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svdh3 Group Title

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).

  • 9 months ago
  • 9 months ago

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  1. svdh3 Group Title
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    |dw:1381448059482:dw|

    • 9 months ago
  2. dave0616 Group Title
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    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

    • 9 months ago
  3. svdh3 Group Title
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    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|

    • 9 months ago
  4. svdh3 Group Title
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    Would the free body diagram look like?: |dw:1381449348982:dw|

    • 9 months ago
  5. dave0616 Group Title
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    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?

    • 9 months ago
  6. dave0616 Group Title
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    but this one you will do the same basic thing as in your part A

    • 9 months ago
  7. svdh3 Group Title
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    I just set up two triangles to find the tension forces right?

    • 9 months ago
  8. dave0616 Group Title
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    |dw:1381449709193:dw|

    • 9 months ago
  9. dave0616 Group Title
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    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)

    • 9 months ago
  10. svdh3 Group Title
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    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)

    • 9 months ago
  11. svdh3 Group Title
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    Wait that is not right, that's not what the question is asking. Sorry!

    • 9 months ago
  12. dave0616 Group Title
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    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.

    • 9 months ago
  13. dave0616 Group Title
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    So we are up to two equations now: \[w=Ta _{y}+Tb_{y}\] and \[Ta _{x}=Tb_{y}\]

    • 9 months ago
  14. svdh3 Group Title
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    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?

    • 9 months ago
  15. dave0616 Group Title
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    I see a mistake, the second equation should be Tax=Tbx. not Tby, that is my fault, im sorry.

    • 9 months ago
  16. svdh3 Group Title
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    That's okay, anything helps!

    • 9 months ago
  17. dave0616 Group Title
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    not exactly, i would leave those components not compiled into one number like that because we are going to use the angles by themselves.

    • 9 months ago
  18. dave0616 Group Title
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    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*

    • 9 months ago
  19. svdh3 Group Title
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    W = 210 cos (60) + 210 cos (45)?

    • 9 months ago
  20. svdh3 Group Title
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    Oh wait, okay I see what you're saying, solve for Ta.

    • 9 months ago
  21. dave0616 Group Title
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    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)

    • 9 months ago
  22. dave0616 Group Title
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    in there i made another slight error, it is actually just Ta and Tb, we are multiplying by

    • 9 months ago
  23. svdh3 Group Title
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    Is Tby cos too?

    • 9 months ago
  24. svdh3 Group Title
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    Well now, Tb.

    • 9 months ago
  25. dave0616 Group Title
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    let me correect myself quick, im sorry for all the errors. i will type it in an equation\[210=T _{a}\cos(60) + T _{b}\sin(45)\]

    • 9 months ago
  26. dave0616 Group Title
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    now solve that for Ta

    • 9 months ago
  27. svdh3 Group Title
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    Okay, one minute

    • 9 months ago
  28. dave0616 Group Title
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    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

    • 9 months ago
  29. svdh3 Group Title
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    Okay, yeah this is for my online homework. I have 5 tries left for part A. Almost finished finding Ta.

    • 9 months ago
  30. dave0616 Group Title
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    mastering physics? haha

    • 9 months ago
  31. svdh3 Group Title
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    420 - (1.4142) x Tb = Ta Now would I plug in this equation into the Ta = Tb ?

    • 9 months ago
  32. svdh3 Group Title
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    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.

    • 9 months ago
  33. dave0616 Group Title
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    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

    • 9 months ago
  34. dave0616 Group Title
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    once you solve it with the symbols, put it up here and ill tell you if its right

    • 9 months ago
  35. svdh3 Group Title
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    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.

    • 9 months ago
  36. dave0616 Group Title
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    haha i feel you there, im retaking physics 230 right now

    • 9 months ago
  37. svdh3 Group Title
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    This is my first physics class ever so it's all new to me.

    • 9 months ago
  38. dave0616 Group Title
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    i hear ya. well im here to help

    • 9 months ago
  39. svdh3 Group Title
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    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

    • 9 months ago
  40. dave0616 Group Title
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    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

    • 9 months ago
  41. svdh3 Group Title
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    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.

    • 9 months ago
  42. dave0616 Group Title
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    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

    • 9 months ago
  43. svdh3 Group Title
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    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!

    • 9 months ago
  44. dave0616 Group Title
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    not a problem at all! its helping me study! i have a test in this next week! haha

    • 9 months ago
  45. svdh3 Group Title
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    The 60 degrees is against the wall.

    • 9 months ago
  46. dave0616 Group Title
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    ok that is how i set it up. it doesnt make a huge differece, just in how you find the components

    • 9 months ago
  47. svdh3 Group Title
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    Well now I am leaving so I will check back on here before midnight again, thanks so much!

    • 9 months ago
  48. dave0616 Group Title
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    not a prob

    • 9 months ago
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