Calculating forces on a rod with bearings leaning against a wall

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SUMMARY

This discussion focuses on calculating the horizontal force (Fh) at the floor end of a zero-weight rod supported by zero-friction bearings, with one end against a wall and the other on the floor, under the influence of a vertical force (Fv). The relationship between Fh and Fv is established as Fh = Fv / tan(α), where α is the angle formed between the rod and the wall. The conversation highlights the importance of understanding the geometry and mechanics involved, particularly the concept of mechanical advantage in force distribution.

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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
 
@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.
 
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
 

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  • #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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