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Suppose the given system consisting of 2 masses(M>m) hung over an ideal frictionless pulley is in a lift accelerating with acceleration a in upward direction. Calculate the tension on two sides of the string in lift frame & in ground frame of references. My solution of the above problem is something like this- Let the tensions on left and right parts of the string be t1 & t2 respectively w.r.t lift frame. Then additional forces of Ma & ma act on the two masses respectively. => M(g+a)-T=Mx (x is downward acceleration of mass M w.r.t lift) T-m(g+a)=mx Solving, we get x=(M-m)(g+a)/(M+m) Let w.r.t ground, tensions in left & right stings be T1 & T2 respectively. Then obviously, T1=t1+Ma & T2=t2+ma =>T1 & T2 are different. But this statement is logically wrong. e.g. consider an infinitesimal part of string vertically above pulley. Tension on it is different on both sides. As the mass of this segment is negligible, this is not possible. Can anyone explain me what Im doing wrong?

Suppose the given system consisting of 2 masses(M>m) hung over an ideal frictionless pulley is in a lift accelerating with acceleration a in upward direction. Calculate the tension on two sides of the string in lift frame & in ground frame of references.


My solution of the above problem is something like this-


Let the tensions on left and right parts of the string be t1 & t2 respectively w.r.t lift frame. Then additional forces of Ma & ma act on the two masses respectively.


=> M(g+a)-T=Mx (x is downward acceleration of mass M w.r.t lift)


T-m(g+a)=mx


Solving, we get x=(M-m)(g+a)/(M+m)


Let w.r.t ground, tensions in left & right stings be T1 & T2 respectively.


Then obviously, T1=t1+Ma


& T2=t2+ma


=>T1 & T2 are different.


But this statement is logically wrong. e.g. consider an infinitesimal part of string vertically above pulley. Tension on it is different on both sides. As the mass of this segment is negligible, this is not possible.


Can anyone explain me what Im doing wrong?


 


 

Grade:12th Pass

1 Answers

Ankit Sharma
30 Points
9 years ago

An ideal frictionless pulley doesnt have mass so tension on boths sides will remain same. T1=T2

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