Solving Equilibrium Problems: FBD Drawings

In summary: For #2, you would find the "moments about the joint" and then work out the tension.#1:MR=(572,100i - 20,000j - 64,000k)NmTA=-7,120j + 3,600kWA=20,580jFA=(0i - 34,540j + 3,600k)N
  • #1
caddyguy109
129
0
Okay, started these problems with just the FBD drawing parts, but need some more clues as to how to go about starting them, i.e., just what equations/order is used to set them up, etc.:

1) Supposed to write the 6 equations of equilibrim, so, for example, is this what I need to do:
Max+May+Maz+Fax+Fay+Faz+T+L+Mwing = 0 ?
http://www.ihostphotos.com/show.php?id=176477"

2) Similar procedure as #1?
http://www.ihostphotos.com/show.php?id=176478"

3) This looks a lot like the moment problems I just got done doing--am I looking for forces, moments, or both at A?
http://www.ihostphotos.com/show.php?id=176479"

4) Again, I'm guessing a similar setup as before (if I started that right?)?
http://www.ihostphotos.com/show.php?id=176480"

Also, for #1,2,4 should the weight "W" be included in the FBD?
 
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  • #2
caddyguy109 said:
Okay, started these problems with just the FBD drawing parts, but need some more clues as to how to go about starting them, i.e., just what equations/order is used to set them up, etc.:

1) Supposed to write the 6 equations of equilibrim, so, for example, is this what I need to do:
Max+May+Maz+Fax+Fay+Faz+T+L+Mwing = 0 ?
http://www.ihostphotos.com/show.php?id=176477"
Your FBD is good. Not the equations, though. You can't just put all the terms together and call that a meaningful equation.

You have to learn how to write out the scalar equations that result from Newton's Second Law (and for balanced torques). This is explained in your text. Read it first.

Also, you will find tutorials at the top of this forum - a couple of which teach you how to deal with Newtonian mechanics problems.

PS : This is introductory physics, not advanced physics - just so you know where to post next time.
 
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  • #3
Thanks.

Actually, this is mid-semester Statics stuff, so I just thought this would be a good place to post it (even though it is, really, just FBD's).

"Not the equations, though. You can't just put all the terms together and call that a meaningful equation."

Yeah, I kinda just saw that--that's more of the "theory" behind the problem to start than anything really meaningful I guess.

So, do I need the W term or not in the FBD's?
 
  • #4
Are you supposed to find the reactions? If so make a resultant-couple system about O to find the moments (decompose the moment vector into components in x,y,z axis). For the airplane problem.
 
  • #5
Cyclovenom said:
Are you supposed to find the reactions? If so make a resultant-couple system about O to find the moments (decompose the moment vector into components in x,y,z axis). For the airplane problem.

Yeah, and for the others too:
#1, supposed to find "the components of reaction where the wing is fixed to the fuselage at A".
#2, find "the components of reaction at the ball and socket joint A and then tension in the supporting cables BC and BD."
#3, find "the reactions at the fixed wall A."

but then
#4 says "if the hoist is in equilibrium, what forces act on the shaft at A? What is the max. mass m in kg that can be lifted?"

For the airplane one, what is the moment vector? r x L?
 
  • #6
Yes
the resultant moment of the forces, now simply descompose in x,y,z axis to find the values of the reaction moment components.

[tex] \vec{M}_{R} = \vec{r}_{1} \times \vec{L} + \vec{r}_{2} \times \vec{T} + \vec{r}_{3} \times m \vec{g} [/tex]
 
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  • #7
Oh, okay, that makes sense.
 
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  • #8
Okay, ended up with this for #1:
MR=(572,100i - 20,000j - 64,000k)Nm

It had said to "write equations of equilibrium--6 eq's and 6 unknowns". Do I really even have to do that, since I basically just found the components of reaction at A?
 
  • #9
You found the components of the Moment vector reaction at A, but there is also a Force vector Reaction at A!
 
  • #10
RIGHT--oops!

So, that would just be each F multiplied by its unit vector?
 
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  • #11
Yes, descompose the forces in unit vectors and work out the equations.
 
  • #12
So, is this right for force L (just one of them):
LA=45,000*[(0-0)i+(0-7)j+(0-0)k]/sqrt((-7^2))
LA=45,000*(-7j/7)= -45,000j??
 
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  • #13
Does this look right? Then I found TA*r and WA*u the same way.

To get:
LA=-45,000j
TA=-7,120j + 3,600k
WA=20,580j
then FA=(0i - 34,540j + 3,600k)N?

Does there being no x(i) value seem wrong? Does to me.
 
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  • #14
Something with the final force components is confusing me...
 
  • #15
Did I calculate the reactive forces for #1 right or not?

Moving on, would #2 be the same thing as this last part of 1, finding the reactive forces about the ball and socket joint? Then how do I get the tension forces?
 
  • #16
Where'd everyone go? Dinner?:tongue2:
 
  • #17
Anyone? Did I calculate the Force components at A correctly or not?
 
  • #18
Your descomposition in unit vectors of the Force L is does not "works" with your defined sign convention because of your set axis.
 
  • #19
Cyclovenom said:
Your descomposition in unit vectors of the Force L is does not "works" with your defined sign convention because of your set axis.

Okay, so? I'm not feeling too well today, sorry--but what exactly do I change then? Seems VERY simple, but something isn't clicking.:frown:
 
  • #20
Well +Z should be the positive so:

[tex] \vec{L} = 45000 \vec{k} [/tex]
 
  • #21
Oh dear god...can't believe I flubbed that one!

So, I forgot, do the forces simply translate--i.e., the upward force of L at a point O is the same as it would be at point A? I would think so.
 
  • #22
So there just ends up being this at point A, for the forces:
FA=L+T+mg
FA=(45,000k + 8,000i -20,580k)N
FA=(8,000i + 24,420k)N

Right?
 
  • #23
Yes but remember to change the signs, because they must be opposite to keep the equilibrium so the resultant force is equal to the null vector. The same for the moment vector, remember when you add them they must give 0.
 
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  • #24
Okay, thanks...that helped a lot! Now I got to work on the other 3...but I guess the same things pretty much hold true for them, just a little different.
 
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  • #25
Okay, finished 1). Now have most all of 2) done, except got stuck at the end.
Here's the problem again:
http://www.ihostphotos.com/show.php?id=177804"

And I got almost to the end with this (MA means sum of Moments about Pt. A):
MA = (1TBC + 1TBD - 35,200)i + (0.67TBC - 0.67TBD)j + (-0.67TBC + 0.67TBD)k = 0
then the unknowns are: TBC, TBD, RAX, RAY, RAZ
But, I'm getting a little confused on how to find these unknowns then, from what I have above? Any help?
 
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  • #26
Anyone got a clue on how to find the unknowns? Should be simple algebra, I would think, but it isn't working out...
 
  • #27
Calculate Moment about the support, to find the tensions, then do a sum of forces to find the reactions.
 
  • #28
Okay, thanks. I'll see where I get.
 
  • #29
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  • #30
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1. What is an FBD drawing?

An FBD (Free-Body Diagram) drawing is a visual representation of all the forces acting on an object in a given system. It includes all the external forces, such as gravity and applied forces, as well as internal forces, such as tension and friction. FBD drawings are commonly used in solving equilibrium problems to help determine the net force and direction of an object.

2. Why are FBD drawings important in solving equilibrium problems?

FBD drawings are important because they help to break down a complex system into simpler components, making it easier to analyze and solve equilibrium problems. By isolating all the forces acting on an object, FBD drawings allow for a more accurate calculation of the net force and direction, which is crucial in determining if an object is in equilibrium or not.

3. How do I create an FBD drawing?

To create an FBD drawing, you first need to identify the object or system you are analyzing. Then, draw a simple outline of the object, leaving out any unnecessary details. Next, add all the external forces acting on the object, such as gravity and applied forces, using arrows to represent the direction and magnitude of each force. Finally, add any internal forces, such as tension and friction, to complete the FBD drawing.

4. Can an FBD drawing be used for non-equilibrium problems?

Yes, FBD drawings can also be used for non-equilibrium problems. In these cases, the net force and direction calculated from the FBD drawing will not be zero, indicating that the object is not in equilibrium. FBD drawings can be used to analyze the motion of an object in non-equilibrium situations, such as when it is accelerating or decelerating.

5. Are there any limitations to using FBD drawings?

While FBD drawings are a useful tool for solving equilibrium problems, they have some limitations. FBD drawings assume that the forces acting on an object are constant and do not change over time. In reality, forces can vary and change direction, making it difficult to accurately represent them in an FBD drawing. Additionally, FBD drawings do not take into account the size or mass of an object, which can affect the magnitude of forces acting on it.

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