# 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

Mentor
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.

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

... 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?

Mentor
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...

Gold Member
@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.

• berkeman
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.

Homework Helper
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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?

#### Attachments

• Rod against wall and grade.pdf
33.3 KB · Views: 20
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

#### Attachments

• pipe repair.pdf
2.6 MB · Views: 19
Homework Helper
Gold Member
g2c
Thanks, It states the formula nut doesn't tell why this is so

Homework Helper
Gold Member
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.