- anonymous

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?

- schrodinger

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- anonymous

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- rajat97

interesting question i am working on it

- anonymous

@rajat97 thanks for your help in advance

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## More answers

- rajat97

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

- rajat97

- anonymous

unfortunately it isn't right

- rajat97

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

- rajat97

just wait for a few minutes

- rajat97

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

- anonymous

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

- rajat97

now i've got the answer as 377.55 N

- anonymous

still incorrect

- rajat97

wow

- anonymous

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

- rajat97

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

- anonymous

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

- anonymous

@AllTehMaffs have any suggestions to this one?

- rajat97

yeah sure

- anonymous

0_0
hmm.....

- anonymous

I know , -.-'

- anonymous

I can't tell what the moment is doing - is it trying to bend the beam down or pushing it to the left? :P

- anonymous

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

- anonymous

oh, derp, because of the cross beam. Gotchya

- anonymous

Ya aha

- rajat97

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

- anonymous

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

- rajat97

@AIITehMaffs what is your ref. point

- anonymous

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

- anonymous

I know that's wrong, actually :P

- 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

- rajat97

so i did the same

- rajat97

do you have any idea ????

- anonymous

|dw:1383944792922:dw|

- 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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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!

- anonymous

Thank you @melaghil

- rajat97

@melaghil you are cool

- rajat97

even i did the same but got the wrong answer

- rajat97

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

- anonymous

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

- anonymous

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

- rajat97

can you please post your calculations @melaghil

- anonymous

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

- anonymous

There are some spelling mistakes there but you can decipher them

- rajat97

thanks a lot man @melaghil

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