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NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice

The friction coefficient between the board and the floor shown in diagram is μ . Find the maximum force that the man can exert on the rope, so that the board does not slip on the floor.

Options

  1. Aμ m + M g 2 + μ
  2. Bμ m + M g 1 + μ
  3. Cμ m + M g 2 - μ
  4. Dμ m + M g 1 - μ

Correct answer

B. μ m + M g 1 + μ

Step-by-step solution

The forces acting on the system are shown in the diagram For vertical equilibrium of the point P T   =   F ....(i) And for the vertical equilibrium of the system R   +   T   =   ( m + M ) g , i.e.,   R   =   ( m + M ) g   -   T ....(ii) Now the system will not move horizontally till   T   <   f L i.e.,   T   <   μ [ m + M g   -   T ]   f L   =   μ R which on simplification gives T < &#

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