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March 11th, 2017, 11:10 AM   #1
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From: England Stoke-on-Trent

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Unusual Equilibrium Problem


I am relatively new to FBD but I do understand the principle, however this problem has thrown me, most the problems I’ve attempted before have had both ropes higher than the load, the fact that in this problem rope 2 is pulling downwards on the load thus increasing the Y component on rope 1 greater than simply Mg. this has me stuck.
I’m not looking for the answer just an explanation of the correct method, I am struggling to decide what way to attack it from?
I’m inclined to treat the top of the bar as a fulcrum and calculate the torque on the bar and from this calculate the tensions, but for this I’d need the angle of the bar.
I’m quite stuck to be honest I have attempted a FBD at either end of the bar and another attempt I found the intersection of the two ropes and did FBD there, both to no avail, If anyone could point me in the right direction I would greatly appreciate it.
Thanks Ted

A uniform bar 3m in length weighing 1kn is hanging on rope 1 (frictionless pulley)
Rope 2 is then used to pull the bottom of the bar out and the bar lowered to its finish position
What are the tensions on rope 1 and rope 2?
What angle of rotation would the bar be at during illustrated point of the transfer?
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File Type: jpg suspended bar example.jpg (64.6 KB, 5 views)
tedmotox is offline  
March 12th, 2017, 09:14 AM   #2
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I'm assuming the beam is in translational equilibrium ... I didn't work out the rotational forces.

My annotated diagram is posted. Barring any mistakes on my part, I get the the upper rope tension as $T_1 \approx 1.224 \, kN$ and the lower rope tension as $T_2 \approx 0.451 \, kN$

equations ...

$T_1\sin{\alpha}-T_2\cos{\beta} = 1 \, kN$

$T_1\cos{\alpha} - T_2\sin{\beta} = 0$

Angle that the beam makes with the horizontal, $\theta \approx 48.2^\circ$

skeeter is offline  
March 13th, 2017, 02:26 AM   #3
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Thank you for the reply Skeeter,
Yes that's right it is in translational equilibrium

I think that solves part of the problem however the thing that puzzling me is the position of the bar left to right can be adjusted by rope 1 and 2 but the bar will rotate counter clockwise the further its pulled over to the right, what I'm trying to do is find the forces on the ropes and in turn what that does to the rotation, you couldn't rotate the bar in one location by pulling on a rope it would rotate the bar to a different location.

hope this explains it a bit better.

Thanks Ted
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