
A uniform metal chain is placed on a rough table such that one end of the chain hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, find the coefficient of static friction.
A. \[\dfrac{3}{4}\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{2}{3}\]
D. \[\dfrac{1}{2}\]
Answer
218.4k+ views
Hint:In order to proceed with this question let’s have a look at static friction. Static friction is defined as the limiting frictional force above which the object starts to slide or move on a surface and it will be equal to the weight of the hanging chain.
Formula Used:
To find the coefficient of static friction the formula is,
\[\mu = \dfrac{F}{N}\]
Where, F is force and N is normal force.
Complete step by step solution:
Consider a uniform metal chain that is placed on a rough table such that one end of the chain hangs down over the edge of the table. When one-third of its (metal chain) length hangs over the edge, the chain starts sliding. Then we need to find the coefficient of static friction.
Initially, we have the length here, which is,
Length of the chain \[ = L\]
And when it becomes one-third, then,
\[{L^1} = \dfrac{L}{3}\]
The chain starts sliding then,
\[\dfrac{L}{3} = \dfrac{{\mu L}}{{\mu + 1}}\]
Here, \[\mu \] is the coefficient of frictional force.
\[\mu + 1 = 3\mu \]
\[\Rightarrow 2\mu = 1\]
\[ \therefore \mu = \dfrac{1}{2}\]
Therefore, the value of the coefficient of static friction is \[\dfrac{1}{2}\].
Hence, Option D is the correct answer.
Note:There is an alternative method to find this solution.
The length of the chain \[ = L\]
The mass of the chain \[ = m\]
When the one-third of its length hangs over the edge then,
\[{L^1} = \dfrac{L}{3}\]
The mass distributed along the length is \[\dfrac{m}{3}\].
When the force pulls the chain downwards due to gravity then, \[F = \dfrac{{mg}}{3}\]
The length of the chain on the table is given by \[\dfrac{{2L}}{3}\].
The normal force on the chain on the table is, \[N = \dfrac{{2mg}}{3}\]
Since, \[\mu \] is the coefficient of frictional force
The frictional force is,
\[F = \mu N\]
\[\Rightarrow \mu = \dfrac{F}{N}\]
Substituting the value of F and N in the above equation, we get
\[\mu = \dfrac{{\left( {\dfrac{{mg}}{3}} \right)}}{{\left( {\dfrac{{2mg}}{3}} \right)}}\]
\[\therefore \mu = \dfrac{1}{2}\]
Therefore, the value of coefficient of static friction is \[\dfrac{1}{2}\].
Formula Used:
To find the coefficient of static friction the formula is,
\[\mu = \dfrac{F}{N}\]
Where, F is force and N is normal force.
Complete step by step solution:
Consider a uniform metal chain that is placed on a rough table such that one end of the chain hangs down over the edge of the table. When one-third of its (metal chain) length hangs over the edge, the chain starts sliding. Then we need to find the coefficient of static friction.
Initially, we have the length here, which is,
Length of the chain \[ = L\]
And when it becomes one-third, then,
\[{L^1} = \dfrac{L}{3}\]
The chain starts sliding then,
\[\dfrac{L}{3} = \dfrac{{\mu L}}{{\mu + 1}}\]
Here, \[\mu \] is the coefficient of frictional force.
\[\mu + 1 = 3\mu \]
\[\Rightarrow 2\mu = 1\]
\[ \therefore \mu = \dfrac{1}{2}\]
Therefore, the value of the coefficient of static friction is \[\dfrac{1}{2}\].
Hence, Option D is the correct answer.
Note:There is an alternative method to find this solution.
The length of the chain \[ = L\]
The mass of the chain \[ = m\]
When the one-third of its length hangs over the edge then,
\[{L^1} = \dfrac{L}{3}\]
The mass distributed along the length is \[\dfrac{m}{3}\].
When the force pulls the chain downwards due to gravity then, \[F = \dfrac{{mg}}{3}\]
The length of the chain on the table is given by \[\dfrac{{2L}}{3}\].
The normal force on the chain on the table is, \[N = \dfrac{{2mg}}{3}\]
Since, \[\mu \] is the coefficient of frictional force
The frictional force is,
\[F = \mu N\]
\[\Rightarrow \mu = \dfrac{F}{N}\]
Substituting the value of F and N in the above equation, we get
\[\mu = \dfrac{{\left( {\dfrac{{mg}}{3}} \right)}}{{\left( {\dfrac{{2mg}}{3}} \right)}}\]
\[\therefore \mu = \dfrac{1}{2}\]
Therefore, the value of coefficient of static friction is \[\dfrac{1}{2}\].
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

