PLEASE HELP ME FRIENDS WITH THIS...

- anonymous

PLEASE HELP ME FRIENDS WITH THIS...

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- katieb

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- anonymous

INCLUDE THE FREE-BODY DIAGRAMS AND EXPLANATORY DIAGRAMS IN THE CALCULATION. PRESENT A CLEAR AND LOGICAL SOLUTION.

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- BAdhi

id start with finding F_(FB) and F_(DB) from the point B

- mom.

mom is here (;

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- anonymous

how to understand that diagram? :( where will I start?

- mom.

|dw:1444390415983:dw|

- ganeshie8

Just posting the picture
|dw:1444390734310:dw|

- anonymous

i don't get it...

- mathmate

Hints:
1. Start by checking that the vertical reactions are correct by taking moments and \(\sum Fy=0\).
(they are correct, I checked).
2. Find the horizontal Rah by \(\sum Fx=0\).
3. From the known forces at A and B, write down the force directions of each member (tension or compression). If you make a mistake (not likely), the force will simply come out as negative.
4. Solve joint by joint, starting with joint A (four forces, two of which are known) using \(\sum Fx=0\) and (\sum Fy=0\)
Note: to solve by joints, you need to make an FBD (free-body diagram) of the joint and isolate the forces. Normally, you have two unknown forces which can be solved with the two given equations.
5. Once forces along AE and AD are known, you will have other joints with only two unknown forces which you can solve by the method of joints.
6. You can also proceed with joint B, which has 3 forces of which 1 is known. Again, only two unknown forces with two equations.

- anonymous

How to apply the moments on that diagram? Is it the same with the problems in equilibrium of concurrent forces?

- mathmate

|dw:1444391438555:dw|
By isolating joint A, you will see that AE is in compression because \(\sum Fy=0\), and that AD is in tension because \(\sum Fx=0\).
|dw:1444391622266:dw|

- BAdhi

the idea is that all the "JOINTS" of the structure is in an equilibrium so that the forces from each of the rod that is connected to the joint should come to and equilibrium

- mathmate

Taking moments can be used when the forces are NOT concurrent.
WHen forces are concurrent, you use the method of joints, as explained above.
For Rah, you need only sum horizontal forces.

- anonymous

aww.. i forgot to tell you that i already forgot how to take moments.. :( sorry..

- anonymous

wait.. on the first # of your hints.. how did you know that vertical reaction is correct?

- mathmate

taking moments:
https://www.youtube.com/watch?v=qSB_RXw1jV8
http://web.mit.edu/emech/dontindex-build/full-text/emechbk_2.pdf
(section 2.2 and up)

- mathmate

by taking moments.
Assuming clockwise moments positive, and take moments about A,we have
500*1.5+400*2.598-Rbv(2.598)=0
Solve for Rbv=298.2
Then sum forces in y-direction,
Rav+Rbv-500=0
gives Rav=201.8.
Btw, we should also check the geometry, because the dimension 2.598=3sqrt(3), which is correct.

- anonymous

hey! how did you checked the vertical reaction?

- mathmate

Answer is right above your question, lol.

- mathmate

|dw:1444392770010:dw|
So ∑Fy=0 means
201.8- Fea sin 60=0
solve for Fea (compression)
∑Fx=0 means
Fed-400-Fea cos(60)=0
Solve for Fed (tension)
Get familiar with the tension compression notations as shown earlier.

- mathmate

You're probably in class now.
We'll continue later.

- anonymous

this is my answer... is it right?

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- anonymous

@ganeshie8 is my answer correct?

- anonymous

@mathmate is my answer correct?

- mathmate

Your answers are close, but not exact, missing by about 3N, starting from Fae.
Can you recalculate Fae=201.8/sin(60)?
I get 233.019N.
and hence recalculate Fad.

- mathmate

I think you had just a bad transcription of the third digit. The rest of the digits are all good. Therefore you need to correct Fae and recalculate Fad.
After that, you can proceed to other joints. Remember to decide on the direction of forces (tension or compression) when you draw the FBD.
Remember in Engineering statics, you need to have correct answers. Check and recheck you answers. Structures fail because of human errors.

- anonymous

for Fae...

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- anonymous

for Fad..

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- mathmate

Perhaps my eyes are not as good as yours, but I read 23\(\color{red}{3}\).018... from
img_20151009_210329.jpg
You also entered 238... in the calculation for Fad.
So if you change anything for Fae, you need to update Fad as well.

- anonymous

AWW.. my bad. :3

- anonymous

my final answers are:
Fae = 233.0185686 N (c)
Fad = 516.5092843 N (t)
Ffb = 338.09631376 N (c)
Fbd = 569.0481588 N (c)
are my answers correct?

- mathmate

1. Ffb and Fbd cannot be both compression. If they are, \(\sum Fx\) will not be zero.
Draw the FBD to determine that. (Fdb should be in tension)
You should determine the direction of the forces BEFORE you set up the equation.
Otherwise you will end up with the wrong values.
2. Ffb is off, again by a few newtons. Fbd is way off. You probably did not set up the equation correctly.
Ffb cos(60)-Fdb=0

- anonymous

Ffb = 338.0963176 N (C)
Fbd = 169.0481558 N (T)

- mathmate

Check if you used vertical force Rbv=298.2, or the angle 60\(^\circ\)
I get Ffb=344.3317005446928, and Fdb=172.1658502723462
(ignore the extra digits, just for the purpose of your checking).
|dw:1444404520132:dw|

- mathmate

For joint P:
if you resolve forces (mentally) perpendicular to CB, you will find that Ffd is downwards to balance the force of 400N. Therefore Ffd is in tension.
By intuition, Fbc is in tension, since both the 400N and Ffd are pulling away from C. So we end up with the FBD as follows:
|dw:1444412954486:dw|
Resolve perpendicular to FB:
400 sin(60)-FDsin(60)=0
means that FD=400.
Resolve parallel to (and along FB):
400cos(60)+FDcos(60)-BF-FC=0
Solve for FC.
You can solve joint E similarly.

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