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GavinxFiasco

  • one year ago

A force F = 6.61×104 N and a couple moment M = 3.41×104 N·m are applied to the steel frame as shown. Assume L1 = 1.19 m, L2 = 3.59 m. The angle of the cable BC is indicated by the small triangle. What is the tension in the cable?

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  1. GavinxFiasco
    • one year ago
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  2. rajat97
    • one year ago
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    interesting question i am working on it

  3. GavinxFiasco
    • one year ago
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    @rajat97 thanks for your help in advance

  4. rajat97
    • one year ago
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    you're welcome and i've got the answer as T=1231.81 N if it is right, please let me know

  5. rajat97
    • one year ago
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    @GavinxFiasco

  6. GavinxFiasco
    • one year ago
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    unfortunately it isn't right

  7. rajat97
    • one year ago
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    oh see what i did was that i balanced all the torques and i think i need to work on it

  8. rajat97
    • one year ago
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    just wait for a few minutes

  9. rajat97
    • one year ago
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    i took the wrong ref. point and now i am trying it with point A

  10. GavinxFiasco
    • one year ago
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    alright, take as much time as you need I'm trying to figure it out as well

  11. rajat97
    • one year ago
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    now i've got the answer as 377.55 N

  12. GavinxFiasco
    • one year ago
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    still incorrect

  13. rajat97
    • one year ago
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    wow

  14. GavinxFiasco
    • one year ago
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    ya, this questions been stumping me for quite a while now

  15. rajat97
    • one year ago
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    i've got something new here and my answer is 521N try try but don't cry lol

  16. GavinxFiasco
    • one year ago
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    Aha wrong also, i'm not stressing it. I'm sure if i take a break from it then come back should be fine

  17. GavinxFiasco
    • one year ago
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    @AllTehMaffs have any suggestions to this one?

  18. rajat97
    • one year ago
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    yeah sure

  19. AllTehMaffs
    • one year ago
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    0_0 hmm.....

  20. GavinxFiasco
    • one year ago
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    I know , -.-'

  21. AllTehMaffs
    • one year ago
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    I can't tell what the moment is doing - is it trying to bend the beam down or pushing it to the left? :P

  22. GavinxFiasco
    • one year ago
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    I believe it is trying to bend the beam downwards or push down on it

  23. AllTehMaffs
    • one year ago
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    oh, derp, because of the cross beam. Gotchya

  24. GavinxFiasco
    • one year ago
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    Ya aha

  25. rajat97
    • one year ago
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    i"ve got another answer and i know its really irritating and my answer is 751.33N

  26. AllTehMaffs
    • one year ago
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    right or wrong at least you have answers - I'm still trying to set up the darn thing :/ ^_^

  27. rajat97
    • one year ago
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    @AIITehMaffs what is your ref. point

  28. AllTehMaffs
    • one year ago
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    that's what I'm trying to figure out :/ I was thinking along the top beam, so the moment is equal to 3L1 cross Ty, but I'm pretty sure that wrong

  29. AllTehMaffs
    • one year ago
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    I know that's wrong, actually :P

  30. rajat97
    • one year ago
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    look if you take the ref. point as the hinged point A, it will be benificial as the torque due to hinge forces will be seen as zero

  31. rajat97
    • one year ago
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    so i did the same

  32. rajat97
    • one year ago
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    do you have any idea ????

  33. AllTehMaffs
    • one year ago
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    |dw:1383944792922:dw|

  34. rajat97
    • one year ago
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    i live in india and its around 2:45 in the morning and i haven't slept for the whole night !! so i think i shall sleep now and this is a mind blowing question and somethings gotta go wrong with openstudy so i think i shall go now

  35. AllTehMaffs
    • one year ago
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    \[M=-\sqrt{L_1^2+L_2^2} \times F +\sqrt{L_1^2+L_2^2} \times T_y\] this is not at all correct even a little - these darn engineering problems....

  36. AllTehMaffs
    • one year ago
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    there should be a Tx in there too, but I don't know how to isolate variables then....

  37. melaghil
    • one year ago
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    \[T _{y} = \frac{ F \times L_{1} + M }{ \sqrt{L_{1}^2 + L_{1}^2 } }\] that gets you the vertical component of the tension. \[T = T_{y}*\frac{ 5 }{ 4 }\] so your answer would be 37267 N

  38. melaghil
    • one year ago
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    Sorry I messed it up \[T = \frac{ F*L_1+M }{ L_1*(4/5) + L_2*(3/5) }\] That will get you the tension resulting in T = 36303 N I just tested this solution and it worked!

  39. GavinxFiasco
    • one year ago
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    Thank you @melaghil

  40. rajat97
    • one year ago
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    @melaghil you are cool

  41. rajat97
    • one year ago
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    even i did the same but got the wrong answer

  42. rajat97
    • one year ago
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    by calculating by @melaghil 's method , i get the previous answer i.e. 377.55 N

  43. melaghil
    • one year ago
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    I am a fourth year dal engineering student. I Capas for my friends. If you email me your netid, B00 number and capa ID I'll do it for ya

  44. melaghil
    • one year ago
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    @rajat97 You must be doing something wrong because 377.55 N is too small. You should be getting something in the thousands

  45. rajat97
    • one year ago
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    can you please post your calculations @melaghil

  46. melaghil
    • one year ago
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    Alright so the key point is choosing a pivot point in the whole frame and sticking to it, It doesn't matter where the pivot point is but you always make sure that it minimizes the unknowns. |dw:1383969861855:dw| There are 3 unknowns T, Ax and Ay. So I chose the pivot to be at A. The couple moment can be moved around and what really matters when we talk about a force is the effective distance so the following drawing will simplify it. (Note in the drawing I flipped Ty and Tx and I dont know how to edit drawings on this website) |dw:1383970015750:dw| So sum of the moments about the pivot = 0 \[0 = -M -F_1*L_1+T_x*L_2 +T_y*L_1\] Also note that \[T_x = T*\frac{ 3 }{ 5 }\] while \[T_x = T*\frac{ 4 }{ 5 }\] So in that equation now all you have to do is solve for T

  47. melaghil
    • one year ago
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    There are some spelling mistakes there but you can decipher them

  48. rajat97
    • one year ago
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    thanks a lot man @melaghil

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