# A flexible chain of weight w hangs between 2 points A and B at the same level. the inclination of chain with horizontal at two points of support is theeta.what is tension in chain at mid point?

Aman Bansal
592 Points
11 years ago

dear student,

using the equation,

2Tcos(theta) = w*g,

therefore w*g*sec(thete)  / 2

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ankitesh gupta
63 Points
9 years ago

SINCE WE HAVE TO FIND TENSION AT THE MIDPOINT THEREFORE MASS OF HALF CHAIN IS W/2 LET THE ANGLE BE ALPHA SINCE WE HAVE TO FIND TENSION AT ''C'' CLEARLY FROM THE F.B.D

T*SIN(ALPHA)=T............................ (1)

BY MISTAKE I DINT SHOW ONE FORCE ON FBD WHICH IS (W/2)g DOWNWARS WHICH IS FORCE BY GRAVITY

THEREFORE  T*COS(ALPHA)=(W/2)g.......................(2)

SINCE YOU HAVE TO FING VALUE OF T DIVIDE (1) AND (2) EQUATION YOU GET...

TAN(ALPHA)=T/(W/2)g  THEREFORE

T=[TAN(ALPHA)](W/2)g]........................................... ANS

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