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