28) [ friction ]

 



This one is a little tricky problem. To find the solution, not only you need to identify the forces acting on the blocks but also figure out the relative motion of the two as well.



As the string is inextensible, whenever the lower end of the string (connected to M) displaces by x units to the right, the other end of the string (connected to block m) lowers down by 2x units. This is to be showed by 'constant length of the string' method as described well in HCV - 1 pg 73 , example 6 in ch. Newton's Laws of Motion



So accl^n of m (vertically downward) = 2 ( accl^n of block M in horizontal right direction )



Let accl^n of M = a



So accl^n of m = 2a.



Also, the block m is always in contact with the larger block M as far as motion in the horizontal direction is concerned. So accl^n of block m in horizontal direction is also a (same as that of M)



 



Now look at the figure.



The blue coloured are the forces on M and red ones act on m



Motion of m:



the forces are:



1) mg downwards



2) R ( contact force by M ) towards right



3) 1R ( frictional force) upwards



4) T ( Tension) upward



 



For horizontal direction.



R = ma



 



In vertical direction



mg - T - 1R = m(2a)



or T = mg - ma(2 + 1)



 



Motion of M:



 



The forces on M are:



1) Mg downwards



2) R ( contact force by m ) towards left



3) 1R ( reaction of frictional force on m ) downwards



4) T ( Tension due to string on the lower end ) towards right



5) N ( contact force by ground ) upwards



6) 2N ( frictional force due to ground ) towards left



7) T ( Tension due to string on pulley attached to M ) towards right



8) T ( Tension due to string on pulley attached to M ) downwards



 



for vertical equilibrium



N = Mg + T + 1R



or N = Mg + T + 1ma {since R = ma}



 



In horizontal direction



2T - R - 2N = Ma



 



Putting values of R, T and N.......



2T - ma - 2(Mg + T + 1ma) = Ma



or (2 - 2)T - ma - 2(Mg + 1ma) = Ma



or (2 - 2)[mg - ma(2 + 1)] - ma - 2(Mg + 1ma) = Ma



or 2mg - 2g(m + M) = 5ma + 2ma1 - 2ma2 + Ma



or a[M + m{5 + 2(1 - 2)}] = g[2m - 2(m + M)]



or a = g[2m - 2(m + M)] / [M + m{5 + 2(1 - 2)}]