\) , 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 edge is \(\frac{\mu_s}{1+\mu_s}\).

Step by step solution

01

Identifying Forces

Consider a rope of length \(l\) and mass \(m\) placed on a table. Let \(x\) be the length of the rope hanging over the edge. The weight of the hanging part, which is the force trying to pull the rope off the table, is \(mgx/l\).
02

Applying Conditions for Equilibrium

The rope will start to slide when the force due to the weight of the hanging part exceeds the maximum static friction force. The friction force is \(\mu_smg\), where \(\mu_s\) is the coefficient of static friction and \(mg\) is the weight of the part of the rope on the table.
03

Setting up the Equation

For equilibrium, the maximum static friction equals the weight of the hanging part: \(mgx/l = \mu_smg(1-x/l)\). This simplifies to \(x/l = \mu_s (1-x/l)\).
04

Solving the Equation

Rearrange the equation: \(x/l + \mu_s x/l = \mu_s\). Simplifying, we have \(x/l (1 + \mu_s) = \mu_s\), and therefore, \(x/l = \mu_s/(1+\mu_s)\).
05

Conclusion

The fraction of the rope that can hang over the edge without sliding is \(\mu_s/(1+\mu_s)\).

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Key Concepts

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

Coefficient of Static Friction
In physics, the coefficient of static friction, commonly symbolized as \( \mu_s \), is a crucial concept that pertains to surfaces in contact. It represents the ratio of the maximum static friction force that can develop between two surfaces before motion initiates. This coefficient is dimensionless and varies depending on the materials in contact. For example, rubber on concrete typically has a higher coefficient of static friction than ice on metal.

Understanding \( \mu_s \) helps predict whether an object will start moving when a force is applied, making it an essential factor in solving equilibrium problems. It’s calculated using the formula: \( F_{s\,max} = \mu_s N \), where \( F_{s\,max} \) is the maximum static friction force and \( N \) is the normal force. In the context of the rope problem, \( \mu_s \) determines how much of the rope can dangle over the table's edge before gravity overcomes the force of static friction between the rope and the table's surface.
Equilibrium Conditions
Equilibrium conditions in this context refer to the state where the forces acting on an object are balanced, meaning no net force is present to cause movement. For an object to remain in equilibrium, the sum of the external forces and the sum of the torques acting on it must both be zero.

In the case of our rope, it will remain stationary on the table, not slipping over the edge, as long as the static friction force equals or exceeds the gravitational pull on the hanging section of the rope. This balance can be expressed mathematically, where the force due to the weight of the hanging rope part \( \frac{mgx}{l} \) equals the static friction \( \mu_s mg (1 - x/l) \). Solving for \( x/l \) gives us our equilibrium condition: \( x/l = \mu_s/(1 + \mu_s) \). This tells us the maximum fraction of the rope that can hang over without causing a slip.
Rope Dynamics
Rope dynamics in the context of this exercise involves understanding how forces interact along the length of a rope, especially when part of it is unsupported. With one section resting on the table and another part hanging, different forces are acting on these sections.

Firstly, the mass of the entire rope creates a weight force \( mg \) that is spread along its length. When a portion of the rope is hanging, the gravitational force becomes \( \frac{mgx}{l} \), acting to pull the rope downwards and potentially off the table. Static friction acting along the pieces of the rope on the table counteracts this, keeping the rope from moving.

Understanding rope dynamics helps us analyze how alterations in its position or the surface characteristics might change behavior. This hints at practical applications, such as determining safe loading limits in real-world scenarios where materials are laid over edges or other objects. Always remember, maintaining equilibrium conditions prevents slipping.

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Most popular questions from this chapter

Lunch Time! You are riding your motorcycle one day down a wet street that slopes downward at an angle of \(20^{\circ}\) below the horizontal. As you start to ride down the hill, you notice a construction crew has dug a deep hole in the street at the bottom of the hill. A Siberian tiger, escaped from the City Zoo, has taken up residence in the hole. You apply the brakes and lock your wheels at the top of the hill, where you are moving with a speed of 20 \(\mathrm{m} / \mathrm{s}\) . The inclined street in front of you is 40 \(\mathrm{m}\) long. (a) Will you plunge into the hole and become the tiger's lunch, or do you skid to a stop before you reach the hole? (The coefficients of friction between your motorcycle tires and the wet pavement are \(\mu_{\mathrm{s}}=0.90\) and \(\mu_{\mathrm{k}}=0.70 .\) (b) What must your initial speed be if you are to stop just before reaching the hole?

Banked Curve I. A curve with a \(120-\mathrm{m}\) radius on a level road is banked at the correct angle for a speed of 20 \(\mathrm{m} / \mathrm{s}\) . If an automobile rounds this curve at 30 \(\mathrm{m} / \mathrm{s}\) , what is the minimum coefficient of static friction needed between tires and road to prevent skidding?

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the platform when it is brought up to a speed of 40.0 rev/min, provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of station between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

Apparent Weight. A 550 -N physics student stands on a bathroom scale in an \(850-k g\) (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 450 \(\mathrm{N}\) . (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads 670 \(\mathrm{N}\) ? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (lanore any friction between the wall and the picture frame.)

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