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Q12) Find the minimum value coefficient of static friction between block of weight W' = 40 N 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 determine the minimum coefficient of static friction (μ) required to keep the system stationary, we will analyze the forces acting on the block with weight W' = 40 N. The setup includes a block on a flat surface connected by a rope to another block hanging vertically.
Forces Acting on the Block
- Weight of the Block (W'): The weight acting downward is 40 N.
- Normal Force (N): This force acts upward, equal to the weight of the block, thus N = W' = 40 N.
Considering the Forces Due to Tension
- The rope makes a 30-degree angle with the ceiling.
- The tension (T) in the rope can be resolved into two components:
- Horizontal Component: T*cos(30)
- Vertical Component: T*sin(30) = T*0.5
Using the vertical component of tension, we can express the normal force:
- N = W' - T*sin(30)
Static Friction Requirement
The frictional force (F_friction) must balance the horizontal component of tension to keep the block stationary:
- F_friction = μ * N
- F_friction = T*cos(30)
Setting the two equations equal gives:
μ * (W' - T*0.5) = T * 0.9
Solving for the Coefficient of Static Friction (μ)
Rearranging the above equation, we can derive μ:
- μ = (T * 0.9) / (W' - T * 0.5)
To find the minimum value of μ, we assume T = W' to maximize the tension:
- μ = (W' * 0.9) / (W' - W' * 0.5)
- μ = (40 * 0.9) / (40 - 20) = 36 / 20 = 1.8
Conclusion
Thus, the minimum coefficient of static friction required to keep the system stationary is μ = 1.8. This indicates a significant requirement for friction to prevent sliding under the described conditions.
To determine the minimum coefficient of static friction (μ) required to keep the system stationary, we will analyze the forces acting on the block with weight W' = 40 N. The setup includes a block on a flat surface connected by a rope to another block hanging vertically.
Forces Acting on the Block
- Weight of the Block (W'): The weight acting downward is 40 N.
- Normal Force (N): This force acts upward, equal to the weight of the block, thus N = W' = 40 N.
Considering the Forces Due to Tension
- The rope makes a 30-degree angle with the ceiling.
- The tension (T) in the rope can be resolved into two components:
- Horizontal Component: T*cos(30)
- Vertical Component: T*sin(30) = T*0.5
Using the vertical component of tension, we can express the normal force:
- N = W' - T*sin(30)
Static Friction Requirement
The frictional force (F_friction) must balance the horizontal component of tension to keep the block stationary:
- F_friction = μ * N
- F_friction = T*cos(30)
Setting the two equations equal gives:
μ * (W' - T*0.5) = T * 0.9
Solving for the Coefficient of Static Friction (μ)
Rearranging the above equation, we can derive μ:
- μ = (T * 0.9) / (W' - T * 0.5)
To find the minimum value of μ, we assume T = W' to maximize the tension:
- μ = (W' * 0.9) / (W' - W' * 0.5)
- μ = (40 * 0.9) / (40 - 20) = 36 / 20 = 1.8
Conclusion
Thus, the minimum coefficient of static friction required to keep the system stationary is μ = 1.8. This indicates a significant requirement for friction to prevent sliding under the described conditions.
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Q12) Find the minimum value coefficient of static friction between block of weight W' = 40 N 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?
Question Description
Q12) Find the minimum value coefficient of static friction between block of weight W' = 40 N 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' = 40 N 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' = 40 N 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' = 40 N 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' = 40 N 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' = 40 N 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' = 40 N 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' = 40 N 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' = 40 N 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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