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Pulling force making an angle θ to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is α. then the magnitude of force required to move the body is equal to
  • a)
    W sin α/g tan (θ-α)
  • b)
    W cos α/cos(θ-α)
  • c)
    W sin α/cos(θ-α)
  • d)
    W tan α/sin(θ-α)
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Pulling force making an angle θ to the horizontal is applied on ...
Explanation:
- Free Body Diagram (FBD):
- Draw the free body diagram of the block.
- Show all forces acting on the block including weight, normal reaction, frictional force, and the applied force at an angle θ to the horizontal.
- Resolving Forces:
- Resolve the forces acting on the block into horizontal and vertical components.
- The horizontal component of the applied force will be Fcosθ, and the vertical component will be Fsinθ.
- Force of Friction:
- The maximum force of static friction is given by fs = μsN, where μs is the coefficient of static friction and N is the normal reaction.
- The normal reaction N is equal to the weight W acting downwards.
- Therefore, the maximum force of static friction is fs = μsW.
- The force of friction is given by f = μkN, where μk is the coefficient of kinetic friction.
- Equilibrium Condition:
- For the block to move, the horizontal component of the applied force must overcome the force of friction.
- The force of friction is given by f = μkN = μkW.
- Therefore, Fcosθ = μkW.
- Solving for F, we get F = μkW/cosθ.
- Angle of Friction:
- The angle of friction α is related to the coefficients of friction by tanα = μs = μk.
- Final Expression:
- Substituting the value of μk from the angle of friction relation, we get F = Wsinα/cos(θ-α).
Therefore, the magnitude of force required to move the body is equal to Wsinα/cos(θ-α), which corresponds to option 'C'.
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Pulling force making an angle θ to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is α. then the magnitude of force required to move the body is equal toa)W sin α/g tan (θ-α)b)W cos α/cos(θ-α)c)W sin α/cos(θ-α)d)W tan α/sin(θ-α)Correct answer is option 'C'. Can you explain this answer?
Question Description
Pulling force making an angle θ to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is α. then the magnitude of force required to move the body is equal toa)W sin α/g tan (θ-α)b)W cos α/cos(θ-α)c)W sin α/cos(θ-α)d)W tan α/sin(θ-α)Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Pulling force making an angle θ to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is α. then the magnitude of force required to move the body is equal toa)W sin α/g tan (θ-α)b)W cos α/cos(θ-α)c)W sin α/cos(θ-α)d)W tan α/sin(θ-α)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Pulling force making an angle θ to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is α. then the magnitude of force required to move the body is equal toa)W sin α/g tan (θ-α)b)W cos α/cos(θ-α)c)W sin α/cos(θ-α)d)W tan α/sin(θ-α)Correct answer is option 'C'. Can you explain this answer?.
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