Highly selective Backlog Qs for JEE MainPhysicsLaws of Motion
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
- Aμ m + M g 2 + μ
- Bμ m + M g 1 + μ
- Cμ m + M g 2 - μ
- 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 < μ m + M g 1 + μ So in the light of equation (i), we get F max = μ m + M g 1 + μ