
A chain of length L starts sliding down from the horizontal surface of a table. If the hanging length is l, the coefficient of friction between the table and chain is:
A. \[\dfrac{l}{L}\]
B. \[\dfrac{L}{l}\]
C. \[\dfrac{l}{L-l}\]
D. \[\dfrac{L-l}{l}\]
Answer
566.4k+ views
Hint: IN this question we have been given that a chain of length L starts sliding down the horizontal surface. We have been asked to calculate the coefficient of friction between the table and chain. Static friction is the limiting frictional force above which the object starts to slide or move on a surface. We know that the maximum static friction will be equal to the weight of the hanging chain. Since the weight of the hanging chain is the force that is opposed by the static friction.
Complete step by step solution:
The total length of the chain is given as L, the length of the chain hanging from the table is given as l as shown in the figure.
Let us assume that the chain has a uniformly distributed mass, say ‘m’. Also, the length of chain on the table will be (L-l).
Therefore,
mass of chain on the table \[=\dfrac{mg(L-l)}{L}\]
Similarly,
mass of chain hanging from the table \[=\dfrac{mgl}{L}\]
Now, we know that
\[F=\mu N\] ……………… (1)
Where, F is the frictional force, N is the normal reaction and \[\mu \]is the coefficient of friction.
We know that
\[N=\dfrac{mg(L-l)}{L}\] ………………… (2)
We also know that, the maximum frictional force is equal to the weight of the hanging chain
Therefore,
\[F=\dfrac{mgl}{L}\] ………………….. (3)
Therefore, from (1), (2) and (3)
We get,
\[\dfrac{mgl}{L}=\mu \dfrac{mg(L-l)}{L}\]
On solving,
We get,
\[\mu =\dfrac{l}{(L-l)}\]
So, the correct answer is “Option C”.
Note: When we try to move an object, we experience a friction force. This force is known as static friction force. The static friction force keeps the object at rest. Therefore, it opposes motion. The friction force experienced when on a moving object is known as dynamic friction force. This force acts opposite to the direction of the motion of the object.
Complete step by step solution:
The total length of the chain is given as L, the length of the chain hanging from the table is given as l as shown in the figure.
Let us assume that the chain has a uniformly distributed mass, say ‘m’. Also, the length of chain on the table will be (L-l).
Therefore,
mass of chain on the table \[=\dfrac{mg(L-l)}{L}\]
Similarly,
mass of chain hanging from the table \[=\dfrac{mgl}{L}\]
Now, we know that
\[F=\mu N\] ……………… (1)
Where, F is the frictional force, N is the normal reaction and \[\mu \]is the coefficient of friction.
We know that
\[N=\dfrac{mg(L-l)}{L}\] ………………… (2)
We also know that, the maximum frictional force is equal to the weight of the hanging chain
Therefore,
\[F=\dfrac{mgl}{L}\] ………………….. (3)
Therefore, from (1), (2) and (3)
We get,
\[\dfrac{mgl}{L}=\mu \dfrac{mg(L-l)}{L}\]
On solving,
We get,
\[\mu =\dfrac{l}{(L-l)}\]
So, the correct answer is “Option C”.
Note: When we try to move an object, we experience a friction force. This force is known as static friction force. The static friction force keeps the object at rest. Therefore, it opposes motion. The friction force experienced when on a moving object is known as dynamic friction force. This force acts opposite to the direction of the motion of the object.
Recently Updated Pages
Master Class 4 Science: Engaging Questions & Answers for Success

Master Class 4 English: Engaging Questions & Answers for Success

Master Class 4 Maths: Engaging Questions & Answers for Success

Class 4 Question and Answer - Your Ultimate Solutions Guide

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

