# Two Beam Problem: Solve 6 Equations, 7 Unknowns

• robot6
In summary, the problem is that I have six equations and seven unknowns. I've drastically simplified the problem for the sake of discussion. The problem is that I end up with six equations and seven unknowns. Let ##R_{AX}## (and similar) be the reaction force at the A node in the rightward direction on the page, and ##R_{AY}## (and similar) be the reaction force at the A node in the upward direction on the page. ##M## is an applied moment and ##M_B## is the reactive moment at node B. For the left beam, I have: ##R_{AX}+R_{BX}+F_1\sin(\alpha) =
robot6
Homework Statement
I have a two beam problem, see illustration below. The problem requires a solution for the left beam and for the right beam.
Relevant Equations
##\Sigma F_x = 0##
##\Sigma F_y = 0##
##\Sigma M = 0##
I've drastically simplified the problem for the sake of discussion. The problem is that I end up with six equations and seven unknowns. Let ##R_{AX}## (and similar) be the reaction force at the A node in the rightward direction on the page, and ##R_{AY}## (and similar) be the reaction force at the A node in the upward direction on the page. ##M## is an applied moment and ##M_B## is the reactive moment at node B.

For the left beam, I have:
##R_{AX}+R_{BX}+F_1\sin(\alpha) = 0##
##R_{AY}+R_{BY}-F_1\cos(\alpha) = 0##
##R_{AX}L_1\sin(\alpha)-R_{AY}L_1\cos{\alpha}+M+M_B=0##

For the right beam, I have:
##R_{BX}+R_{CX}-F_2\sin(\alpha)=0##
##R_{BY}+R_{CY}-F_2\cos(\alpha)=0##
##R_{CX}L_2\sin(\alpha)+R_{CY}L_2\cos{\alpha}+M_B=0##

So I have six equations and seven unknowns, namely, ##R_{AX}, R_{AY}, R_{BX}, R_{BY}, R_{CX}, R_{CY}, M_B##. What am I missing?

#### Attachments

• tm1-02.png
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The suspense is killing me. Pun intended...

Hm. I think that ##M_B## may not exist... because it's accounted for in ##R_{A}, R_{B}, ## and ##R_{C}##.

OK I have convinced myself that ##M_B## should be removed. Does it make sense, then, that M should appear in both the third and sixth equations, or should it only appear in one of them?

I am unclear what M and MB represent in the diagram. Is M an external torque applied only to the left hand beam, while MB is the torque the beams exert on each other?
If the joint is a hinge, yes, MB can be dropped as the beams cannot exert torques on each othe.

The signs look inconsistent in your first equation in post #1.

Always make it clear what axis you are using for torque equations. Why do the F forces not appear in those?

M is an external torque applied (called a "moment" in this notation, from a different country.)

Sign for equation 1: RAX points to the right, RBX points to the right, and F points to the right, so all should be positive, in my opinion. How is it incorrect?

AH! since they drew M the way I drew it, it looks like M is positive. I-hat (index finger) x j-hat (middle finger) should point out of the paper (thumb). Pointing out of the paper (thumb) means counterclockise (way fingers go). At least that's how my right hand looks... Did I do something wrong?

Oh the axis for the torque equation is k-hat.

I finally did get a 6x6 matrix with det=29 so I think I solved it. Thanks to Octave!

robot6 said:
Sign for equation 1: RAX points to the right, RBX points to the right, and F points to the right, so all should be positive, in my opinion. How is it incorrect?
You are right, the problem is with the fourth and fifth equations. If the right-hand beam exerts a force Fx on the left one then the left exerts a force -Fx on the right.
robot6 said:
the axis for the torque equation is k-hat.
Unless an origin is specified, that is only a direction, not an axis. Are you taking B as origin?

Yes, B is origin, which makes it simpler because M is about B.

By the way it is so helpful that you point out that I need to review equations 4 and 5.

I think they are correct, too, because

in equation 4, RBX is positive to the right, RCX is positive to the right, but F2 is pointing towards the left, so it is negative;

in equation 5, RBY is positive up, RCY is positive up, but F2 is pointing down, so it is negative as well.

i am sorry I did not label the R's, but they are all pointing the same direction, + to the right and + up. F's are not pointing in this direction, though.

robot6 said:
in equation 4, RBX is positive to the right
That would be fine if you were defining RBX as a force left beam exerts on right beam, but in equation 1 you defined it as a force right beam exerts on left.
So the force left exerts on right is -RBX.

Last edited:

## What is the "Two Beam Problem"?

The Two Beam Problem is a mathematical problem that involves solving six equations with seven unknown variables. It is commonly used in physics and engineering to analyze the behavior of two intersecting beams of light or particles.

## Why is it important to solve 6 equations with 7 unknowns in the Two Beam Problem?

Solving 6 equations with 7 unknowns allows us to accurately predict the behavior of two intersecting beams. This is crucial in understanding and designing various systems, such as optical instruments, particle accelerators, and communication networks.

## How do you solve the Two Beam Problem?

The Two Beam Problem can be solved using various mathematical methods, such as substitution, elimination, or matrix operations. It requires a deep understanding of algebra, trigonometry, and calculus.

## What are some real-life applications of the Two Beam Problem?

The Two Beam Problem has numerous applications in science and engineering, such as designing optical systems, analyzing particle collisions, and optimizing communication networks. It is also used in fields like astronomy, medical imaging, and robotics.

## Are there any limitations to the Two Beam Problem?

One limitation of the Two Beam Problem is that it assumes ideal conditions, such as perfectly straight and parallel beams. In reality, there may be factors like diffraction, refraction, and interference that can affect the behavior of the beams. Additionally, the problem becomes more complex when considering multiple intersecting beams instead of just two.

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