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Chapter 6: Problem 60

A chain lies on a rough horizontal table. It starts sliding when one-fourth of its length hangs over the edge of the table. The coefficient of static friction between the chain and the surface of the table is (a) \(\frac{1}{2}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{5}\)

Short Answer

Expert verified
(b) \(\frac{1}{3}\)

Step by step solution

01

Understand the Problem

We need to determine the coefficient of static friction for a chain on a table that begins to slide when one-fourth of its length is hanging off the table. The static friction holds the chain in place until the force due to gravity on the overhanging part causes it to move.
02

Define Forces Involved

Let the total length of the chain be \(L\) and its weight per unit length be \(\lambda\). The chain of length \(L/4\) is hanging. The weight of the hanging part is \((\lambda L/4)g\). The rest of the chain's weight is \((3\lambda L/4)g\) and is on the table.
03

Set Up the Equations

The force of static friction, which prevents sliding, is \(f_s = \mu_s \times \text{Normal Force}\). The normal force here is equal to the weight of the part of the chain on the table, \((3\lambda L/4)g\). Therefore, \(f_s = \mu_s (3\lambda L/4)g\).
04

Balance the Forces at the Point of Sliding

At the point when the chain is just about to slide, the gravitational force on the hanging part will be equal to the maximum static friction force: \( \frac{\lambda L}{4}g = \mu_s \frac{3\lambda L}{4}g\).
05

Solve for Coefficient of Static Friction \(\mu_s\)

Divide both sides by \(\lambda g\) and simplify the equation to find \(\mu_s\): \[ \frac{1}{4} = \mu_s \frac{3}{4} \]This simplifies to \(\mu_s = \frac{1}{3}\).
06

Identify the Correct Answer

The correct answer from the given options is (b) \(\frac{1}{3}\).

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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 static friction, denoted as \( \mu_s \), is a crucial factor in determining when an object will start to move on a surface. In this problem, it helps us quantify the resistance that arises between the chain and the table's surface before the chain begins to slide over the edge.
  • The coefficient of static friction is a dimensionless number indicating how much frictional force exists between two non-moving surfaces.
  • It tells us how much force needs to be overcome to initiate motion.
  • Each material combination鈥攍ike metal on wood, or wood on wood鈥攈as its own unique coefficient.
For the chain on the table, static friction counteracts the gravitational pull due to the hanging part of the chain. The formula used to determine \( \mu_s \) in this case is derived from knowing the point at which the chain starts slipping, revealing that the sufficient static friction coefficient \( \mu_s \) balances the gravitational force of the chain's overhanging section.
Physics Problem Solving
Approaching physics problems systematically is vital for finding accurate solutions. In this exercise, identifying each force involved and understanding their interplay leads to finding the correct coefficient of static friction.
  • First, we identify the problem's requirements: determine at what \( \mu_s \) the chain starts sliding as one-fourth hangs off.
  • Next, we define all the forces acting on the system, including gravitational force due to the hanging portion and frictional force holding the chain on the table.
  • We write equations representing the conditions when motion begins and solve these equations step-by-step.
  • This mathematical breakdown translates the physical scenario into a solvable format, relying on carefully balancing forces.
By following these methodical steps, anyone can break down even complex physics puzzle into manageable parts, ensuring they can navigate from problem statement to solution confidently.
Forces and Motion
Understanding forces and their effect on motion is fundamental to solving this chain problem. At play here are primarily two forces: gravitational force and static frictional force.
  • Gravitational Force: This force acts downward on the entire chain due to gravity, with more on the chain's part hanging over the table.
  • Static Frictional Force: This is the opposing force that prevents motion between the chain and the table until enough force overcomes it.
  • Motion occurs when the gravitational force from the hanging part of the chain exceeds the static friction holding the rest of the chain in place.
This problem is a classic example of tension and balance of forces. The motion initiates precisely when the pulling force of gravity on the overhanging chain equals the maximum frictional force from the table. By calculating when these forces balance, we can determine the conditions for motion onset.

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Most popular questions from this chapter

A rocket with a lift-off mass \(20000 \mathrm{~kg}\) is blasted upwards with an initial acceleration of \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). Calculate the initial thrust (force) of the blast. [NCERT] (a) \(3 \times 10^{5} \mathrm{~N}\) (b) \(2 \times 10^{8} \mathrm{~N}\) (c) \(4 \times 10^{5} \mathrm{~N}\) (d) \(5 \times 10^{5} \mathrm{~N}\)

A block \(B\) is pushed momentarily along a horizontal surface with an initial velocity \(v\). If \(\mu\) is the coefficient of sliding friction between \(B\) and the surface, block \(B\) will come to rest after a time [UP SEF 2007] (a) \(\frac{v}{g \mu}\) (b) \(\frac{g \mu}{v}\) (c) \(\frac{g}{v}\) (d) \(\frac{v}{g}\)

A body weighs \(8 \mathrm{~g}\) when placed in one pan and \(18 \mathrm{~g}\). when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, the true weight of the body is. (a) \(13 \mathrm{~g}\) (b) \(12 \mathrm{~g}\) (c) \(15.5 \mathrm{~g}\) (d) \(15 \mathrm{~g}\)

A block of mass \(2 \mathrm{~kg}\) is placed on the floor. The coefficient of static friction is \(0.4\). If a force of \(2.8 \mathrm{~N}\) is applied on the block parallel to floor, the force of friction between the block and floor (Taking \(\left.g=10 \mathrm{~ms}^{-2}\right)\) is (a) \(2.8 \mathrm{~N}\) (b) \(8 \mathrm{~N}\) (c) \(2 \mathrm{~N}\) (d) zero

Two blocks of masses \(m_{1}=4 \mathrm{~kg}\) and \(m_{2}=2 \mathrm{~kg}\) are connected to the ends of a string which passes over a massless, frictionless pulley. The total downward thrust on the pulley is nearly (a) \(27 \mathrm{~N}\) (b) \(54 \mathrm{~N}\) (c) \(0.8 \mathrm{~N}\) (d) zero
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