The formula Hitchman is using computes the tension on one of the suspension ropes (which makes sense when you're interested in the threat of one of these ropes breaking.) After much weeping and gnashing of teeth it was established that that formula isOriginally Posted by DGrav
tension = (W/2)/sin(angle)
where W is the weight. You can punch this up on any calculator. With W = 200 lbs and angle = 45 degrees you get tension = 141.4 lbs.
Grizz the teary-eyed
I think this one is a good illustration:
http://whiteblaze.net/forum/vbg/show...imageuser=1908
Thanks, Smee!
The whiteblaze thread where Smee posted this is a hoot.
http://www.whiteblaze.net/forum/showthread.php?t=18685
Starts with similar questions about stresses, and spins wildly out of control, due in no small part to some of the posters in the hammockforums.
The mind staggers at the possibilities.
Grizz, restraining self
Last edited by GrizzlyAdams; 06-07-2007 at 06:09. Reason: wish I knew English as a first language
Ahhhhh ok. I did not realize he was only accounting for half the system. That is where I got confused. It looked like the tension would be less than the initial weight after 30*.
Dear Hammock Physics 102 class,
A number of you have come to my office hours asking about Professor TeeDee's cryptic solution to midterm question number 42. Recall that the problem asks for you to give the tension on a ridgeline, given inclination angle alpha on the suspension ropes above the ridgeline, inclination angle beta on the suspension ropes below the ridgeline, and weight w. Professor TeeDee's solution is illustrated below
The key to understanding this solution is to think about the forces on the join point J where the ridgeline attaches to the suspension ropes.
Slightly abusing notation by reversing direction, the force F2 is the restorative force due to stretching, the force F1 is a force due to gravity on the hammock/user along the rope, and force P is the tension on the ridgeline. We know how to compute F2; we know that the force due to gravity is equal in magnitude to the restorative force on that line. Importantly, we know that in vector arithmetic P + F2 + F1 = 0; J is stationary, so the forces cancel.
Remember the geometry of vector addition : to compute F2+F1 we can draw vector F2, reposition a vector with the same length and angle as F1 at the head of F2, and the sum is the vector extending from F2's base to the head of the repositioned F1. This is illustrated below.
The first thing to note is that the direction of the vector F2+F1 has to be opposite to the direction of P. This is handy, because it means we can use right triangles in the analysis.
Now you need to recall some geometric identities, also illustrated in this picture, of angles equal to alpha and beta. You also need to recall a little bit of trigonometry, that says the length of the adjacent side is the cosine of the angle times the length of the hypotenuse. Knowing the length F2 and the angle alpha, we can compute the length of line segment ac; knowling length F1 and the angle beta, we can compute the length of line segment bc. The difference between lengths of ac and bc gives the length of ab, which is the magnitude of P, and is equal to Professor's TeeDee's solution.
Professor TeeDee reminds you all that the final exam will ask this same question, without giving you angles alpha and beta, only the spring constants and lengths of the suspension ropes and ridgeline.
Please note that June 15 is the last day you can drop this class without penalty. There are still plenty of spaces available in Professor JustJeff's popular survey course on Hammock Camping.
Hammock Physics 102 TA
Last edited by GrizzlyAdams; 06-07-2007 at 10:07. Reason: how could I possibly get such a long post right the first time?
May I continue to audit your class? My plate is kinda full right now with the 'Whittling, Rocking, Whistling and Spitting' class I just signed up for. It is independent study, but it may take up a lot of my time.
I am planning on signing up, that way if I ace the tests I'll stay in. I just need to find out the last day to drop without taking an incomplete.May I continue to audit your class? My plate is kinda full right now with the 'Whittling, Rocking, Whistling and Spitting' class I just signed up for. It is independent study, but it may take up a lot of my time.
Is that too much to ask? Girls with frikkin' lasers on their heads?
The hanger formly known as "hammock engineer".
Please note that June 15 is the last day you can drop this class without penalty. There are still plenty of spaces available in Professor JustJeff's popular survey course on Hammock Camping.
Hammock Physics 102 TA[/QUOTE]
**** you Grizzly...and I thought that I would finally stop having those dreams about having forgotten to go to class all semester and didn't study for the exam. Problem is, I've been out of college for 35 or so years and still occasionally have that dream...lol
Michael
qpens
Dgrav, here's my delayed response - you probably got the answer in the meantime.
At 45deg , a 200lb load would produce 141 lb tension on each rope. If the hammock is level the load is equally split by the two ends, which is my assumption. Being a symmetric system I just have to look at one half, meaning that half the vertical load is sustained by each end. I know this because if I tie only one end the hammock won't stay horizontal
Pretty impressive stuff. Being a lazy engineer I try to simplify things first until I get a very simple formula like: w*2*0.5=w
Of course, you might say that a black box may not be as comfortable as a hammock with a ridgeline. That's why engineers solve small problems one at a time while physicists are still trying to solve one big problem: how does the universe work?
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