Basics of Equilibrium
Basics of Equilibrium
(OP)
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....
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....





RE: Basics of Equilibrium
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
RE: Basics of Equilibrium
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.
RE: Basics of Equilibrium
RE: Basics of Equilibrium
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.
RE: Basics of Equilibrium
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 ?
RE: Basics of Equilibrium
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?
RE: Basics of Equilibrium
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 ??
RE: Basics of Equilibrium
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.
RE: Basics of Equilibrium
RE: Basics of Equilibrium
@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).
RE: Basics of Equilibrium
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
RE: Basics of Equilibrium
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).
RE: Basics of Equilibrium
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.)
RE: Basics of Equilibrium
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.
RE: Basics of Equilibrium
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.
RE: Basics of Equilibrium
RE: Basics of Equilibrium
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.
RE: Basics of Equilibrium
RE: Basics of Equilibrium
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/
RE: Basics of Equilibrium
RE: Basics of Equilibrium
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Basics of Equilibrium
This has been a rather interesting problem because of the nature of the solution set provided.
RE: Basics of Equilibrium
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/