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The pulleys A and C are fixed while the pulley B is movable . A mass M_(2) is attached to pulley B , while the string has masses M_(1) "and " M_(3) at the two ends . Find the acceleration of each mass . |
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Answer» Solution :The force acting on masses `M_(1),M_(2)"and " M_(3)` are SHOWN in figure . Let `a_(1) "and" a_(3)` be upward acceleration of masses `M_(1)"and" M_(3) "and" a_(2)` the downward acceleration of mass `M_(2)`. For motion of mass M_(1) `T-M_(1) G=M_(1)a_(1)`....(1) For motion of mass `M_(3)` `T-M_(3)g=M_(3) a_(3)`....(2) For motion of mass `M_(2)` `M_(2)g-2T=M_(2)a_(2)`....(3) Let `l_(1),l_(2),l_(3)` be the lengths of vertical portions of the string between PULLEY at any instant . `M_(1) ` goes up through `x_(2) M_(2)` goes down through y and `M_(3)` goes up through z. Then `l_(1)+2l_(2)+l_(3)=(l_(1)-x)+2(l_(2)+y)+l_(3)-z` `rArr x+z =2y rArr (d^(2)x)/(dt^(2))+(d^(2)z)/(dt^(2))=2(d^(2)y)/(dt^(2))` `a_(1)+a_(3)=2a_(2) rArr a_(2)=(a_(1)+a_(3))/2`....(4) From (1), a_(1)=T/M_(3)-g`....(5) From (2),` a_(3)=T/M_(3)-g`....(6) substituting `a_(1) "and" a_(3)` in (4) , we get `a_(2)=T/2(1/M_(1)+1/M_(3))-g`....(7) From (3), `T=(m_(2)(g-a_(2))/2`....(8) substituting this value in (7), we get `a_(2)=M_(2)/4 (g-a_(2))(1/M_(1)+1/M_(3))-g "or" a_(2)=(M_(2)(g-a_(2))(M_(3)+M_(1)))/(4M_(1)M_(3))-g` or `4M_(1)M_(3)a_(2)-M_(2)(M_(1)+M_(3))g-M_(2)(M_(1)+M_(3)a_(2)-4M_(1)M_(3)g` or `[4M_(1)M_(3)+M_(2)(M_(1)+M_(3))]a_(2)=(M_(1)+M_(3))M_(2)g-4M_(1)M_(3)g` `a_(2)=((M_(1)+M_(3))M_(2)g-4M_(1)M_(3)g)/(4M_(1)M_(3)+M_(2)(M_(1)+M_(3)))a_(2)=(M_(1)M_(2)+M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)+4M_(1)M_(3))....(9) Substituting value of `a_(2)` in above equation , we get `a_(1)=(-M_(1)M_(2)+3M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)+4M_(1)M_(3))g`....(10) similarly from (6)and (9) `a_(3)=(3M_(1)M_(2)-M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)-4M_(1)M_(3))`....(11) 
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