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Question

Three blocks A,B and C of equal masses m each are placed one over the other on a frictionless table. The coefficient of friction between any two blocks is μ. Find the maximum value of mass of block D so that the blocks A,B and C move without slipping over each other.


A
3mμμ+1
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B
3m(1μ)μ
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C
3m(1+μ)μ
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D
3mμ(1μ)
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Solution

The correct option is D 3mμ(1μ)
Taking blocks B and C as a system, and applying the condition of no slip between A and B.
The FBD of the system is as shown

Let the mass of block D be mD
from the FBD we have
N1=2mg
f=μN1=μ(2mg)

Since the system should move without slipping over each other, then the common acceleration of the system is given by
ac=(Supporting force - Opposing Force)Total mass
ac=mDg3m+mD........(i)

Also, from equation of force we have
f=2mac
2mac=μ2mg ac=μg....(ii)
From eq (i) & (ii) we get
μg=mDg3m+mD
mD(1μ)=3mμ
mD=3mμ(1μ)

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