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anonymous
 3 years ago
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
 3 years ago
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).

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381448059482:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first 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
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Would the free body diagram look like?: dw:1381449348982:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0basically, 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
 3 years ago
Best ResponseYou've already chosen the best response.0but this one you will do the same basic thing as in your part A

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I just set up two triangles to find the tension forces right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381449709193:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is 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
 3 years ago
Best ResponseYou've already chosen the best response.0Not 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
 3 years ago
Best ResponseYou've already chosen the best response.0Wait that is not right, that's not what the question is asking. Sorry!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0those 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
 3 years ago
Best ResponseYou've already chosen the best response.0So we are up to two equations now: \[w=Ta _{y}+Tb_{y}\] and \[Ta _{x}=Tb_{y}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0I see a mistake, the second equation should be Tax=Tbx. not Tby, that is my fault, im sorry.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's okay, anything helps!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not exactly, i would leave those components not compiled into one number like that because we are going to use the angles by themselves.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0those 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
 3 years ago
Best ResponseYou've already chosen the best response.0W = 210 cos (60) + 210 cos (45)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh wait, okay I see what you're saying, solve for Ta.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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
 3 years ago
Best ResponseYou've already chosen the best response.0in there i made another slight error, it is actually just Ta and Tb, we are multiplying by

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let 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
 3 years ago
Best ResponseYou've already chosen the best response.0now solve that for Ta

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but 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
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, yeah this is for my online homework. I have 5 tries left for part A. Almost finished finding Ta.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mastering physics? haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0420  (1.4142) x Tb = Ta Now would I plug in this equation into the Ta = Tb ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 3 years ago
Best ResponseYou've already chosen the best response.0mmmmm 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
 3 years ago
Best ResponseYou've already chosen the best response.0once you solve it with the symbols, put it up here and ill tell you if its right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah 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
 3 years ago
Best ResponseYou've already chosen the best response.0haha i feel you there, im retaking physics 230 right now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is my first physics class ever so it's all new to me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i hear ya. well im here to help

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks, 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
 3 years ago
Best ResponseYou've already chosen the best response.0hahaha 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
 3 years ago
Best ResponseYou've already chosen the best response.0Thank 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
 3 years ago
Best ResponseYou've already chosen the best response.0not a problem at all! its helping me study! i have a test in this next week! haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 60 degrees is against the wall.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok that is how i set it up. it doesnt make a huge differece, just in how you find the components

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well now I am leaving so I will check back on here before midnight again, thanks so much!
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