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Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9?
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Q12) Find the minimum value coefficient of static friction between blo...
Understanding the Problem
To find the minimum coefficient of static friction between the block (weight W' = 40 N) and the tabletop, we need to analyze the forces acting on the system.
Forces Acting on the System
- Weight of the Block: The weight W' = 40 N acts vertically downward.
- Tension in the Rope: The rope creates tension, which affects the equilibrium of the block.
Components of Forces
- The tension in the rope can be resolved into two components:
- Vertical Component: T * sin(30) = 0.5T
- Horizontal Component: T * cos(30) = 0.9T
Equilibrium Conditions
- For the block to remain stationary:
- The vertical forces must sum to zero:
W' = 0.5T
- The horizontal forces must also balance:
Friction = Horizontal Component of Tension
Calculating Tension
- Rearranging the vertical equilibrium condition:
T = W' / 0.5 = 40 N / 0.5 = 80 N
Frictional Force Requirement
- The frictional force must counter the horizontal component of tension:
Friction = 0.9T = 0.9 * 80 N = 72 N
Static Friction and Normal Force
- The maximum static friction (fs) is given by:
fs = μ * N, where μ is the coefficient of static friction and N is the normal force.
- Here, normal force N = W' = 40 N.
Calculating Coefficient of Static Friction
- Setting the friction equal to the static friction:
72 N = μ * 40 N
- Solving for μ gives:
μ = 72 N / 40 N = 1.8
Conclusion
The minimum coefficient of static friction required to keep the system stationary is 1.8. This value ensures that the static friction can effectively counter the forces acting on the block.
To find the minimum coefficient of static friction between the block (weight W' = 40 N) and the tabletop, we need to analyze the forces acting on the system.
Forces Acting on the System
- Weight of the Block: The weight W' = 40 N acts vertically downward.
- Tension in the Rope: The rope creates tension, which affects the equilibrium of the block.
Components of Forces
- The tension in the rope can be resolved into two components:
- Vertical Component: T * sin(30) = 0.5T
- Horizontal Component: T * cos(30) = 0.9T
Equilibrium Conditions
- For the block to remain stationary:
- The vertical forces must sum to zero:
W' = 0.5T
- The horizontal forces must also balance:
Friction = Horizontal Component of Tension
Calculating Tension
- Rearranging the vertical equilibrium condition:
T = W' / 0.5 = 40 N / 0.5 = 80 N
Frictional Force Requirement
- The frictional force must counter the horizontal component of tension:
Friction = 0.9T = 0.9 * 80 N = 72 N
Static Friction and Normal Force
- The maximum static friction (fs) is given by:
fs = μ * N, where μ is the coefficient of static friction and N is the normal force.
- Here, normal force N = W' = 40 N.
Calculating Coefficient of Static Friction
- Setting the friction equal to the static friction:
72 N = μ * 40 N
- Solving for μ gives:
μ = 72 N / 40 N = 1.8
Conclusion
The minimum coefficient of static friction required to keep the system stationary is 1.8. This value ensures that the static friction can effectively counter the forces acting on the block.
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Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9?
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Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9?.
Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9?.
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Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9?, a detailed solution for Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9? has been provided alongside types of Q12) Find the minimum value coefficient of static friction between block of weight W' = 40N and the table top, so as to keep the system stationary. One rope connects the two blocks to each other and it goes horizontal and then vertical. A light rod connects the rope to ceiling at 30degree angle as shown. sin 30 is 0.5 and cos 30 is 0.9? theory, EduRev gives you an
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