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Question

Two blocks A and B of mass mA=1 kg and mB=3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F (in newtons) that can be applied on B horizontally, so that the block A does not slide over the block B is [Take , g=10 m/s2]

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Solution

Acceleration a of system of blocks A and B is
a=Net forceTotal mass=Ff11mA+mB
where, fl1 = limiting friction between B and the surface
=μ(mA+mB)g
So,
a=Fμ(mA+mB)g(mA+mB)...........(i)
Here, μ=0.2,mA=1 kg,mB=3 kg,g=10 ms2
Substituting the above values in Eq. (i), we have
a=F02(1+3)×101+3
a=F84.............(ii)
Due to acceleration of block B, a pseudo force F' acts on A
This force F' is given by
F=mAa
where, a is acceleration of A and B caused by net force acting on B.
For A to slide over B; pseudo force on A, i.e. F' must be greater than limiting friction between A and B.
mAafl2
We consider limiting case,
mAa=fl2mAa=μ(mA)g
a=μg=0.2×10=2 ms2......(iii)
Putting the value of a from Eq. (iii) into Eq. (ii). we get
F84=2
F=16 N

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