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Grade: 12th pass

                        

Two blocks A and B of mass 6kg and 3g respectively are placed on a smooth surface of an elevator in such a way that B is placed above A.The coefficient of friction between the two blocks is 1/4 and the elevator is ascending up with acceleration g/3.A horizontal force 10N is applied on A. Find the accelerations of the blocks w.r.t. the elevator.

6 years ago

Answers : (4)

Pushkar Aditya
71 Points
							First draw the diagram by yourself according to the the given conditions and draw FBD.
Mass of A= 6Kg, Mass of B= 3Kg  
As elevator is accelerating upwards with acceleration of g/3,value of Normal Reaction(N) will change
N-Mg=Mg/3
N=4Mg/3
Hence value of  N between 2 blocks will be = 4*3*9.8/3= 39.2Newton
Now common acceleration of 2 blocks in horizontal direction = 10/(3+6) =10/9 m/s2

*Now check whether there will be slipping between 2 blocks or not
Force required by B to move with common acceleration in horizontal direction = 3*10/9 = 10/3 Newton
Maximum friction between blocks = µ*N = 1/4*39.2 = 9.8 Newton
As 9.8>10/3,
There will be no slipping between the blocks.
They will move with common acceleration in horizontal direction that is 10/9 m/s2


Hence acceleration of both blocks w.r.t. the elevator is = (10/9 i cap + g/3 j cap) m/s2....
						
6 years ago
Harsh Srivastava
34 Points
							Sorry my final answer is w.r.t Earth.

Hence acceleration of both blocks w.r.t. the elevator is = (10/9 i cap ) m/s2....
						
6 years ago
sivasankari sadasivan
18 Points
							Is it 3g or 3kg.....
						
6 years ago
Nirmal Singh.
askIITians Faculty
44 Points
							
201-757_Untitled.png
Thanks & Regards,
Nirmal Singh
Askiitians Faculty
6 years ago
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