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?
dear student,
using the equation,
2Tcos(theta) = w*g,
therefore w*g*sec(thete) / 2
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AMAN BANSAL
Approved Tutor Answer13 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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