Calculating forces on a rod with bearings leaning against a wall

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Discussion Overview

The discussion revolves around calculating the horizontal force exerted at the floor end of a zero weight rod with frictionless bearings, which is leaning against a wall at an angle alpha while a vertical force is applied at the wall end. The participants explore the mechanics involved in this setup, including the derivation of formulas and the implications of different angles of inclination.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes the setup and asks how to calculate the horizontal force at the floor end given a vertical force at the wall end.
  • Another participant suggests that the problem resembles a schoolwork question and requests a sketch or Free Body Diagram (FBD) for clarity.
  • A participant expresses confusion about the concept of a zero weight rod resting on the floor and leaning against a wall.
  • One participant proposes a formula for the horizontal force, suggesting it could be fh = fv * tan(alpha), but admits uncertainty about deriving it.
  • Another participant indicates that the horizontal force at the floor end may be zero based on the initial description, but later acknowledges that the floor end is constrained horizontally, indicating missing information.
  • A participant discusses how the horizontal force depends on the angle of the rod, proposing that for different angles, the relationship between the vertical and horizontal forces changes, potentially involving a cotangent function.
  • One participant requests further details or literature on the foundational aspects of the proposed results.
  • Another participant mentions the need for anchorage to prevent buckling due to forces at connection points.
  • There is a discussion about mechanical advantage and the relationship between vertical and horizontal forces based on geometry.

Areas of Agreement / Disagreement

Participants express various viewpoints on the mechanics of the setup, with no consensus reached on the correct formula or the implications of the zero weight rod. Multiple competing views remain regarding the calculations and the physical interpretation of the forces involved.

Contextual Notes

Some participants note that additional information is needed to fully understand the forces at play, particularly regarding the constraints on the rod and the implications of its zero weight. There are also unresolved questions about the derivation of the proposed formulas and the conditions under which they apply.

g2c
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TL;DR
Deriving the force in a given direction from force in in a other
Hello, id appreciate your help for the following case: in a room, a zero weight rod has zero friction bearings at its extremities. One of his ends lies on the floor, the other is against a wall, forming with it an angle alpha. A verical force fv is applied to the 'wall end'. How to calculate the resulting horizontal force fh at the 'floor end'. Thank a lot
 
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g2c said:
TL;DR Summary: Deriving the force in a given direction from force in in a other

Hello, id appreciate your help for the following case: in a room, a zero weight rod has zero friction bearings at its extremities. One of his ends lies on the floor, the other is against a wall, forming with it an angle alpha. A verical force fv is applied to the 'wall end'. How to calculate the resulting horizontal force fh at the 'floor end'. Thank a lot
Sounds a bit like a schoolwork question. Is this for school or self-study, or part of a project you are working on?

Show us a sketch please, and your Free Body Diagram (FBD) for the setup. Thanks.
 
berkeman said:
Show us a sketch please, and your Free Body Diagram (FBD) for the setup. Thanks.
what he said (very small).jpg
 
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@berkeman - I used intuitively this setup to repair a broken pvc pipe and it did generate a large enough horizontal force but i don't know how to derive the formula. I "guess" though it could be fh=fv*tg(alpha). I dont know what is fdb and how to build it. Do you mean a drawing?
 
g2c said:
... a zero weight rod has zero friction bearings at its extremities. One of his ends lies on the floor, the other is against a wall, forming with it an angle alpha. ...
That "zero weight rod" really confuses me.
How is it possible to rest a zero weight rod on the floor, or lean on a wall?
 
g2c said:
Do you mean a drawing?
Yes, a drawing would be a big help. Or a picture, since it sounds like you actually built it.

User the "Attach files" link below the Edit window to upload PDF or JPEG files...
 
g2c said:
@berkeman - I used intuitively this setup to repair a broken pvc pipe and it did generate a large enough horizontal force but i don't know how to derive the formula. I "guess" though it could be fh=fv*tg(alpha). I dont know what is fdb and how to build it. Do you mean a drawing?
From the opening post, the way it was written, at floor end the force horizontal = 0, same for horizontal force from the wall onto the rod.

But,
From this post, it appears that the floor end is constrained horizontally.
Some information is missing.
 
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For Fv, applied downwards at the wall end.
Fh, along the floor, will depend on the rod angle, α.

For a near horizontal rod, α = 0; Fh = ∞
For a diagonal rod, α = 45°; Fh = Fv
For a near vertical rod, α = 90°; Fh = 0

That appears to be a cotangent function. Fh = Fv / tan( α ).
The rod will fail, probably due to buckling, for high axial forces at low α.
 
g2c said:
... i don't know how to derive the formula. I "guess" though it could be fh=fv*tg(alpha). I dont know what is fdb and how to build it....
Could it be something like what is represented in the attached PDF file?
 

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  • #10
@Lnewqban - exactly! Can you please detail the fondement of this result? Or send a link to literature dealing with such kind of forces problems? Below a picture of the worksite
1- fulcrums
2- long rod
3- short rod
4- clamps for adjusting the short rod length so as to have short + long = a bit longer than current visible length of pipe
 

Attachments

  • #12
Thanks, It states the formula nut doesn't tell why this is so
 
  • #13
g2c said:
Thanks, It states the formula nut doesn't tell why this is so
The reason is a mechanical advantage generated by the geometry.
Work or energy in = Work or energy out

The pushing vertical/horizontal forces ratio is proportional to the ratio of the horizontal/vertical distances covered by the ends/middle of the pipe.

Please, see:
https://en.m.wikipedia.org/wiki/Mechanical_advantage
 

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