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Grade: 12
        

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?

8 years ago

Answers : (2)

Aman Bansal
592 Points
							

dear student,

using the equation,

2Tcos(theta) = w*g,

therefore w*g*sec(thete)  / 2

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AMAN BANSAL

8 years ago
ankitesh gupta
63 Points
							

  

  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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7 years ago
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