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Two blocks of mass M and 3M (in kg) are connected by an inextensible light thread which passes over a light frictionless pulley. The whole system is placed over a fixed horizontal table as shown.The coefficient of friction between the blocks and table surface is mu. The pulley is pulled by a string sttached to its centre and accelerated to left with acceleration a((m)/(s^2)). Assume that gravity acts with constant acceleration g((m)/(s^2)) downwards through the plane of table. Then: (a) Find the forizontal accelration of both the blocks (Assume that both the blocks are moving.) (b) What is the maximum acceleration a, for which the block of mass 3 M will remain stationary? |
Answer» Solution :  (a) Let the ACCELERATION of blocks of mass M and 3 M are `a_1` and `a_2` RESPECTIVELY. From constraint RELATION `a_(pulley)=(a_1+a_2)/(2)=a`.(i) F.B.D. of BLOCK of mass M `T-muMg=Ma_1`.(ii) F.B.D. of block of mass 3 M `T=(-3mu)/(MG)=3Ma_2` .(iii) From (ii) and (iii): `-muMg=3muMg=Ma_1-3Ma_2` `implies2mug=a_1-3a_2` From (i) and (iv) `2mug=(2a-a_2)-3a_a` `impliesa_2=(a-mug)/(2)` From (i) and (v) `a_1=2a-a_2=2a-((a-mug)/(2))=(3a)/(2)+(mug)/(2)=(3a+mug)/(2)` (b) Since `a=(a_1+a_2)/(2)` Acceleration a will be maximum when `a_1` is maximum (since`a_2=0`) Acceleration `a_1` is maximum when tension in the string is maximum for block of mass 3 M to be stationary `Tle3muMg` `T_(max)=3muMg` For block of mass M `T_(max)-muMg=Ma_1` `implies3muMg-muMg=Ma_1` `implies2mug=a_1` So maximum acceleration `a=(a_1+a_2)/(2)=(2mug+0)/(2)=mug` |
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