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A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is _________ m.
Correct answer is '2'. Can you explain this answer?
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A uniform chain of 6 m length is placed on a table such that a part of...
Given:
- Length of the uniform chain = 6 m
- Coefficient of static friction between chain and table surface = 0.5
To find:
- Maximum length of the chain hanging from the table
Explanation:
When a part of the chain hangs over the edge of the table, the weight of the hanging part of the chain creates a torque that tries to rotate the chain around the point where it touches the table. The static friction between the chain and the table prevents the chain from slipping and maintains its equilibrium.
To determine the maximum length of the chain hanging from the table, we need to consider the conditions for equilibrium.
Conditions for equilibrium:
1. Sum of all the forces acting on the chain should be zero.
2. Sum of all the torques acting on the chain should be zero.
Force analysis:
Since the chain is at rest, the only external force acting on the hanging part of the chain is its weight (mg), which acts vertically downward.
Torque analysis:
The torque due to the weight of the hanging part of the chain creates a clockwise rotation. The static frictional force between the chain and the table surface creates a counterclockwise torque.
Equations:
1. Sum of vertical forces: Tension in the chain (T) - Weight of the chain (mg) = 0
2. Sum of torques: Torque due to weight - Torque due to static friction = 0
Calculations:
Let's assume that the length of the chain hanging from the table is 'x'.
1. Sum of vertical forces: T - mg = 0
T = mg
2. Sum of torques: (Weight of hanging chain) × (Distance from the pivot) - (Static frictional force) × (Distance from the pivot) = 0
(mg) × (x/2) - (μN) × (x) = 0
(mg) × (x/2) - (μmg) × (x) = 0 (Since normal force (N) = mg for equilibrium)
x/2 - μx = 0
x(1/2 - μ) = 0
x = 0/(1/2 - μ)
x = 0/(-0.5)
x = 0
From the above equation, we see that x = 0. This means that there is no length of the chain hanging from the table in equilibrium. However, we know that the chain length is 6 m. Therefore, the maximum length of the chain hanging from the table is 6 - 0 = 6 m.
Conclusion:
The maximum length of the chain hanging from the table is 6 meters.
- Length of the uniform chain = 6 m
- Coefficient of static friction between chain and table surface = 0.5
To find:
- Maximum length of the chain hanging from the table
Explanation:
When a part of the chain hangs over the edge of the table, the weight of the hanging part of the chain creates a torque that tries to rotate the chain around the point where it touches the table. The static friction between the chain and the table prevents the chain from slipping and maintains its equilibrium.
To determine the maximum length of the chain hanging from the table, we need to consider the conditions for equilibrium.
Conditions for equilibrium:
1. Sum of all the forces acting on the chain should be zero.
2. Sum of all the torques acting on the chain should be zero.
Force analysis:
Since the chain is at rest, the only external force acting on the hanging part of the chain is its weight (mg), which acts vertically downward.
Torque analysis:
The torque due to the weight of the hanging part of the chain creates a clockwise rotation. The static frictional force between the chain and the table surface creates a counterclockwise torque.
Equations:
1. Sum of vertical forces: Tension in the chain (T) - Weight of the chain (mg) = 0
2. Sum of torques: Torque due to weight - Torque due to static friction = 0
Calculations:
Let's assume that the length of the chain hanging from the table is 'x'.
1. Sum of vertical forces: T - mg = 0
T = mg
2. Sum of torques: (Weight of hanging chain) × (Distance from the pivot) - (Static frictional force) × (Distance from the pivot) = 0
(mg) × (x/2) - (μN) × (x) = 0
(mg) × (x/2) - (μmg) × (x) = 0 (Since normal force (N) = mg for equilibrium)
x/2 - μx = 0
x(1/2 - μ) = 0
x = 0/(1/2 - μ)
x = 0/(-0.5)
x = 0
From the above equation, we see that x = 0. This means that there is no length of the chain hanging from the table in equilibrium. However, we know that the chain length is 6 m. Therefore, the maximum length of the chain hanging from the table is 6 - 0 = 6 m.
Conclusion:
The maximum length of the chain hanging from the table is 6 meters.
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Community Answer
A uniform chain of 6 m length is placed on a table such that a part of...
Mass per unit length = λ
N = mg = λ(L - x)g
fsmax = μSN
fsmax = (0.5)(λ)(L - x)g
Also, fsmax = mxg
0.5λ(L - x)g = λxg
L - x / 2 = x
So,
N = mg = λ(L - x)g
fsmax = μSN
fsmax = (0.5)(λ)(L - x)g
Also, fsmax = mxg
0.5λ(L - x)g = λxg
L - x / 2 = x
So,
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A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is _________ m.Correct answer is '2'. Can you explain this answer?
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A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is _________ m.Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is _________ m.Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is _________ m.Correct answer is '2'. Can you explain this answer?.
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