# Tensions and Weights!

1. Jan 27, 2010

### jegues

1. The problem statement, all variables and given/known data
See figure.

2. Relevant equations
$$\sum F = ma$$

3. The attempt at a solution
By just looking at the picture I was able to generate 2 equations but I seem to have 3 unknowns so I'm sorta stuck right now. Here are my equations:

$$50 = T_{ba}sin(130) + T_{bc}sin(\alpha)$$ and,

$$40 + T_{bc}sin(180+\alpha) = T_{cd}sin(50)$$

Any ideas of anything I'm missing? Am I taking the correct approach to this problem? Is there a simplier way!?

Thanks,

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2. Jan 27, 2010

### tiny-tim

Hi jegues!

hmm … two vertical equations

now i wonder where you could get a third equation from?

3. Jan 27, 2010

### jegues

I think I may have figured out how to generate the extra equations I need.

I know the the rings where all the cables are attached are in equilibrium in the x and y directions.

If I apply $$\sum F_{x} = 0$$ I should be able to generate enough equations to solve this problem.

Is this the correct approach?

EDIT: $$T_{bc}cos(\alpha) + T_{ba}cos(130) = 0$$

4. Jan 27, 2010

### tiny-tim

That's it!

(though it might be easier in this case to use Tba and Tcd instead. )

5. Jan 27, 2010

### jegues

So we just "ignore" the section in the middle? (BC) It is all in equilibrium so it shouldn't affect our answers anyways correct?

So,

$$T_{ba}cos(130) = T_{cd}cos(50)$$,

Correct?

6. Jan 27, 2010

### tiny-tim

That's right!

(except for the sign … why do you persist in using angles > 90º ? …

you can make mistakes doing that! )

7. Jan 27, 2010

### jegues

I run into sign errors whenever I've done it any other way. I think our professor showed us a way to correct it either using sin or cos for the given angle but I've always found it easier to simply use cosines for the x components and sines for the y component.

I'm a little lost at the best way to solve the 3 equation 3 variable system, it's always easier when I don't have to solve for an angle.... ( $$\alpha$$ in this case)

Any more tips!?

8. Jan 27, 2010

### jegues

I'm left with the following three equations:

(1) : $$T_{ba}sin(130) + T_{bc}sin(\alpha) = 50$$

(2) : $$T_{bc}sin(180+\alpha)-T_{cd}sin(50) = -40$$

(3) : $$T_{ba}cos(130) - T_{cd}cos(50) = 0$$

I'm getting a headache trying to solve them :s

EDIT: Doh! I just reliazed I had 4 unknowns. I'm going to need another equation aren't I?

9. Jan 27, 2010

### jegues

So solving for 4 unknowns,

(1) : $$T_{ba}sin(130) + T_{bc}sin(\alpha) = 50$$

(2) : $$T_{bc}sin(180+\alpha)-T_{cd}sin(50) = -40$$

(3) : $$T_{ba}cos(130) - T_{cd}cos(50) = 0$$

(4) : $$T_{bc}cos(\alpha) + T_{ba}cos(130) = 0$$

EDIT: Is this really what I'm stuck with doing? This is going to be an algebraic nightmare...

10. Jan 27, 2010

### tiny-tim

eek! :yuck: sorry, but i'm with your professor on this …

don't be so inflexible, choose cos or sin to be appropriate for the diagram, not to make you feel at home!
hmm … I've just noticed you have four unknowns, not three …

looks like you're going to need that Tbc and Tba equation anyway.

Tips? No, just use common-sense, and keep going.

EDIT: if you used 50 instead of 130, it wouldn't look so intimidating. And sin(180º + α) = -sinα.

11. Jan 27, 2010

### jegues

How is this done? He went over it really fast in class so I couldn't really understand how he was determining to either place a cos or a sin for the given angle.

Thank again, and am I actually stuck with that algebraic nightmare!? :yuck:

12. Jan 27, 2010

### tiny-tim

It's always cos of the angle between the force and the direction.

It's never sin (except, of course, that sometimes the angle marked in the diagram is the "wrong" angle … so if they tell you that the angle to the horizontal is α, but you need cos of the angle to the vertical, then since it's always cos, that means it's cos(90º - α), which is sinα )

13. Jan 27, 2010

### sezw

instead of making things complicated using T[c] why dont u just name them T1.T2,T3? easier that way so u dont confuse them

14. Jan 27, 2010

### jegues

So if I'm looking at angles based from the horizontal axis I use cos(a) and if its from the vertical sin(a)? How do you know which sign to assign to these?

Also,

?

15. Jan 27, 2010

### sezw

16. Jan 27, 2010

### jegues

Still wondering if I'm actually stuck with solving a system of 4 equations and 4 unknowns... :yuck: ???

17. Jan 28, 2010

### tiny-tim

(just got up :zzz: …)
If the force is in (approximately) the same direction as the positive axis, then it's +, if it's the other way, its -.

And I repeat, always use cos.
Well, for a start, it's only 3, because you immediately know that Tab = Tcd.

Get on with it!