Calculating forces on a rod with bearings leaning against a wall

[g2c]

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

 

[berkeman]

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.

 

[phinds]

Show us a sketch please, and your Free Body Diagram (FBD) for the setup. Thanks.

 

[g2c]

@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?

 

[Baluncore]

... 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?

 

[berkeman]

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...

 

[256bits]

@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.

 

[Baluncore]

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.

 

[Lnewqban]

... 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?

 

[g2c]

@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

 

[Lnewqban]

This may help in this specific case:

https://www.engineeringtoolbox.com/toggle-joint-d_2077.html

You may need some anchorage to ground in order to prevent the pipe from bucling back up or sideways by forces at points of connection in normal operation.

 

[g2c]

Thanks, It states the formula nut doesn't tell why this is so

 

[Lnewqban]

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