Calculating forces on a rod with bearings leaning against a wall

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  • #1
g2c
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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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  • #2
berkeman
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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.
 
  • #3
phinds
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Show us a sketch please, and your Free Body Diagram (FBD) for the setup. Thanks.
what he said (very small).jpg
 
  • #4
g2c
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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?
 
  • #5
Baluncore
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... 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?
 
  • #6
berkeman
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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...
 
  • #7
256bits
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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?
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.
 
  • #8
Baluncore
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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.
 
  • #9
Lnewqban
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... 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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  • Rod against wall and grade.pdf
    33.3 KB · Views: 20
  • #10
g2c
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@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

  • pipe repair.pdf
    2.6 MB · Views: 19
  • #12
g2c
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Thanks, It states the formula nut doesn't tell why this is so
 
  • #13
Lnewqban
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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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