Solving 5 Unknowns with 6 Equations 1

[newreynolds]

Probably a stupid question but I am doing a static force analysis using a Method of Joints. There are 3 joints of interest, so 6 equations but I only need to solve for 5 unknowns.

The last unknown is in the last 2 equations so I solve one and then do a check with the final equation but it's never zero. My question is then, should that last equation be satisfied since I only really needed 5 equations?

I believe it should but after looking over my algebra and not finding an error, I'm questioning things.

 

[DRWeig]

Tell us your equations.

Good on ya,

Goober Dave

 

[desertfox]

post up the actual problem

 

[newreynolds]

Thanks guys, the equations are as follows:

1: 0 = F2*cos(phi1 + theta6) - F1*cos(theta6) + To*cos(lamda4) - To*cos(lamda5)
2 : 0 = F2*sin(phi1 + theta6) - F1*sin(theta6) + To*sin(lamda4) + To*sin(lamda5)
3 : 0 = - F2*cos(phi1 + theta6) - F3*cos(phi2 + theta7) - Fs*cos(psiL)
4 : 0 = - F2*sin(phi1 + theta6) - F3*sin(phi2 + theta7) - Fs*sin(psiL)
5 : 0 = F3*cos(phi2 + theta7) - F4*cos(theta7) + To*cos((3*pi)/2 + lamda2) + To*cos(pi/2 - lamda3)
6 : 0 = To*sin(pi/2 - lamda3) + F3*sin(phi2 + theta7) - F4*sin(theta7) + To*sin((3*pi)/2 + lamda2)

The variables are F1, F2, F3, F4, Fs and each equation pair is the force summation at the three respective joints. i.e. equation 3 is the sum of the forces in the x-direction and equation 4 is the sum of the forces in the y-direction for Joint B

 

[desertfox]

hi

Your post is meaningless without a diagram of the problem.

 

[zekeman]

Eq 3 and 4 are redundant.
So you can use eq 1,2,3,5,6 and get your answer.

When you get F2 and F3 from these equations, either eq3 or eq 4 will yield F6 uniquely. Try it.

 

[zekeman]

Correction

use eq 1,2 5,6 to get

F1, F2, F3, F4
Then use either eq 3 or Eq 4 to get F5

 

[TheTick]

Plenty of matrix solvers out there waiting for input.

 

[electricpete]

Looking at the math without knowing the underlying problem, I don't see Equations 3 and 4 as being dependent. They seem independent to me.

Let's say we have 3 vectors A, B, and C, represented by complex numbers with magnitudes -F2, -F3 and -Fs respectively and angles in the complex plane of phi1+theta6, phi2+theta7 and psiL, respectively.
A = -F2* {cos(phi1+theta6) + i * sin(phi1+theta6)}
B = - F3*{cos(phi2 + theta7) + i * sin(phi2 + theta7)
C = - Fs*{cos(psiL) + i * sin(psiL)}

We can see that equation 3 tells us the real part of the vector sum (A+B+C) is 0. In contrast equation 4 tells us the imaginary part of the vector sum (A+B+C) is 0. The vector sum has two independent coordinates, there is no reason that one has to be zero if the other one is zero.


=====================================
(2B)+(2B)' ?

 

[zekeman]

"Plenty of matrix solvers out there waiting for input."

Right, just solve the 5x5.

 

[rb1957]

give us a pic ...

it looks like you're assuming an applied force (To) to be the toatl joint reaction in a direction ?

not so sure i agree that eqns 3 & 4 are redundant (ie equivalent, a multiple of one another ... looks to like sum of forces in orthogonal directions.

3 joints = 6 equations (3*sum forces in two direction).
how many members ? 4? (F1 to F4) and Fs and To ??

 

[zekeman]

"Looking at the math without knowing the underlying problem, I don't see Equations 3 and 4 as being dependent. They seem independent to me.

Let's say we have 3 vectors A, B, and C, represented by complex numbers with magnitudes -F2, -F3 and -Fs respectively and angles in the complex plane of phi1+theta6, phi2+theta7 and psiL, respectively.
A = -F2* {cos(phi1+theta6) + i * sin(phi1+theta6)}
B = - F3*{cos(phi2 + theta7) + i * sin(phi2 + theta7)
C = - Fs*{cos(psiL) + i * sin(psiL)}

We can see that equation 3 tells us the real part of the vector sum (A+B+C) is 0. In contrast equation 4 tells us the imaginary part of the vector sum (A+B+C) is 0. The vector sum has two independent coordinates, there is no reason that one has to be zero if the other one is zero."

If A, B, C were arbitrary vectors, you would be right.
However, if you look at it in the context of a joint, these are not arbitrary vectors but sum to zero.

Given that, if you know 2 of the vectors, then the amplitude of the third requires only one equation.

 

[zekeman]

Another way of looking at this is the OP eqs 3 and 4 are precisely the statement that
A+B+C=0

 

[electricpete]

Another way of looking at this is the OP eqs 3 and 4 are precisely the statement thatA+B+C=0

Yes. That is a 2-D vector equation, which is equivalent to 2 scalar equations.

If A, B, C were arbitrary vectors, you would be right.However, if you look at it in the context of a joint, these are not arbitrary vectors but sum to zero.

Two scalar equations are required to make those vectors sum to zero. Just because you know they sum to zero doesn't change the fact that these 2 equations must be satisfied to make them sum to zero.

So, my conclusion remains that Equations 3 and 4 by themselves constitute 2 independent scalar equations in 3 scalar unknowns (F2, F2, Fs), assuming all the angles are known.

HOWEVER, assuming there is a solution, I agree with you there must be redundancy somewhere among the 6 equations. It certainly may be that equation 3 or equation 4 can be dropped. But as for myself, I don't think that can be determined without looking at the other four equations.

I notice Equation 5 and 6 have the same form as equations 3 and 4 (vectors sum to 0). However, equations 1 and 2 are different in that the magnitude T0 appears twice. So it is only a vector sum to 0 equation if there are included two different vectors with exactly the same magnitude T0, but with different directions (described by lambda4 and lambda5). Does that part make sense for the physical problem?

=====================================
(2B)+(2B)' ?

 

[electricpete]

I notice Equation 5 and 6 have the same form as equations 3 and 4 (vectors sum to 0). However, equations 1 and 2 are different in that the magnitude T0 appears twice. So it is only a vector sum to 0 equation if there are included two different vectors with exactly the same magnitude T0, but with different directions (described by lambda4 and lambda5). Does that part make sense for the physical problem?

Is it possible there should be another unknown magnitude substituted for one of those T0's? (That would give 6 equations in 6 unknowns)

I'm just asking. I don't know the answer. I'm not real familiar with the joint method.

=====================================
(2B)+(2B)' ?

 

[hokie66]

I agree with those who say we need the actual problem rather than a set of equations.

 

[ione]

Are you trying to solve a truss? Post a drawing and you’ll get an answer.

From a strictly mathematical point of view you have a system of equations and the number of equations exceeds the number of unknowns so:

1. If you don’t have two equations linearly dependent amongst your five, your system is impossible
2. Upon condition you have two equations linearly dependent amongst your five, if your equations are not compatible the system is impossible
3. Upon condition you have two equations linearly dependent amongst your five, and your equations are compatible, the system has, at least, one solution.
4. If your equations are compatible and you have more than two equations linearly dependent amongst your five, the system has infinite solutions.

 

[electricpete]

I agree that having a drawing describing the problem is better than trying to guess from the equations. By guessing I made a mistake assuming “To” was an unknown. Therefore disregard my message 17 Oct 11 6:00.

If we attempt to analyse in absence of drawing, we notice:
Eq1 and 2 have unknowns: F1, F2
Eq2 and 3 have unknowns F2, F3, Fs
Eq5 and 6 have unknowns F3, F4

ASSUMING the equations are correct as written, and the unknowns are F1, F2, F3, F4, Fs, then if we wanted to solve it manually, we could manually solve two ways (below). The first way suggests we can ignore either equation 1 or 2. The second way suggests we can ignore either equation 5 or 6. I could not find any straightforward way to suggest we can ignore equations 3 or 4:

FIRST WAY:
Solve Eq1 and Eq2 For F2 and F1.
Use results of above (F2) to solve Eq3 and Eq4 for F3 and Fs.
Now we are left with 2 equations (Eq5 and Eq6) with only one unknown (F4). In this case (assuming equations are correct) I think it is safe to ignore 5 or 6 (use the ignored one as a double-check).

SECOND WAY:
Solve Eq5 and Eq6 For F3 and F4.
Use results of above (F3) to solve Eq3 and Eq4 for F2 and Fs.
Now we are left with 2 equations (Eq1 and Eq1) with only one unknown (F1). In this case (assuming equations are correct) I think it is safe to ignore 1 or 2 (use the ignored one as a double-check).


=====================================
(2B)+(2B)' ?

 

[electricpete]

Add the following way:
THIRD WAY:
Use 1, 2, 5, 6 to solve F1, F2, F3, F4.
Then we have left equations 3 and 4 with only one unknown (Fs).
Ignore 3 or 4.

This suggests we can pick any of the 6 equations and ignore it (assuming the equations are correct and define a unique solution)


=====================================
(2B)+(2B)' ?

 

[electricpete]

My apologies zekeman for disagreeing with you on that point.

=====================================
(2B)+(2B)' ?