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A particle of small mass m is joined to a very heavy body by a light string passing over a light pulley. Both bodies are free to move. The total downward force in the pulley is
- a)Mg
- b)2Mg
- c)4Mg
- d)Can not be determined
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A particle of small mass m is joined to a very heavy body by a light s...
The heavy body will fall due to accelaration due to gravity.The accelaration of the lighter body wil be 'g' in upward direction.Taking the equlibrium of the small body
T-mg=ma
but a=g
Therefore
T-mg=mg
T=2mg
reation at the pulley
2T=4mg
Most Upvoted Answer
A particle of small mass m is joined to a very heavy body by a light s...
Explanation:
When a small mass m is joined to a very heavy body by a light string passing over a light pulley, both bodies are free to move. Let's analyze the forces acting on each body separately.
For the small mass m:
- There is a downward force due to gravity acting on the small mass, which can be represented as mg.
- As the small mass is connected to the heavy body by a string passing over a pulley, there is a tension force in the string acting upwards.
- The net force on the small mass is the difference between the tension force and the gravitational force, which can be represented as T - mg.
For the heavy body:
- There is a downward force due to gravity acting on the heavy body, which can be represented as Mg.
- As the heavy body is connected to the small mass by a string passing over a pulley, there is a tension force in the string acting downwards.
- The net force on the heavy body is the difference between the gravitational force and the tension force, which can be represented as Mg - T.
Since the system is in equilibrium, the net force on both bodies must be zero. Therefore, we can equate the net forces on the small mass and the heavy body and solve for T:
T - mg = Mg - T
Simplifying the equation, we get:
2T = Mg + mg
Dividing both sides of the equation by 2, we get:
T = (Mg + mg)/2
The total downward force in the pulley is the sum of the tension forces on both sides of the pulley. Since the tension force on one side is T, the total downward force in the pulley is 2T.
Therefore, the correct answer is option C: 4Mg.
When a small mass m is joined to a very heavy body by a light string passing over a light pulley, both bodies are free to move. Let's analyze the forces acting on each body separately.
For the small mass m:
- There is a downward force due to gravity acting on the small mass, which can be represented as mg.
- As the small mass is connected to the heavy body by a string passing over a pulley, there is a tension force in the string acting upwards.
- The net force on the small mass is the difference between the tension force and the gravitational force, which can be represented as T - mg.
For the heavy body:
- There is a downward force due to gravity acting on the heavy body, which can be represented as Mg.
- As the heavy body is connected to the small mass by a string passing over a pulley, there is a tension force in the string acting downwards.
- The net force on the heavy body is the difference between the gravitational force and the tension force, which can be represented as Mg - T.
Since the system is in equilibrium, the net force on both bodies must be zero. Therefore, we can equate the net forces on the small mass and the heavy body and solve for T:
T - mg = Mg - T
Simplifying the equation, we get:
2T = Mg + mg
Dividing both sides of the equation by 2, we get:
T = (Mg + mg)/2
The total downward force in the pulley is the sum of the tension forces on both sides of the pulley. Since the tension force on one side is T, the total downward force in the pulley is 2T.
Therefore, the correct answer is option C: 4Mg.
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A particle of small mass m is joined to a very heavy body by a light string passing over a light pulley. Both bodies are free to move. The total downward force in the pulley isa)Mgb)2Mgc)4Mgd)Can not be determinedCorrect answer is option 'C'. Can you explain this answer?
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