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View Full Version : Help! Statics and Dynamics specific question!

Cubsfan
07-31-2002, 05:45 AM
Ok. I am hoping that some of you have the same book that I do which seems to be the standard for SnD classes. IT is "Vector Mechanics for Engineers" 5th ed. Beer/Johnston.

Its gonna be hard to explain my question unless you can look at the diagram yourself, so please refer to the diagram on problem 3.45, pg 85. It should be an L shaped pipe suspended by a single cable thru a ring.

My question is "How do I solve for the tension in the cable with the downward force P given". The force at the corner. For instance, if the force P was given as 1000N, or any force for that matter, how do I translate that into the tension in the cable? I don't need a solution, I just want a method because I need to be able to do this.

If it is a rigid body, which it is, does that mean that I can move the force P to any point in the rigide body and retain all of the same characteristics? For some reason I am thinking that the tension in the cable is = P. I say this because the force is only in one direction. If I am not right, which I suspect I am not, then do I have to solve for the components of the force on the pieces of cable on each side of the rings and then add them?

I am doing this class independent study and don't have a way to check my method until I meet with my instructor which isnt for a week. This is holding me back from solving other pieces of the problem.

BTW, I know the cable tension is given in the problem, but I need to know how to solve for the situation that I gave you, so disregard the cable tension that is given.

zut
07-31-2002, 06:29 AM
If you don't get an answer to your question soonish, it might be a good idea to at least try to describe the figure. I suspect that the pool of people with the fifth edition of Beer and Johnson is pretty limited (I have the fourth edition, for example). Seems like such a problem would be pretty easy to describe generally, but I suspect your quick description above is missing some details, since an object suspended by one cable and nothing else is pretty trivial.

Cubsfan
07-31-2002, 06:43 AM
Ok, how about this. When I get home from work in 2 hours I will scan the diagram and post it on my website for you to look at. I will flub it up if I try to describe it.

Check back in 2 hoursish and I will post it on my site.

Thank you for looking. This class is fricking hard. Visualizing the forces is the tough part for me.

Cubsfan
07-31-2002, 09:00 AM
Ok. I am back. I scanned the figure I am referring to out of the book and put it on my website. I am trying to figure out how to calculate the tension in the cable based on the measurements and a given force at P. In case you cannot tell it is one cable going from G thru a ring at B and back to H. The pipe is weightless.

www.beefchips.com/statics

Any help would be greatly appreciated. I know it is something simple I am missing.

There should only be a Y component of force on the cable correct? Since the P force pushes straight down? Under that assumption I calculated the components using P and units vectors going from BG and BH and added them together. I don't think this is right but I had to try something.

FWIW I tried a few different values of P, but for a value of 2000 N I got 2976N tension in the cable.

Cubsfan
07-31-2002, 09:05 AM
Good grief. I dont know why that link didnt work. if you type it in your address bar it works.

http://www.beefchips.com/statics

Sorry.

zut
07-31-2002, 09:23 AM
I can't get your link to work at all, either typing or clicking.

zut
07-31-2002, 09:46 AM
OK, I got it. The link you wanted was:

http://www.beefchips.com/statics.jpg

The method used to calculate tension in the cable, given force P, is to balance all the forces and moments in the pipe structure; I think this procedure is more advanced than the section of the book that you're in right now. In any case, you can think of the pipe ABC as a lever, so that the upwards force at B must be P*800/400 = 2P. Since the cable is continuous, the same tension is in both BG and BH. Using geometry, take the z-component of the tension for both BG and BH, these must add to 2P.

Chronos
07-31-2002, 01:19 PM
The way I would do that problem, is I would first re-draw the diagram such that my eye is along the axis AD, so I now have a simple lever. Then, I'd use the pivot point of the lever as my axis for torque, so the torque from the wire must equal the torque from force P. This lets me find the upward component of the force on the ring. From there, I'd find the angles the cable is from vertical, and use that and some trig to get the total tension.

Cubsfan
08-01-2002, 02:33 AM
Ok, I must be totally confused. The force points straight down the Y axis. There is no Z component. As far as it being a lever goes, it is hinged at points A and D, so it isn't, as far as I can tell, a lever that operates only on the Y axis, but also the X and Z.

zut
08-01-2002, 06:31 AM
Oh crud; my mistake. I just assumed that the z-axis was vertical. Substitute "Y" for "Z" in the post above.

zut
08-01-2002, 07:10 AM
And further: The way to solve this problem, as completely as possible, is to sum all the forces and moments on the free body. I think, looking at the arrangement of concepts in my book, that this idea might be introduced in the next chapter? Anyway, the idea is that, since the pipe is not moving, you know that the sum of all the forces in any direction must equal zero, and (since the body is not spinning, either) the sum of all moments is zero.

Now, for this problem, you have one known force, P, acting on the pipe. You also have two rope forces; they are both same magnitude (but different, known, directions); we'll call the tension in the rope T. Assuming that the hinges at A and D can be rotated in any direction, there are also two reaction forces at the hinges. Unlike the force from the rope, you don't know the direction of the reaction force. The most convenient way to express the reaction forces, then, is to break them down into their vector components; you can think of the reactions as being three seperate forces, Fax in the x-direction, Fay in the y-direction, and Faz in the z-direction at point A, and three forces, Fdx in the x-direction, Fdy in the y-direction, and Fdz in the z-direction at point D. So, it turns out, there are seven unknown forces on the pipe.

Having seven unknown forces is unfortunate for an example problem like this, because summing the forces and moments in all directions gives you six equations: SFx=0, SFy=0, SFz=0, SMx=0, SMy=0, and SMz=0. Thus, this problem is indeterminate, i.e., we can't solve for all unknowns. This might make intuitive sense from looking at the picture: suppose the pipe had some "spring" in it so that it was pushing on points A and D along line AD; that fact would change the final result, but you can't represent it in the original problem.

Anyway, all is not lost, because, even if all the unknown can't be solved for, some can:

SMx=0 (summing moments around point C) -> 600mmFdy = 0 -> Fdy = 0 (all other forces have a moment arm of zero)
SMz=0 (summing moments around point B) -> 400mm(Fay) - 400mmP = 0 -> Fay = P (remembering that Fdy = 0)
SFy=0 -> Fdy + Fay + P - (the z-components of the tension T) = 0

Since Fdy = 0 and Fay = P, you now have T as a function of P, so a little geometry gets you T.

Again, if you haven't seen this methodology before, this is probably an unfortunately complex (and indeterminate) problem to introduce them in.

Cubsfan
08-01-2002, 08:44 AM
Ok. I think I follow most of what you are saying. Bottom line is that the tension in that cable is equal to 2P?

Thanks for taking the time to help me on this. This is the last time (hopefully) I ever indepent study an engineering class again.

zut
08-01-2002, 09:04 AM
Well, the force on the pipe at point B is the vector addition of the two tensions (both T). The y-component of this force is 2P. So you still have to do some geometry to see what magnitude T is, such that the y-component of T along BG plus the y-component of T along BH equals 2[b]P/b].

Chronos
08-01-2002, 04:16 PM
Wow, engineers call torques "moments"? You guys are weird. What do you call moments?