Question

Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the two blocks is µ1 and that between the bigger block and the ground is µ2.Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "

Answer 1

Solution :
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Let the acceleration of the block M is 'a' towards right.
So the block 'm' must go down with an acceleration '2a'
As the block m is in contact with the block 'M', it will also have acceleration'a' towards right . So it will experience two internal forces as shown in the free body diagram.

From the free body diagram:
Considering the forces in the vertical direction on block m. mg-T- µ1.ma =m.2a 
→ T= mg- µ1.ma -2ma

Now consider the vertical forces on the larger block M, See FBD.  
Downward forces are weight Mg and friction by smaller block m = µ1.ma
and Tension T at the pulley.
So total downward force =Mg+µ1.ma+T=Normal force by the ground R' 
So friction force =  µ2(Mg+µ1.ma+T)    its direction will be opposite to 'a'.

Now consider the horizontal forces on the larger block
M2T-R-µ2(Mg+µ1.ma+T)= Ma
 → 2mg-µ1.2ma-4ma-ma-µ2(Mg+µ1.ma+mg- µ1.ma -2ma) =Ma   
(putting the value of T)
→ 2mg-µ2Mg -µ2.mg+ µ2* 2ma  =Ma+5ma+ µ1.2ma
→ [2m-µ2(M+m)]g=Ma+5ma+µ1.2ma -µ2.2ma   
→ [2m-µ2(M+m)]g=[M+m{5+2(µ1 -µ2)}]a  
→a =  [2m-µ2(M+m)]g/[M+m{5+2(µ1 -µ2)}]