Solving Tensions & Weights Homework Problem

[jegues]

Homework Statement

See figure.

Homework Equations

$$\sum F = ma$$

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 sort of 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,

 

[tiny-tim]

Hi jegues!

hmm … two vertical equations …

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

 

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

 

[tiny-tim]

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

That's it!

(though it might be easier in this case to use T_ba and T_cd instead. )

 

[jegues]

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

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?

 

[tiny-tim]

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?

That's right!

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

you can make mistakes doing that! )

 

[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!?

 

[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?

 

[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...

 

[tiny-tim]

I've always found it easier to simply use cosines for the x components and sines for the y component.

eek! 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!

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)

hmm … I've just noticed you have four unknowns, not three …

looks like you're going to need that T_bc and T_ba 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.

 

[jegues]

choose cos or sin to be appropriate for the diagram

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!?

 

[tiny-tim]

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.

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 )

 

[sezw]

instead of making things complicated using T[c] why don't u just name them T1.T2,T3? easier that way so u don't confuse them

 

[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,

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

?

 

[sezw]

maybe instead of the horizon u should use a bearing instead?

 

[jegues]

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

 

[tiny-tim]

(just got up :zzz: …)

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?

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.

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

Well, for a start, it's only 3, because you immediately know that T_ab = T_cd.

Get on with it! ​