## GavinxFiasco 2 years 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?

1. GavinxFiasco

2. rajat97

interesting question i am working on it

3. GavinxFiasco

@rajat97 thanks for your help in advance

4. rajat97

you're welcome and i've got the answer as T=1231.81 N if it is right, please let me know

5. rajat97

@GavinxFiasco

6. GavinxFiasco

unfortunately it isn't right

7. rajat97

oh see what i did was that i balanced all the torques and i think i need to work on it

8. rajat97

just wait for a few minutes

9. rajat97

i took the wrong ref. point and now i am trying it with point A

10. GavinxFiasco

alright, take as much time as you need I'm trying to figure it out as well

11. rajat97

now i've got the answer as 377.55 N

12. GavinxFiasco

still incorrect

13. rajat97

wow

14. GavinxFiasco

ya, this questions been stumping me for quite a while now

15. rajat97

i've got something new here and my answer is 521N try try but don't cry lol

16. GavinxFiasco

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

@AllTehMaffs have any suggestions to this one?

18. rajat97

yeah sure

19. AllTehMaffs

0_0 hmm.....

20. GavinxFiasco

I know , -.-'

21. AllTehMaffs

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

I believe it is trying to bend the beam downwards or push down on it

23. AllTehMaffs

oh, derp, because of the cross beam. Gotchya

24. GavinxFiasco

Ya aha

25. rajat97

i"ve got another answer and i know its really irritating and my answer is 751.33N

26. AllTehMaffs

right or wrong at least you have answers - I'm still trying to set up the darn thing :/ ^_^

27. rajat97

@AIITehMaffs what is your ref. point

28. AllTehMaffs

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

I know that's wrong, actually :P

30. rajat97

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

so i did the same

32. rajat97

do you have any idea ????

33. AllTehMaffs

|dw:1383944792922:dw|

34. rajat97

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

$M=-\sqrt{L_1^2+L_2^2} \times F +\sqrt{L_1^2+L_2^2} \times T_y$ this is not at all correct even a little - these darn engineering problems....

36. AllTehMaffs

there should be a Tx in there too, but I don't know how to isolate variables then....

37. melaghil

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

38. melaghil

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

Thank you @melaghil

40. rajat97

@melaghil you are cool

41. rajat97

even i did the same but got the wrong answer

42. rajat97

by calculating by @melaghil 's method , i get the previous answer i.e. 377.55 N

43. melaghil

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

@rajat97 You must be doing something wrong because 377.55 N is too small. You should be getting something in the thousands

45. rajat97

can you please post your calculations @melaghil

46. melaghil

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

There are some spelling mistakes there but you can decipher them

48. rajat97

thanks a lot man @melaghil