Please help!
Find the tension in each cord in the figure if the weight of the suspended object is w = 210N
A) Find the tension in the cord A for system (b).
B) Find the tension in the cord B for system (b).

- anonymous

- katieb

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- anonymous

|dw:1381448059482:dw|

- anonymous

first start by drawing a free body diagram of the knot (connection of all three strings).
next find the components of each tension.
is that something you can do? or do you need help setting that up?
just as an fyi i like to help people answer the question themselves when i answer questions, not just give answers, because then you dont learn anything. just wanted to let you know why i am asking questions the way i am

- anonymous

Well there was a part A, where I had to find the tension forces and I was able to do that, but this example is throwing me off a bit. The previous one I solved look like:
|dw:1381449221117:dw|

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## More answers

- anonymous

Would the free body diagram look like?:
|dw:1381449348982:dw|

- anonymous

basically, but you have to put arrow heads, to show where the forces are going. can you find the x and y components of those forces?

- anonymous

but this one you will do the same basic thing as in your part A

- anonymous

I just set up two triangles to find the tension forces right?

- anonymous

|dw:1381449709193:dw|

- anonymous

is what your free body diagram would look like.
and if you see, your y components of each tension will be equal to weight because the object is not moving (it is in equilibrium)

- anonymous

Not sure if this is correct but this is what I came up with:
Ta
X comp: 210 sin 60
Y comp: 210 cos 60
Tb
X comp: 210 sin 45
Y comp: 210 cos 45
If this is correct then I think I would sum these and then use the equation F = sqrt ( (X)^2 + (Y)^2)

- anonymous

Wait that is not right, that's not what the question is asking. Sorry!

- anonymous

those are correct, but that isnt exactly how you would solve this problem. if you were looking for total tension that would be correct, but you are looking for tension of each string.
so you will have:
\[F _{net}=Ta _{x}+Tb\]
if we do a little bit of plugging in:
\[0=Tax+Tbx\]
it is equal to zero because the object is in equilibrium.
from there we can get that:
Tax=Tbx
I know you would expect a negative sign in there, but there isnt because there is no such thing as a negative tension.

- anonymous

So we are up to two equations now:
\[w=Ta _{y}+Tb_{y}\]
and
\[Ta _{x}=Tb_{y}\]

- anonymous

So I got:
Ta
X comp: -181.87
Y comp: -105
Tb
X comp: 148.49
Y comp: 148.49
Would I plug these values in and solve for w first?

- anonymous

I see a mistake, the second equation should be Tax=Tbx. not Tby, that is my fault, im sorry.

- anonymous

That's okay, anything helps!

- anonymous

not exactly, i would leave those components not compiled into one number like that because we are going to use the angles by themselves.

- anonymous

those equations are right tho, just mentally adjust them with the now caught error.
the first thing you are going to do is solve the first equation (the one with w) for Ta .that will give us the magnitude of the tension in string a. *note: this is the step why you want to keep your components not multiplied by weight just quite yet*

- anonymous

W = 210 cos (60) + 210 cos (45)?

- anonymous

Oh wait, okay I see what you're saying, solve for Ta.

- anonymous

you got it kinda right. 210 is the weight of the block, so that would be w. right now all we have is
210=Tay cos(60) + Tby(45)

- anonymous

in there i made another slight error, it is actually just Ta and Tb, we are multiplying by

- anonymous

Is Tby cos too?

- anonymous

Well now, Tb.

- anonymous

let me correect myself quick, im sorry for all the errors. i will type it in an equation\[210=T _{a}\cos(60) + T _{b}\sin(45)\]

- anonymous

now solve that for Ta

- anonymous

Okay, one minute

- anonymous

but we still arent completely done, so if you are doing online homework, and only have a limited amount of trys (as i do) dont go ahead and type that answer you get in right away

- anonymous

Okay, yeah this is for my online homework. I have 5 tries left for part A. Almost finished finding Ta.

- anonymous

mastering physics? haha

- anonymous

420 - (1.4142) x Tb = Ta
Now would I plug in this equation into the Ta = Tb ?

- anonymous

Yeah, I have been so busy this whole week and this is my last problem on it due by midnight, just didn't get a chance to go get help from my professor.

- anonymous

mmmmm thats not quite right. try solving it with symbols first, and then plugging values in once you have the equation solved for Ta.
haha i hear you there, we use that website too. i come on here when im stuck too, and it seems like no one ever wants to answer my questions haha

- anonymous

once you solve it with the symbols, put it up here and ill tell you if its right

- anonymous

Yeah I overlooked your question, nothing I could help with unfortunately, lol. Physics is definitely not my strong subject! Okay, I will try to reconfigure.

- anonymous

haha i feel you there, im retaking physics 230 right now

- anonymous

This is my first physics class ever so it's all new to me.

- anonymous

i hear ya. well im here to help

- anonymous

Well I was working along with my work from the previous Part A, and I entered in my answer, and it's wrong. So I guess I'm just lost. This is all I have gotten to, and idk if it is right.
210 = Ta cos (60) + Tb sin (45)
210 = Ta x (.5) + Tb (.7071)
210 - (.7071) x Tb = Ta x (.5)
420 - (1.4142) x Tb = Ta

- anonymous

ok let me actually solve this problem. and i will post a picture of how i did it etc. because through this, i myself am getting a tad lost. give me about 10 minutes

- anonymous

Thanks, I appreciate the help. I've just got other homework tonight due by tomorrow too so I will be giving up on this problem soon! lol.

- anonymous

hahaha i hear ya. like i said, ten minutes. and if it really gets to it, i may or may not have the actual number in my own mastering physics, so if it comes to it we might be able to just go that route haha :p

- anonymous

Thank you for helping me! Not a problem if you can't find it or anything, no big deal! I'm leaving my laptop for about 10 minutes and will be back to check afterwards to see if there is a response. That again for all the help, I really do appreciate it!

- anonymous

not a problem at all! its helping me study! i have a test in this next week! haha

- anonymous

The 60 degrees is against the wall.

- anonymous

ok that is how i set it up. it doesnt make a huge differece, just in how you find the components

- anonymous

Well now I am leaving so I will check back on here before midnight again, thanks so much!

- anonymous

not a prob

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