Alignment force

[SC83]

I have a cup (see figure) that will be pushed on a conical tip with a force F.
The cup is not perfectly aligned with the conical tip and touches it in point 'A'. The cup is flexible supported.
In the X-direction it can translate via a spring with a stiffness K_x. The cup can also rotate via a torsion spring with a stiffness K_theta.
Via the vertical force F, the cup will align with the conical tip.

How can I calculate how big the force F at least must be? Via the torque equilibrium around point 'A'?

 

[GregLocock]

you'd need the friction coefficient at A I suspect.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[SC83]

Thank you for your answer GregLococ.

Yes the friction force at point A plays also a role.
But as the cup must translate in the X-direction and rotates around the Z-axis while the cup moves downwards (negative Y-direction) also K_x and K_thate defines F, right?

 

[Sparweb]

The contact force between the cup and the bar touching it on top is NORMAL/PERPENDICULAR to the surface.
Any force component pointing side-to-side on the top comes from friction at their contact.
You should draw some free-body diagrams of each component separately to sort out the rest.

No one believes the theory except the one who developed it. Everyone believes the experiment except the one who ran it.
STF

 

[rothers]

What's the stiffness of the K_x spring in the y direction ?

How does the torsion spring interact with forces in the y direction ?

 

[chicopee]

This is what I think. Draw a free body diagram and don't forget the weight of the cup and pin. Breaking down the components of F, you'll have Fy an unknown that includes the weight of the pin and Fx the spring reaction (=Kx) on F. At A, Ax and Ay will be part of the free body diagram summation of the forces in the X and Y directions. The direction of F is not quiet accurate because on that pin when a FBO is completed Fx will be the spring force whereas Fy will be an unknown value that will include the weight of the pin.