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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?

Physics
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interesting question i am working on it
@rajat97 thanks for your help in advance

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Other answers:

you're welcome and i've got the answer as T=1231.81 N if it is right, please let me know
unfortunately it isn't right
oh see what i did was that i balanced all the torques and i think i need to work on it
just wait for a few minutes
i took the wrong ref. point and now i am trying it with point A
alright, take as much time as you need I'm trying to figure it out as well
now i've got the answer as 377.55 N
still incorrect
wow
ya, this questions been stumping me for quite a while now
i've got something new here and my answer is 521N try try but don't cry lol
Aha wrong also, i'm not stressing it. I'm sure if i take a break from it then come back should be fine
@AllTehMaffs have any suggestions to this one?
yeah sure
0_0 hmm.....
I know , -.-'
I can't tell what the moment is doing - is it trying to bend the beam down or pushing it to the left? :P
I believe it is trying to bend the beam downwards or push down on it
oh, derp, because of the cross beam. Gotchya
Ya aha
i"ve got another answer and i know its really irritating and my answer is 751.33N
right or wrong at least you have answers - I'm still trying to set up the darn thing :/ ^_^
@AIITehMaffs what is your ref. point
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
I know that's wrong, actually :P
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
so i did the same
do you have any idea ????
|dw:1383944792922:dw|
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
\[M=-\sqrt{L_1^2+L_2^2} \times F +\sqrt{L_1^2+L_2^2} \times T_y\] this is not at all correct even a little - these darn engineering problems....
there should be a Tx in there too, but I don't know how to isolate variables then....
\[T _{y} = \frac{ F \times L_{1} + M }{ \sqrt{L_{1}^2 + L_{1}^2 } }\] that gets you the vertical component of the tension. \[T = T_{y}*\frac{ 5 }{ 4 }\] so your answer would be 37267 N
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!
Thank you @melaghil
@melaghil you are cool
even i did the same but got the wrong answer
by calculating by @melaghil 's method , i get the previous answer i.e. 377.55 N
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
@rajat97 You must be doing something wrong because 377.55 N is too small. You should be getting something in the thousands
can you please post your calculations @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
There are some spelling mistakes there but you can decipher them
thanks a lot man @melaghil

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