I am trying to model the dynamics of some system.

In summary, the system consists of one horizontal linear actuator that drives an object along a planar surface and one vertical actuator that brings the object in contact with the planar surface.

The planar surface (plate) deforms as the vertical motion of the object comes into contact with it. After the plate deforms, the object is dragged along the surface, giving arise to frictional force that opposed the horizontal motion.

My question is, does the normal reaction force from the elastic plate depend on the frictional force?

Initially I thought that because the two forces (normal and frictional/tangential force) should not interfere with each other as they are described by unit vectors perpendicular to each other. But placing an encoder on the vertical actuator, I found that moving the object tangentially along the deformed surface (along x-axis) affects the vertical motion (displacement/velocity along y-axis).


  • $\begingroup$ It's hard to visualize exactly what you are doing, but why don't you expect that, approximately, $F = \mu N$ where $F$ is the friction force, $N$ is the normal force, and $\mu$ is the coefficient of (kinetic) friction? $\endgroup$
    – alephzero
    Mar 22, 2019 at 10:00
  • $\begingroup$ "does the normal reaction force from the elastic plate depend on the frictional force?" Can you explain or elaborate what the "normal reaction" meant, do you mean reaction normal to the plate, or else? $\endgroup$
    – r13
    Aug 9, 2021 at 17:55

1 Answer 1


If your object has to move through a depression created by vertical force, then the force F is the sum of two forces.

Friction force required to pull it horizontally and the viscosity, and plastic deformation of the pit on the support platform. The first part

$$ F= \mu F_y $$

The tricky part is force needed to deform the plate and keep rolling this depression.

You need material properties on the support.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.