# Bouyancey + Tension Problem

1. Nov 15, 2008

### emoseman

1. The problem statement, all variables and given/known data
A dirigible is moored to two wires. The hydrogen capacity of the craft was 200,000 m3, and together with its cabin, access spaces, and navigation surfaces it displaced 205,000 m3 of air. Its fully-loaded weight, including hydrogen gas, engines, diesel fuel, crew, 72 passengers, and 76 kg of luxury items such as caviar, was 1.95×105 kg. When the ship was safely docked, it was held down by two angled lines, each 20.6 meters long, at a height of 14.2 m above their attachment points on two mooring towers. What was the tension on each mooring line? Assume that the density of air is 1.23 kg/m3

2. Relevant equations
q=m/V
F=qVg

3. The attempt at a solution
Density of airship:
q = m/V = 9.5122e-1kg/m3

Force of air displaced:
qVg = 1.23 * 205000 * 9.8 = 2,471,070N
Weight of ship:
mg = 1.95e5 * 9.8 = 1,911,000N

Upward force due to bouyancy:
2,471,070N - 1,911,000N = 2,279,970N

I'm pretty confident in that so far, the next part, which is what I have a mental block all semester. I have already posted a question on the topic of force body diagrams and have had a great response, but it still isn't making sense.

This is the diagram that I've come up with for this problem:
http://img235.imageshack.us/img235/4369/fbd14122wu4.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

I have found the angle at which the two lines to be 46.4237degrees by way of cos-1(14.2/20.6).

The next part is what I'm not getting, still... The force being exerted on the wire is vertical, so am I finding the force of tension on the wire that is now the resultant ratio of the hypotenuse to the adjacent line?

Assuming that my earlier calculation for bouyant force is correct, the force I came up with is 785,824.902N. Which is wrong.
The equation I used is F = (F cos 46.4327) / 2

So, did I setup the FBD correctly?
Am I correct in that the tension of the wire is a fraction of the force because it is longer than the actual vertical force? The force is being exerted over a longer distance, therefore it is less? And the length is represented by the ratio of adjacent / hypotenuse?
Am I correct in using cosine because of the information I have available, which is what I perceive to be the adjacent side and the hypotenuse side?

--
Evan

Last edited by a moderator: May 3, 2017
2. Nov 15, 2008

### PhanthomJay

correct this subtraction error!
yes, correct
You mean the ratio of the hypotenuse over the adjacent line (20.6/14.2) times 1/2 the net upward force on the ship is the wire tension.
you have used F twice here to represent different forces. Use T to represent tension, and F to represent ther net upward force on the ship. Which is which?
No, this is not correct.
yes, correct. Draw the FBD of the joint where the line wire meets the ship. One half the upward force on the ship acts up on the joint, which must be balanced by the vertically downward component of the wire tension force.

3. Nov 15, 2008

### emoseman

Why is it Hypotenuse over adjacent and not adjacent over hypotenuse? I thought that cosine = adjacent / hypotenuse?

4. Nov 15, 2008

### PhanthomJay

Cosine does equal adjacent over hypotenuse. The hypotenuse represents the wire tension, T. The adjacent side represents 1/2 the upward force on the ship. So since cos theta =adj/hyp, then cos theta = (F/2)/T, or rearranging, T= (F/2)/cos theta, which is T=(F/2)/(adj/hyp), which is T = (F/2)(hyp/adj). No matter how you cut it.

5. Nov 15, 2008

### emoseman

I've never seen this solved that way, but it makes more sense. I've been taught to create a table containing the x and y forces and then solve for T, in this case I thought it would be T = Fcos theta. Which is obviously wrong.
It turns out to be T = F * (inverse sin) theta
Have I been taught wrong? Or am I missing fundamentals?

6. Nov 15, 2008

### PhanthomJay

No, it 's T=F/cos theta, where F is half the upward force on the ship.
I don't think you've been taught wrong, you're just getting confused with geometry, trig, and Newton's 1st law.

7. Nov 16, 2008

### emoseman

Do you know of a good source on the topic of Free Body Diagrams? A book perhaps that teaches this in detail? Or maybe even a reference on the internet?

Thanks very much for all of your help, I really do appreciate it!

--
Evan

8. Nov 17, 2008

### PhanthomJay

I suppose you can google on "Free Body Diagrams" and get thousands of results. FBD's are useful in identifying all forces (weight and contact forces) and direction of those forces acting on an object or objects or parts of an object or objects, or particles and joints. Once these forces are identified, then Newton's laws may be applied to determine the unknown values.
In your problem, you should understand that there are 2 guy wires holding down the airship that prevent it from floating up. One of these wires is attached at the left of the ship, and the other at the right. Your diagram doesn't show that, and the problem didn't specifically state it, so I just want to be sure you understand the setup. Drawing a FBD of the ship, you should be able to see, from symmetry, that the net upward force on the ship, applied at its center, must be balanced by vertical downward forces at the left and right attachment points, each equal to F/2, where F is the net upward force you already calculated (after correcting for the math error). This comes from Newton 1. (There must also be horizontal forces at these points, equal and opposite at the left and right attachments, respectively). Now turn your attention to the attachment joint at the left, and isolate the joint in a FBD. From Newton 3, there must be, in the vertical direction, a vertical force, F/2, acting up. Per Newton 1, this must be balanced by the vertical component of the wire tension force, Tcos theta, acting down. So thus, Tcos theta = F/2, from which, T = (F/2)/cos theta. Clearer?