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March 17th, 2017, 01:00 AM   #1
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From: England Stoke-on-Trent

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translational equilibrium

See attached drawing,

I'd like to suspend the bar on ropes 1 and 2, rope 1 on one end and rope 2 attached to the other end. once the bar is in equilibrium id like the attachment point of rope1 to remain as drawn,

so my question is: what angle would the bar rotate to in order to be held in equilibrium?,
then I assume after I know what angle of the bar I can draw in rope 2 an from that calculate the tensions on each rope by using:
sum Fy=0=t1cos(-) - t2sin(-)=1KN
sum Fx=0

Basically where would rope2 intercept the bar for it to be in equilibrium

Thank you
Ted
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