Forces in Equilibrium-Tension Forces at Different Angles

[312213]

Homework Statement

Gravity pulls the sign with 490N. Determine the tension in the cables.

Shown as in this picture:
http://img380.imageshack.us/img380/9891/forcesinequilibriumpf0.jpg

Homework Equations

I am not sure.

The Attempt at a Solution

I do not fully understand what I am supposed to do to solve this problem, so I tried different methods that I thought of trying, to see if the answer look like it might fit. This is what I did.

For FT1:
45°+60° = 105° /45° = 2.3333
490/2.3333 = 210
210/sin 45° = 297N = FT1

For FT2:
45°+60° = 105° /60° = 1.75.
490/1.75 = 280
280/sin 60° = 323N = FT2

This is what I did, but I am not sure if this is done correctly to achieve the correct answer. Could someone check the answer and also show me the correct setup to do this kind of problem? The part I don't get is how to determine force using different angles.

 

[rock.freak667]

What you should do is split the tensions into x and y components, can you do that?

EDIT: then use the conditions for equilibrium

 

[312213]

I do not understand the x and y components very well but I guess this might be it:

cos 45° = FT1x/FT1
FT1×cos 45° = FT1x

sin 45° = FT1y/FT1
FT1×sin 45° = FT1y------x---------------y
-FT1×cos 45° | FT1×sin 45°
FT2×cos 45° | FT2×sin 60°
-------------------490N

This is more or less how I remember seeing the x and y.

I guess then it would be:
FT1×sin 45° + FT2×sin 60° - 490N = 0N
FT1×sin 45° + FT2×sin 60° = 490N

Here is where I get confused on what to do next. I would divide the sin off but then I don't know how to do it so that the FT1 and FT2 are left alone, because I see dividing both sides by sin 45° and sin 60° would get the original sins off the FT1 and FT2, but hen it would be divided by the other sin. What do I do next?

 

[rock.freak667]

Well now you know that the sum of the forces in the y direction is 0 and similarly, the sum in the x is 0 as well.


Now you have

T2cos60-T1cos45=0

T2sin60+T1sin45=490


Solve simultaneously now.

 

[wesDOT]

I am having a similar problem. I get to what you said rockfreak on my own, but I am not how to go about solving for the separate tension force

I have

-T1cos(9.6)+T2cos(17.8)=0

and

T1sin(9.6)+T2sin(17.8)-5.5=0

I know that the answers are approximately T1=11.4 and T2=11.8 but I'm not sure how to get those answers.

 

[rock.freak667]

I know that the answers are T1=11.4 and T2=11.8 but I am not sure how to get those answers.

Do you know how to solve simultaneous equations?

example: 3x+y=4
2x+y=1

can you solve those two?

If so, it is the same concept, just calculate the values for the sines and cosines of the angles.

 

[wesDOT]

i would just use substitution. I don't know why i was thinking that i couldn't just evaluate for the values of the sines and cosines. That is where I was stuck. I'll try it. Thanks for the help.

 

[312213]

I don't fully understand simultaneous equations either so just to check, I would do this:

Equations:
T2cos60-T1cos45=0
T2sin60+T1sin45=490

Works:
T2cos60-T1cos45 + T1cos45=0 + T1cos45 --> T2cos60 = T1cos45
T2cos60/cos45= T1cos45/cos45 --> T2cos60/cos45 = T1

By dividing cos60 with cos45, I get 0.707 (sqrt of 0.5) meaning T2×0.5=T1
Substitute T1 with T2×0.5 in T2sin60+T1sin45=490 --> T2sin60+(T2×0.707)sin45=490

T2sin60+(T2×0.707)sin45=490-->T2sin60+T2×0.5=490

Then sin60 is about 0.866, so I assumed that the first term is 0.866th of T2 and the second term is half of T2. Adding them together (0.866 + 0.5 = 1.366) gets T2×1.366 = 490

I assumed that its safe to divide both sides by 1.366, resulting in T2 = 358.7N.

Working the same way for T1, I would get 253.64N. (edited from 346.48N)

I think did something wrong there.

Edit: Wrong output for T1 before edit. Now it looks correct. At least, the numbers substituting the equation for T1 and T2 would result in 490.

 

[rock.freak667]

Your value for T2=358.7N is correct. Put that back into the first equation to find T1.

 

[312213]

The equations works out so thank you for your help.