Question

The friction coefficient between the two blocks shown in figure(6-E9) is µ but the floor is smooth (a) what maximum horizontal force F can be applied without disturbing the equilibrium of the system? (b) suppose the horizontal force applied is double of that found in part(a) , find the acceleration of the two masses..Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "

Answer 1

Solution :
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Both upper block and lower block will have same same acceleration a=2m/s2
R1=mg--------------(1)
F=μR1-T=o
T-μR1=0
and F-μmg=T-------(2)
T=μmg
Putting T =μmg
F=μmg+umg=2μmg

b) 2F=T-μmg-ma=0--------(1)
T-ma-μmg=0[ since R=μg]
T=ma+μmg
Put the value of T in (1) we have,
2F-Ma-μmg-μmg-ma=0
putting the value of F: 
F=2μmg2(2μmg) -2μmg=a(M+m)
4μmg-2μmg=a(M+m)
a=2μmg/M+m

Both the blocks will move this same acceleration 'a' in opposite direction.

Answer 2

Solution :

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Both upper block and lower block will have same same acceleration a=2m/s2

R1=mg--------------(1)

F=μR1-T=o

T-μR1=0

and F-μmg=T-------(2)

T=μmg

Putting T =μmg

F=μmg+umg=2μmg

b) 2F=T-μmg-ma=0--------(1)

T-ma-μmg=0[ since R=μg]

T=ma+μmg

Put the value of T in (1) we have,

2F-Ma-μmg-μmg-ma=0

putting the value of F:  

F=2μmg2(2μmg) -2μmg=a(M+m)

4μmg-2μmg=a(M+m)

a=2μmg/M+m

Both the blocks will move this same acceleration 'a' in opposite direction.