# What is the last relation in this tension problem?

My problem is explained from the below yahoo answers link. I know I'm suppose to follow the template, but my question is kind of long to write, and being so that I already wrote it up on Yahoo Answers, I thought it would be easier if I just list the link.

My problem is not on finding the actual tension, but is just the missing relation in order to solve the tension. I understand how to solve the tension, just that I am currently being hindered by not knowing the last relation.

Edit: Also you can skip the first paragraph that I wrote on Yahoo Answers for I have provided the picture to which the paragraph describes below:
http://www.webassign.net/serpop/p4-24alt.gif

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PhanthomJay
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The 3rd equation you are looking for can be derived by taking a free body diagram of the hanging mass, identify the forces acting on that mass, and use newton 1 to solve for the unknown tnsion force T3 acting on that mass.

The 3rd equation you are looking for can be derived by taking a free body diagram of the hanging mass, identify the forces acting on that mass, and use newton 1 to solve for the unknown tnsion force T3 acting on that mass.
I drew a free body diagram but I see nothing new, just the gravitational force going downwards, and the tension force going upwards; is there something I'm not seeing?

PhanthomJay
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Solve for T3 using Newton 1. But I just realized your first equation is wrong. In your first 2 equations, you are drawing a free body diagram of the joint. In so doing, the massweight does not enter into the equation. Watch your plus and minus signs also.

SammyS
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So, it's easier for you to post a couple of links than it is to cut & paste & figure out how to show an image!

I suppose you want help with solving your problem too.

See, it's not all that difficult!

Solve for T3 using Newton 1. But I just realized your first equation is wrong. In your first 2 equations, you are drawing a free body diagram of the joint. In so doing, the massweight does not enter into the equation. Watch your plus and minus signs also.
actually, the equations I wrote did include the mass weight as I set the combined vertical tensions equal to the gravitational force of the mass weight

SammyS
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actually, the equations I wrote did include the mass weight as I set the combined vertical tensions equal to the gravitational force of the mass weight

T3=-mg

and

T1sin40° + T2sin50° + T3 = 0

T3 = -mg

and

T1sin40° + T2sin50° + T3 = 0
Wait how can T3 alone be the reciprocal force of the gravitational force? Shouldn't the reciprocal force of the gravitational force be equal to the vertical combination of the tension forces opposite to the gravitational force?

T1sin40° + T2sin50° + T3 = -mg

Because T3=-mg alone means that T1sin40° + T2sin50° has to be equal to mg for the equation you wrote to equal zero.

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PhanthomJay
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Wait how can T3 alone be the reciprocal force of the gravitational force? Shouldn't the reciprocal force of the gravitational force be equal to the vertical combination of the tension forces opposite to the gravitational force?

T1sin40° + T2sin50° + T3 = -mg
No. If you look at the entire system, T3 is internal to the system, and does not enter into the equilbrium equation. Only the support forces and gravity force are external to the system.
Because T3=-mg alone means that T1sin40° + T2sin50° has to be equal to mg for the equation you wrote to equal zero.
Yes.

As a note, at the risk of confusing you, when you draw free body diagrams of part of the system, the internal forces become external to that portion of the system when you apply Newton's laws, Fnetexternal = ma (where a =0 in this problem).

SammyS
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Wait how can T3 alone be the reciprocal force of the gravitational force? Shouldn't the reciprocal force of the gravitational force be equal to the vertical combination of the tension forces opposite to the gravitational force?

T1sin40° + T2sin50° + T3 = -mg

Because T3=-mg alone means that T1sin40° + T2sin50° has to be equal to mg for the equation you wrote to equal zero.

T3=-mg

and

T1sin40° + T2sin50° + T3 = 0

As I look back on what I wrote, it would have been better if I had written things a bit differently. What I posted should give the correct answer, but the following will explain things better (I hope.)

A Free Body Diagram of the hanging mass, m1, has force T3 upward, and the force of gravity, m1g pointing downward.

This gives T3 - m1g = 0  →  T3 = m1g . (Yes, the sign is different than in my previous post, but indicates that as far as m1 is concerned, T3 is upward.)

As PhanthomJay points out, a Free Body Diagram of the joint where the three cables meet, does not include m1 directly. It has the three cables with tensions T1, T2, and T3, meeting at the given angles. The vertical components of the forces involved gives:

T1sin40° + T2sin50° - T3 = 0  →  T1sin40° + T2sin50° - m1g = 0  →  T1sin40° + T2sin50° = m1g

I hope that is clearer.
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Ohh I see now, so T3 in a way had no affect on the distribution of the tensions from the first two cords, and should have been treated as just an extension of the same reciprocal gravitational force. Thank you both for taking your time to explain this to me!