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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?
 5 months ago
 5 months 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?
 5 months ago
 5 months ago

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rajat97Best ResponseYou've already chosen the best response.0
interesting question i am working on it
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
@rajat97 thanks for your help in advance
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
you're welcome and i've got the answer as T=1231.81 N if it is right, please let me know
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
unfortunately it isn't right
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
oh see what i did was that i balanced all the torques and i think i need to work on it
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
just wait for a few minutes
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
i took the wrong ref. point and now i am trying it with point A
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
alright, take as much time as you need I'm trying to figure it out as well
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
now i've got the answer as 377.55 N
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
ya, this questions been stumping me for quite a while now
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
i've got something new here and my answer is 521N try try but don't cry lol
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
Aha wrong also, i'm not stressing it. I'm sure if i take a break from it then come back should be fine
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
@AllTehMaffs have any suggestions to this one?
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
I can't tell what the moment is doing  is it trying to bend the beam down or pushing it to the left? :P
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
I believe it is trying to bend the beam downwards or push down on it
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
oh, derp, because of the cross beam. Gotchya
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
i"ve got another answer and i know its really irritating and my answer is 751.33N
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
right or wrong at least you have answers  I'm still trying to set up the darn thing :/ ^_^
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
@AIITehMaffs what is your ref. point
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
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
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
I know that's wrong, actually :P
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
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
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
do you have any idea ????
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
dw:1383944792922:dw
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
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
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
\[M=\sqrt{L_1^2+L_2^2} \times F +\sqrt{L_1^2+L_2^2} \times T_y\] this is not at all correct even a little  these darn engineering problems....
 5 months ago

AllTehMaffsBest ResponseYou've already chosen the best response.0
there should be a Tx in there too, but I don't know how to isolate variables then....
 5 months ago

melaghilBest ResponseYou've already chosen the best response.2
\[T _{y} = \frac{ F \times L_{1} + M }{ \sqrt{L_{1}^2 + L_{1}^2 } }\] that gets you the vertical component of the tension. \[T = T_{y}*\frac{ 5 }{ 4 }\] so your answer would be 37267 N
 5 months ago

melaghilBest ResponseYou've already chosen the best response.2
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!
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
Thank you @melaghil
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
@melaghil you are cool
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
even i did the same but got the wrong answer
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
by calculating by @melaghil 's method , i get the previous answer i.e. 377.55 N
 5 months ago

melaghilBest ResponseYou've already chosen the best response.2
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
 5 months ago

melaghilBest ResponseYou've already chosen the best response.2
@rajat97 You must be doing something wrong because 377.55 N is too small. You should be getting something in the thousands
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
can you please post your calculations @melaghil
 5 months ago

melaghilBest ResponseYou've already chosen the best response.2
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
 5 months ago

melaghilBest ResponseYou've already chosen the best response.2
There are some spelling mistakes there but you can decipher them
 5 months ago

rajat97Best ResponseYou've already chosen the best response.0
thanks a lot man @melaghil
 5 months ago
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