Two Situations of Tension Same Thing?

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The discussion revolves around calculating the tensions in two different diagrams of a box hanging from two wires. The user initially believes both diagrams represent the same situation, leading to confusion over differing answers from their calculations and the answer key. After working through the equations, they find T1 = 25N and T2 = 20N, but suspect an error due to the answer key showing different values. Other participants confirm that T2 should be greater than T1, emphasizing the importance of accuracy in calculations and the relationship between angles and tension. Ultimately, the user is encouraged to double-check their substitutions and rounding errors, as these may have led to discrepancies in their results.
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[SOLVED] Two Situations of Tension... Same Thing?

Homework Statement



A box is hanging with two wires. (Refer to diagrams). Find the two tensions.
My first question is, if the two diagrams are different situations that will lead to different answers? Personally I don't think they would... since the x coordinates still have to equal zero, with the same angles... but I'm not exactly sure, can someone please clarify the concept behind this?

Diagram 1
http://img252.imageshack.us/img252/329/yole7.jpg

Diagram 2
http://img339.imageshack.us/img339/5839/yo1yw5.jpg

Homework Equations



T1*sin56 + T2*sin45 - 35N = 0

-(T1*cos56) + T2*cos45 = 0

The Attempt at a Solution



Prepared for substitution
T2 = T1(cos56)/cos45

T1(sin56) + T1(sin45)(cos56)/(cos45) - 35N = 0
T1(sin56 + sin45*cos56/cos45) = 35N

Solved for T1 = 25N.

Plugged in and got T2 = 20N.

So based on those equations I get T1 = 25N, and T2 = 20N... but the answer key shows a way different answer so I am confused. Could it be that the answer key is wrong? Or maybe it is because of the two different situations I've shown above? I'm use to doing problems like diagram 2, but for this question it is shown like diagram 1?

Please let me know if I made a mistake, this is driving me crazy.
 
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The situations are the same. This is because the assumption that the mass is a point is made. Further more, any vector (and so, any force) can be moved parallel to itself and it will remain effectively the same vector.

Looking at your working out, it looks fine to me, I can't see were you went wrong, unless you made some silly error.

EDIT: T2>T1, you have made an error in finding T2, perhaps T1 is correct.
I can see that T2>T1 because it is opposite the larger angle in the triangle. Also from this:
T2 = T1(cos56)/cos45
cos56>cos45>0, so cos56/cos45>1, hence T2>T1
 
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Ok wait. So if I label the left tension as T1, and the right tension as T2, did I do the problem correctly?
 
Can anyone please help me? I don't understand the error I made.
I think I made an error here

-(T1*cos56) + T2*cos45 = 0

Please help me!
 
Hell I don't know either, it looked ok to me

That equation should be ok, it doesn't matter where the negative goes, you could divide both sides by -1, which moves it to the other term, and 0/-1 is still 0, so whichever

You DID use more accuracy than 25 and 20, right? I got like 25.stuff and 20.stuff
 
Are you getting 25 for the left and 20 for the right?
 
Look, the string opposite the larger angle MUST be bigger (from trigonometry), if T1 (the string opposite the smaller angle) is 25N, T2 MUST be larger. As far as I can tell, you got T1 correct, you must be substituting incorrectly- sorry I don't have a calculator on me.
 
Crud, I made a mistake: cos56<cos45

You made rounding off errors:

T1 = 25.21
T2 = 19.94

Totally disregard what I said earlier- I was making a fool of myself.
 
Yeah thanks for confirming... no idea why this simple problem is causing so much problems.
 
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