Basics of Equilibrium

[CSCPE24]

Hi All,

Just a trivial question. I'm trying to prepare for a PE exam and came across a sample question that seems to escape me. I just need a nudge in the correct direction.

I've enclosed the diagram to the problem. It is asking me to find the magnitude, direction, and distance from point A of the force required to act on the rigid body shown to hold it in equilibrium.

Summing forces in X and Y directions solves for the Magnitude and Direction of the force. Summing moments about point A also gives me 13Y + 14X - 468 =0. However, what am I missing to find another equation with respect to X and Y so I can solve for them? X & Y are the distances from point A respectively.

Any help is greatly appreciated.

Thanks in Advance....

 

[Cockroach]

I'm not sure I follow your logic.

Summing the forces in either horizontal or vertical direction will give you one equation. Summing the moments about a point will give you another equation. You will have two equations in two unknowns, hence closed form solution set.

You can also sum the moments at a differing location to use the solution set from the above process. Upon subsitution, the result will be the null solution, 0=0. This means that you're arithmetic is correct, basically validifying the solution set.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[CSCPE24]

Cockroach,

Summing forces in the horizontal and vertical direction does not give me an equation with the required distances X & Y.

I only have 3 (Sum Fx, Sum Fy, & Sum Moment at point A) equations with 4 unknowns.

Am I still missing something?

Thanks Again.

 

[rb1957]

like 'roach posts a free body is in equilibrium for forces and moments. in your 2D example you have 2 force equations (Fx and Fy) and one moment (Mz) ... 3 equations, 3 unknowns (magnitude, direction ... simply as x- and y- components, and position (pick a point)

 

[CSCPE24]

rb1957, to be sure I understand this, are you saying that from the moment equation in which I got 13Y + 14X -468 = 0, X & Y can be arbitrary as long as it satisfies the solution?

So if X = 0, then Y = 36....etc?

The answer provided in the sample is X=30.9", and y=2.77" from the lower left corner which is point A. These numbers obviously satisfy the equation 13Y + 14X - 468 =0 but I didn't think it was arbitrary.

I'm sorry I'm not getting this...maybe I'm just dense today.
Thanks for the patience.

 

[rb1957]

yes, you're being dense (today) ...

1st sum Fx ... reaction = 13 to the right
2nd sum Fy ... reaction = 14 acting down
3rd sum moments about A. if the reaction point is on the line AB then Fx has no moment (neither does the load at A or at C) then (CW +ve) Fy*a-10*24-15*12-16*12+12*12 = 0 a = 33 6/14" (to the right of A)

clear as mud ?

 

[CSCPE24]

rb1957, what you just calculated was what I obtained if Y=0, then X=468/14, or 33.43" to the right of point a.(from eq. 13Y + 14X-468=0) which is what you have.

However, this is not the answer provided which is why I'm scratching my head to begin with.

It states that the force on the solid body such that an equilibrium is established is when it is located at point x=30.9" and y=2.77" from point a.

If this is the case, would you agree the answer is incorrect?

 

[rb1957]

run the sums yourself, with the point given and the forces we've calc'd ... it makes sense that if the reaction point is above AB, then x < 33.43.

there are an infinite number of place that the reaction point can be at ... (off the top of my head) along the line of action of the reaction, through the point (33.43,0)

is there some constraint on where the reaction point is in the question ? maybe they're just giving you one solution ??

 

[CSCPE24]

rb1957, I came to the same conclusion too. Unless there exist another set of requirements, the location of the force is almost infinite.

For this case, I would think that all external forces acting on a solid body should be confined to within the boundaries of the solid body such that the point will satisfy 0<X<24 and 0<Y<12.

Thanks again for the sanity check. I appreciate you help.

And yes, I think they only provided one solution.

 

[Compositepro]

The way I look at the problem, you add all the force vectors to get the resultant force vector. The answer is an equal and opposite force vector, which is applied to the solid body at a point on its surface where that vector intersects the surface. You will notice that all the given forces are applied at the surface of the object. I think this last point is what you are missing. I guess that there will actually be two points where the force could be applied (as a push or a pull).

 

[rb1957]

@CSC ... somewhere on the physical part would be a good practical solution (the solution given is off the part, no?)

@composite ... not really, the vector sum of the applied forces only gives you the vector of the reaction (sum Fx, sum Fy) ... you need sum moments to locate the vector on the part (the vector summation doesn't account to the different position of the forces).

 

[hydtools]

You cannot assume Fx acts along the line AB. Just like you did not assume Fy acts along line AC.
You need another sum of moments for the fourth equation to solve for x and y, the point of application of reaction force F. There are four unknowns, Fx, Fy, x, and y.

Ted

 

[rb1957]

sorry, there are many places on the part where the reaction can be applied and the part is stable; there is no one unique point.

once you've established one point, then anywhere along the line of action of the reaction is going to produce equilibrium ('cause it'll produce the same moment about any fixed point).

 

[Wrenchbender]

CSC,
We all figured out the required force to keep this whatever-it-is in eqbm, is Fe = -13i + 14j.

For its magnitude I get sqrt(13^2 + 14^2), or |Fe| = 19.1

The moment to hold this whatever in eqbm is M = -468.

How far from point 'A' would you place the force vector to get this moment? 468/19.1 or 24.5 (perpendicular distance).
Where to physically place this force the the line established, would depend. No?

Sly problem. Good luck on the exam. (Glad I don't have to take it.)

 

[CSCPE24]

hydtools, you are correct. In order to derive a specific answer to the problem, (hence the given answer) there will have to be an additional condition that allows the formulation of a 4th equation to solve for X & Y.

However, there is no such condition presented in the book thus the form -13i + 14j - 468 = 0 leads us to the conclusion that there are multiple scenarios that can satisfy equilibrium.

That is why rb1957 mentioned that you could literally just pick a condition to satisfy this scenario. I also agree that a practical answer would be a force acting on the body itself and not some force that is acting outside of the body as portrayed by the given answer.

Bestwrench, thanks for the wishes.

 

[zekeman]

Think about the answer. If a single vector force can put this thing in equilibrium, then that force is not confined to a single point, so in your case it is on the line

13y+14x+480=0

So what does it mean? It means that the vector force lies on this line and any combination of x and y satisfying that equation is the answer.

 

[rb1957]

exactly as zeke says ...

 

[CSCPE24]

rb1957, zeke, I think we're in agreement for sure.

13y+14X+468=0 as you guys pointed out is:

y=1.08x + 36.92

I think the book should have given the answer as an equation like that.

 

[rb1957]

except there should be at least 1 -ve sign ...

 

[IDS]

Looking at the answer given, it seems that they have resolved forces and taken moments about A for the horizontal and vertical forces separately, then the X and Y dimensions from A are:

Moment due to vertical forces/resultant vertical force
= 432/14 = 30.9"

Moment due to horizontal forces/resultant horizontal force
= 36/13 = 2.8"

Nothing wrong with that, in fact it is probably the simplest way to do it, but as pointed out, any other equal and opposite force at any point on the line of the resultant applied force is equally valid.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/