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Question

Given the set up shown in the figure. Blocks A, B and C have masses mA=M & mB=mC=m. The strings are assumed massless and unstretchable and the pulley is frictionless. There is no friction between block B and the support table, but there is friction between blocks B and C donated by a given coefficient μ. Find the accelaration of block A.


A
(M+μmM2m)g
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B
(M+2μmM2m)g
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C
(M+2μmM+2m)g
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D
(M2μmM+2m)g
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Solution

The correct option is D (M2μmM+2m)g
Apply constraint equation on strings, i.e., length of strings is constant. Differentiate twice to get relation between acceleration. Let acceleration of blocks A, B and C be a,b and c respectively.


l1+l2= constant

¨l1+¨l2=0

Also, l3+l4=constant

¨l3+¨l4=0

From which we get, a=b=c

Applying Newton's law on block A,

MgT=Ma -------(1)

on block B,

TT1μ mg=ma ------ (2)

on block C,

T1μ mg=ma ......... (3)

Adding eq(1) and (2) we get,

T=2ma+2μmg=2m(a+μg)

Substituting the value of T in eq(1) we get,

a=(M2μ mM+2m)g

Hence , option (d) is the correct answer.

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