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A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is?
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A force Pis equal to 3 m g acts on the block of mass m at an angle of ...
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A force Pis equal to 3 m g acts on the block of mass m at an angle of ...
Understanding the Forces Acting on the Block
When a force P is applied at an angle, both the vertical and horizontal components of this force must be considered, along with the weight of the block and the frictional force.
Components of the Applied Force
- The applied force P = 3mg.
- Angle θ = 30 degrees.
- Horizontal component (P_x) = P * cos(θ) = 3mg * cos(30).
- Vertical component (P_y) = P * sin(θ) = 3mg * sin(30).
Weight of the Block
- The weight (W) of the block = mg.
Net Vertical Force
The net vertical force acting on the block combines the weight and the vertical component of the applied force:
- Net Vertical Force (F_net, vertical) = P_y - W = 3mg * sin(30) - mg.
Since sin(30) = 0.5:
- F_net, vertical = 3mg * 0.5 - mg = 1.5mg - mg = 0.5mg.
Normal Force Calculation
The normal force (N) is the force exerted by the surface to support the weight of the block, modified by the vertical component of the applied force:
- Normal Force (N) = W - P_y = mg - 3mg * sin(30) = mg - 1.5mg = -0.5mg (which indicates the block is lifted).
However, since the block cannot be lifted, we consider the effective normal force:
- Effective Normal Force (N) = mg + P_y = mg + 1.5mg = 2.5mg.
Frictional Force Calculation
The frictional force (F_friction) is given by:
- F_friction = μ * N.
Where μ is the coefficient of friction. Thus,
- F_friction = μ * 2.5mg.
This represents the frictional force acting on the block when the force P is applied at an angle of 30 degrees.
When a force P is applied at an angle, both the vertical and horizontal components of this force must be considered, along with the weight of the block and the frictional force.
Components of the Applied Force
- The applied force P = 3mg.
- Angle θ = 30 degrees.
- Horizontal component (P_x) = P * cos(θ) = 3mg * cos(30).
- Vertical component (P_y) = P * sin(θ) = 3mg * sin(30).
Weight of the Block
- The weight (W) of the block = mg.
Net Vertical Force
The net vertical force acting on the block combines the weight and the vertical component of the applied force:
- Net Vertical Force (F_net, vertical) = P_y - W = 3mg * sin(30) - mg.
Since sin(30) = 0.5:
- F_net, vertical = 3mg * 0.5 - mg = 1.5mg - mg = 0.5mg.
Normal Force Calculation
The normal force (N) is the force exerted by the surface to support the weight of the block, modified by the vertical component of the applied force:
- Normal Force (N) = W - P_y = mg - 3mg * sin(30) = mg - 1.5mg = -0.5mg (which indicates the block is lifted).
However, since the block cannot be lifted, we consider the effective normal force:
- Effective Normal Force (N) = mg + P_y = mg + 1.5mg = 2.5mg.
Frictional Force Calculation
The frictional force (F_friction) is given by:
- F_friction = μ * N.
Where μ is the coefficient of friction. Thus,
- F_friction = μ * 2.5mg.
This represents the frictional force acting on the block when the force P is applied at an angle of 30 degrees.
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A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is?
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A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is?.
A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A force Pis equal to 3 m g acts on the block of mass m at an angle of 30 degree with the horizontal the frictional force between the block in ground( assuming mue is equal to coefficient of friction between the contact in surface) is?.
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