
An iron 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 surface of the table is
A. \[\dfrac{1}{2}\]
B. \[\dfrac{1}{3}\]
C. \[\dfrac{1}{4}\]
D. \[\dfrac{1}{5}\]
Answer
466.8k+ views
Hint: In this question we have been given that an iron chain on a table starts sliding down when one fourth of its length hangs over the edge of the table. We have been asked to calculate the static friction between the surface of the table and the 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 answer:
It is given that one fourth of the length of chain is hanging from the side of the table as shown in the figure below.
Now, let us assume that the chain has uniformly distributed mass. Therefore, we can say that one fourth of the mass of the chain is hanging from the side of the table.
Therefore,
Weight of chain hanging from the table \[=\dfrac{mg}{4}\]
Similarly,
Weight of chain on the table \[=\dfrac{3mg}{4}\]
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
\[N=\dfrac{3mg}{4}\] ………………….. (2)
We also know that, the maximum frictional force is equal to the weight of the hanging chain
Therefore,
\[F=\dfrac{mg}{4}\] …………… (3)
Therefore, from (1), (2) and (3)
We get,
\[\dfrac{mg}{4}=\mu \dfrac{3mg}{4}\]
On solving
We get,
\[\mu =\dfrac{1}{3}\]
Therefore, the correct answer is option B.
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 answer:
It is given that one fourth of the length of chain is hanging from the side of the table as shown in the figure below.

Now, let us assume that the chain has uniformly distributed mass. Therefore, we can say that one fourth of the mass of the chain is hanging from the side of the table.
Therefore,
Weight of chain hanging from the table \[=\dfrac{mg}{4}\]
Similarly,
Weight of chain on the table \[=\dfrac{3mg}{4}\]
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
\[N=\dfrac{3mg}{4}\] ………………….. (2)
We also know that, the maximum frictional force is equal to the weight of the hanging chain
Therefore,
\[F=\dfrac{mg}{4}\] …………… (3)
Therefore, from (1), (2) and (3)
We get,
\[\dfrac{mg}{4}=\mu \dfrac{3mg}{4}\]
On solving
We get,
\[\mu =\dfrac{1}{3}\]
Therefore, the correct answer is option B.
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 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

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

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

State the laws of reflection of light
