1T=2mg*cos(theta)
AB is a string whose one end A is fixed at the highest point of a ring placed in vertical plane. C is the center of the ring. At other end B of the ring, is attached a bead of mass m , wich can slide along the ring.If the bead is in equilibrium at B nd the string is taut then fInd the tension in the string .
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6 Answers
23We get
Tsin\theta=mgcos2\theta
and ans is given as T = 2mgcos\theta
isnt it wrong ??? how are they not considering the case sin\theta = 0 ??
and i feel sin\theta = 0 is the only possible case.
1ohk now i got which sin @=0 u were talking of.......see in thth case
T=0
but N=mg....whats the problem??
23@ learning , dont u think at sin theta = 0 ,
A,C,B will be on the same line ?? why are u taking the bead outside the circle ??? there is no point in doing that.
@ anirudh tnx for posting solution , i realized my mistake , ( didnt consider Normal force ) [4]