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In the pulley system shown in figure . P and Q are fixed pulleys while A, B and C are movable pulleys each of mass 1kg . Thestrings are vertical and inextensible . Find the tension in the string and acceleration of frictionless pulleys A,B and C |
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Answer» Solution :A single string whose ends are tied to centres of A and B, passes over all the pulleys, so tension at each point of string is same equal to T WEIGHT of each pulley A,B andC is `mg=1g `Newton. Let `y_(A),y_(B)` and y_(C)`be the distance of centres of pulleys A,B and Cfrom fixed pulleys at any time t. Following the string STARTING from end A and reaching upto end B, we have `(y_(B)-y_(A))+y_(B) +2y_(A)+y_(C)+y_(C)-y_(B)=L`= constant Differentitating twice with respect to t, we get `(d^(2)y_(A))/(dt^(2))+(d^(2)y_(B))/(dt^(2))+2(d^(2)y_(C))/(dt^(2))=0` i.e., `a_(A)+a_(B)+2a_(C)=0` where `a_(A),a_(B)`, and `a_(C)` are acceleration of pulleys A,B,C respectively. Now equations of MOTION of pulleys A,B and C are `mg+T-2T=ma_(A) rArr mg-T=ma_(A)`....(2) `mg+T-2T=ma_(B) rArr mg-T=ma_(B)`....(3) and `mg-2T=ma_(C)`....(4) From (2) and(3) it is obvious that and `a_(A)=a_(B) =(g-T/M) `.....(5) and from (4), `a_(C) =g-(2T)/m`.....(6) substituting `a_(A),a_(B)` and `a_(C) ` in (1) , we get (g-T/m)+(g-T/m)+2(g-(2T)/m)=0` `4g-(6T)/m=0 rArr T=2/3 mg=2/3xx1xx9.8=6.5N` `:. a_(A)=a_(B)=(g-T/m)=9.8-(6.5)/1=3.3 m//s^(2)` From (1), `ac=-a_(A)= -3.3m//s^(2)` |
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