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NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice

A wedge of mass m , lying on a rough horizontal plane, is acted upon by a horizontal force F 1 and another force F 2 , inclined at an angle θ to the vertical. The block is in equilibrium, then the minimum coefficient of friction between it and the surface is

Options

  1. AF 2 sinθ ( mg + F 2 cosθ )
  2. B( F 1 cosθ + F 2 ) mg - F 2 sinθ
  3. C( F 1 + F 2 sinθ ) ( mg + F 2 cosθ )
  4. D( F 1 sinθ - F 2 ) ( mg - F 2 cosθ )

Correct answer

C. ( F 1 + F 2 sinθ ) ( mg + F 2 cosθ )

Step-by-step solution

We know f = μN and N = mg + F 2 cosθ Now, since wedge is in equilibrium, ( F 1 + F 2 sinθ ) = μ ( mg + F 2 cosθ ) μ = F 1 + F 2 s i n θ m g + F 2 c o s θ

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