1
$\begingroup$

I'm trying to compute the torque and current requirements on a motor powering a three-stage linear slide. I have included a diagram of the system below:

Three stage linear slide

The first stage (right-most) is fixed to the ground, while the second (center) and third (left) stages move. When the string is pulled, first the left stage goes up completely, then the second stage. This is different from other types of linear slide which require more torque but have all stages move at the same time.

From my understanding, this setup is essentially equivalent to the luff tackle, a three-pulley block and tackle pulley system. As such, this pulley system divides the torque necessary to pull the first stage up by 3, plus force lost to the pulleys (if we include moment of inertia), and any frictional loss. I understand that the load is divided between the three lengths of string before the string which gets pulled by the motor. Although I understand all of these things, I wanted to confirm this understanding by actually drawing out the free-body diagrams and writing out the equations necessary. However, I can't seem to figure out a way to draw them in a way that makes the math work out.

Here are my free-body diagrams, from the attach point on the left stage, through all three pulleys: Free-body diagrams of a three-stage linear slide

I'm having a hard-time reconciling the fact that the load is balanced across all three parts of the string. I can simply say that:

$$F_{LS} = T_1 + T_2 + T_3$$

How can I support that the load from the final stage is equal to the sum of the tensions in the string, using free-body diagrams and some simple substitutions? What ends up happening is I usually end up getting $F_{LS} = T_1$ with $T_1 = T_2 = T_3$ which is incorrect.

My hunch is that I'm missing something which relates the two pulleys on the second stage, since the forces pulling down on the top pulley are counteracted by the forces pulling up on the bottom. It could also be that my understanding of the system as a whole is flawed, and this is not actually equivalent to a luff tackle.

$\endgroup$
  • $\begingroup$ In a luff tackle, some pulleys move giving mechanical advantage. When only the left segment is moving, all pulleys are fixed. They are not analogous. $\endgroup$ – BowlOfRed Sep 22 '17 at 1:21
  • $\begingroup$ So, in fact, there is no mechanical advantage, $$F_{LS} = T_1 = T_2 = T_3 = T_4$$? $\endgroup$ – ajp Sep 22 '17 at 1:52
0
$\begingroup$

I have assumed that the mass of one single stage slide is m.

The force required to do the needed action is maximum during the last stage.

Here, the net mass of the slides to be lifted is 2m, and hence the tension required is 2mg.

Diagrammatic representation

Thus, you have the maximum force. A motor that provides this force is enough for whole process.

You can find the torque and current requirements which suits your needs , from the site below.

http://www.magtrol.co.in/support/motorpower_calc.html

$\endgroup$
  • $\begingroup$ Yes this is correct, my assumption about the luff tackle was incorrect: the fundamental difference is that the second stage in the linear slide does not move and does not aid in supporting the weight of the first, hence each pulley actually only serves in changing the direction of the force, not in carrying the load. With a linear slide which is setup to move all stages at the same time, the torque requirement goes up, you would have to carry twice the mass of the first stage plus the mass of the second stage, for a total of 3*mg. This is, of course, ignoring friction or moments of inertia. $\endgroup$ – ajp Sep 22 '17 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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