# Roller Screw drive - axial movement instead of friction

I need an equation or a some hints to solve the following problem.

Imagine a roller screw drive. I apply a torque of T to translative move my load mass M. I assume my screw has an efficiency of 90%. Now an additional axial force affects my mass in the opposite moving direction. Is this force completely transformed into torque (of course considering the efficiency) or is it possible, that my whole roller screw is moving, because it is not fixed? I just found papers/books/articles for movable slides/loads, but fixed shafts. But in my case motor and shaft are part of an osciallation system.

I'm not a mechanical engineer, so I'm sorry if the answer may is trivial.

I made a little sketch now The process force Fp is pushing my mass, most of the force is transformed into a load torque Tp which acts against my drive torque TD. Some of the energy is lost by friction. The question is, if there is also a partial force Tp? which is affecting the bearing and therefore exciting my chassis.

• What do you mean "because it is not fixed?" The roller screw drive should affect the motion between two rigid assemblies, and will be affected by any forces between those two assemblies. If neither of those assemblies is nailed down to the ground, then the motion gets more complex. Which gets back to -- what do you mean by "because it is not fixed?" – TimWescott Sep 5 '13 at 21:43
• exactly that, neither of the two is "nailed" to the ground. The shaft is mounted to a bearing, which itself is part of an oscillating system. And I need to model this complex motion. The question is whether also a translative force is affecting the bearing or if the losses are completely consisting of friction. – thewaywewalk Sep 5 '13 at 22:37
• Oi. Now there's not enough information. But to get back to basics: if the ball screw (roller screw -- whatever) is exerting axial force or has force exerted on it, then that force will show up as torque. Note this isn't the same as a plain old jack screw drive -- in that case, when you try to back drive it, the force will be lost to friction in the nut. – TimWescott Sep 6 '13 at 18:21
• I'm having trouble figuring out exactly what you are asking. A diagram, screenshot, or pictures of the setup would help in understanding your problem. – ddevaz Sep 6 '13 at 19:59
• okay I added a sketch (I was logged out, don't be confused) - I hope my question is clearer now. – thewaywewalk Sep 8 '13 at 16:18

OK. as drawn, ignoring mass and accelerations, the force $F_p$ will appear as a torque on your ball screw.
• Further comments from @thewaywewalk: The point where you get stuck, is that torque is a form of energy while force is a potential. Applying Newton's third law, as suggested, $F_p$ is affecting the bearing with 100% - but $F_p$ is also moving the the mass/screw by $x$ and therefore introducing the energy $F_p \cdot x$, which is transformed into the torque $T_p$ ~ $F_p \cdot x$, which finally brakes the motor. – Ian Sep 12 '13 at 16:57