Real Life Problem I Need Help With

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In summary, the gymnast's horizontal bar would result in a resultant force in the x direction at the wheel of the main landing gear, which would be the weight of the plane plus the braking force.
  • #1

Homework Statement

This is mostly a statics problem but I'm sure this forum can help me out. Mostly all in the picture. I am trying to find the force in the x direction at point B resulting from loads P1 and P2 at point A. Point A is at the center of the cross bar so therefore it is at equal distance of the legs attached to the hinges. Additionally, both legs are of the same length,z.

Homework Equations

ƩFx = 0
ƩFy = 0
ƩFz = 0

ƩMa = 0

The Attempt at a Solution

I know a moment will be created at point A and also at the other hinge. I just want to know the resultant force in the x direction at one of the hinges which will be equal to the force in the other direction.

Any help?

Thanks in advance!


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  • #2
Interesting question. It looks to me like a gymnast's horizontal bar, where point B and the other equivalent point are where the structure is touching the floor, and the gymnast would be at A, his weight providing the load. And he would be swinging, which would mean the load wasn't just vertically downward.

Anyway, if we say that the linear forces must sum to zero, then the forces on the point B and the other equivalent point (added together) must be equal and opposite to the load. By symmetry, I would guess that the force at B and at the other equivalent point would be the same. So then the force at B would simply be half the load.

Then if we think about moments, if we calculate the moments around the axis of the horizontal bar (through A), then obviously there is no contribution from the load. So there would just be the contribution from the forces at point B and its equivalent point. Since I said these were the same, they won't cancel, so the total torque would be nonzero. This suggests that the whole thing must topple over!

In conclusion, I don't know the answer. Maybe it would work if the structure were attached to the floor by a surface or line (instead of just at a point)?
  • #3
Ok you're right I did forget to draw the line for the ground but let's just assume that from now on. Here's a little more insight: what I am trying to do is resolve a force at the wheel of the main landing gear produced when braking the aircraft. P1 is the dynamic load produced by the aircraft when braking while P2 is the force measured from the retract actuator of the landing gear during braking. I am trying to simplify this as much as possible and I am not taking any friction force into effect yet produced from the sliding of the tire on the ground. I do suspect that I will need that but right now I am just focused on this.


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  • #4
It looks like the kind of problem which I would make a guess at, but in real life, the answer would be more complicated...

I would guess that the total force on the wheels would be the weight of the plane added to the braking force. And these two forces would be perpendicular to each other.

What do you think the answer would be? Also, are you trying to find out what the stress on the wheels would be, to figure out when they would break?
  • #5

I would first suggest reviewing the equations and concepts related to statics, specifically the principles of equilibrium and moments. It is important to understand these principles in order to accurately solve this problem.

Next, I would recommend creating a free body diagram of the entire structure, including the forces acting at point A and the hinges. This will help to visualize the problem and determine which equations to use.

From the given information, it seems that the force in the x direction at point B can be found by using the principle of moments. The moment at point A can be calculated by multiplying the force P1 by the distance from point A to the hinge, and the moment at the other hinge can be calculated by multiplying the force P2 by the distance from that hinge to point A. These two moments should be equal and opposite in order to maintain equilibrium.

Using the equation ƩMa = 0, you can set up an equation to solve for the force in the x direction at point B. This will involve using the distances and angles given in the diagram, as well as the forces P1 and P2.

It is also important to note that the forces at the hinges will have both an x and y component, so you may need to use the equations ƩFx = 0 and ƩFy = 0 to solve for those components as well.

I hope this helps and good luck with your problem!

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