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A block of mass m is supported on the front surface of a bigger block of mass 3m.The frictional coefficient between the surfaces of two blocks is mu. Find the minimum magnitude of horizontal force F required to be applied so that m will remain at rest with respect to 3m?
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A block of mass m is supported on the front surface of a bigger block ...
Problem Statement: A block of mass m is supported on the front surface of a bigger block of mass 3m. The frictional coefficient between the surfaces of two blocks is mu. Find the minimum magnitude of horizontal force F required to be applied so that m will remain at rest with respect to 3m?

Solution:
To solve this problem, we will use the concept of friction and equilibrium of forces.
Let's consider the forces acting on the block of mass m:
- Force of gravity acting downwards (mg).
- Normal force exerted by the bigger block (N).
- Frictional force acting in the opposite direction of motion (f).

As the block is at rest, the net force acting on it is zero. Therefore, the force of gravity is balanced by the normal force and the frictional force.
Hence, we can write the following equations:
- N = mg
- f = muN = mumg

Now, let's consider the forces acting on the bigger block of mass 3m:
- Force of gravity acting downwards (3mg).
- Normal force exerted by the ground (N').
- Frictional force acting in the opposite direction of motion (f').

As the bigger block is also at rest, the net force acting on it is zero. Therefore, the force of gravity is balanced by the normal force and the frictional force.
Hence, we can write the following equations:
- N' = 3mg
- f' = muN' = 3mumg

Now, let's apply the horizontal force F on the smaller block. This will create a force of friction f on the smaller block and a force of friction f' on the bigger block.
As the blocks are in equilibrium, the net force acting on both the blocks in the horizontal direction must be zero.
Therefore, we can write the following equation:
- F - f = f'
- F - mumg = 3mumg
- F = 4mumg

Hence, the minimum magnitude of horizontal force F required to be applied so that m will remain at rest with respect to 3m is 4mumg.

Conclusion:
In this problem, we used the concept of friction and equilibrium of forces to find the minimum magnitude of horizontal force required to be applied so that m will remain at rest with respect to 3m. It is important to understand the basic concepts of physics and use them in problem-solving.
Community Answer
A block of mass m is supported on the front surface of a bigger block ...
Acceleration of complete system
a = F/(3m + m) = F/(4m)

for vertical equilibrium of m,
μN = mg
N = a * m= m F/4m
N = F/4

so μ > 4mg/F
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A block of mass m is supported on the front surface of a bigger block of mass 3m.The frictional coefficient between the surfaces of two blocks is mu. Find the minimum magnitude of horizontal force F required to be applied so that m will remain at rest with respect to 3m?
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A block of mass m is supported on the front surface of a bigger block of mass 3m.The frictional coefficient between the surfaces of two blocks is mu. Find the minimum magnitude of horizontal force F required to be applied so that m will remain at rest with respect to 3m? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass m is supported on the front surface of a bigger block of mass 3m.The frictional coefficient between the surfaces of two blocks is mu. Find the minimum magnitude of horizontal force F required to be applied so that m will remain at rest with respect to 3m? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass m is supported on the front surface of a bigger block of mass 3m.The frictional coefficient between the surfaces of two blocks is mu. Find the minimum magnitude of horizontal force F required to be applied so that m will remain at rest with respect to 3m?.
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