
Rope Tension  problem  
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Posted by: hhari ^{®} 01/19/2007, 16:07:51 Author Profile eMail author Edit 
Hello all,
I have small statics/dynamics problem and I need help solving it. A guy with mass 'm' is standing on a roof and freely falls a height 'h' as shown in the attached .pdf file. A freebody diagram is also shown in the .pdf file. A cable is hooked up to 2 posts and another cable is tied in the middle of cable. A mass 'm' is attached at the end of small cable with a spring constant 'k1'. I am interested in finding the support reactions at the points 1,2,3 and 4. Also I would like get the tension in the long cable which has spring constant of 'k2'. Length of longer cable is '2L' Length of short cable is 'S' I hope you understand what am I talking about here. Any help on this problem  even an approach is truly appreciated.  This is not a school assignment problem. Thank you

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Re: Rope Tension  problem  
Re: Rope Tension  problem  hhari  Post Reply  Top of thread  Forum 
Posted by: hhari ^{®} 01/22/2007, 19:08:13 Author Profile eMail author Edit 
Hey zekeman, Thank you very much for the reply.
I want to know is my approach correct? Or am I missing anything? Thank 
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Re: Re: Rope Tension  problem  
Re: Re: Rope Tension  problem  hhari  Post Reply  Top of thread  Forum 
Posted by: zekeman ^{®} 01/22/2007, 22:27:02 Author Profile eMail author Edit 
You have 3 different ropes, with 3 different tensions. If the mass is not constrained laterally, then after an initial pendulum type motion, the mass rests directly under the s rope and
k1d = mg But the energy is stored in 3 ropes. the s rope and each half of the diagonal rope. If each half has a spring constant = k2, then I get mgh=1/2 k1d^2+ 1/2k2(X2^2+X3^2) where x2 and x3 are the elongations of each of the 1/2 rope. This differs from your assumption. Now you can write the remaining equations of equilibrium at the connecting point, but you must find the angles that the diagonal rope makes with the posts after the deflections and they depend on the geometry as well as the forces. Hoever, since the deflection angles are assumed to be very small, the remaining equation can be the obtained by equating the horzontal components of the diagonal rope using the initial angles between the diagonal rope and the posts. 
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Re: Re: Re: Rope Tension  problem  
Re: Re: Re: Rope Tension  problem  zekeman  Post Reply  Top of thread  Forum 
Posted by: hhari ^{®} 01/22/2007, 23:22:33 Author Profile eMail author Edit 
Zekeman, I understand what you say, but in my case the mass is not allowed to swing as pendulum because I have lateral obstruction shown in the drawing (which is not shown in the FBD). I think it is easy to calculate the tension (F) in the vertical cable then apply that tension (F) to calculate the stretch in the horizontal cable using the equation you suggested (mgh=1/2 k1d^2+ 1/2k2(X2^2+X3^2)), but what I think is, the load will be at rest at this instance so I can ignore the "h" term from your equation and use
Another case, If I treat the short rope and the mass as one entity and assume that the load directly falls hanging from the long cable in the middle (Example: want to find then tension in the cable when a load "m" falls directly in the middle of the cable from a height "h") then can I write:

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Re: Re: Re: Re: Rope Tension  problem  hhari  Post Reply  Top of thread  Forum 
Posted by: zekeman ^{®} 01/23/2007, 11:00:14 Author Profile eMail author Edit 
This problem is not so simple as you see at first blush.
Now, your assumption that the load can be in the center cannot be sustained since the left horizontal rope would remain slack at all times; however, if you put the mass offset as the force diagram shows, it will work and further your assumption of ignoring the stretch in the vertical cable is a good one since the elasticity of the horizontal rope is orders of magnitude greater, i.e. Keff< Finally, the tension in the s rope is simply the vector sum of the two "horzontal" rope forces. 
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Re: Rope Tension  problem  
Re: Rope Tension  problem  hhari  Post Reply  Top of thread  Forum 
Posted by: zekeman ^{®} 01/22/2007, 14:24:33 Author Profile eMail author Edit 
Looks like a school problem to me but I will give you the benefit of the doubt and one hint anyway.
You equate the energy lost due to potential mgh to the energy absorbed in the cables 1/2k1x1^2+1/2k2x2^2. You need i more equation to solve it. I'll leave that to you. Good luck and I would be interested in the application if it is not academic. 
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