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I am designing a linear telescopic system that gets elevated with a step motor, but to find the compatible motor, I need to work out the torque that is required for the elevation. I have uploaded my CAD model to this thread with how cables are connected. Could someone please teach me how do I go about calculating the torque that is required? it is even more difficult with a pulley that is also moving up in the system. Thank you in advance...

enter image description here

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  • $\begingroup$ That string arrangement isn't going to work. The last length of cable (on the right) does nothing. The second last section pulls the last section down. With the thread tensioned the middle section will be fully up and the upper section fully down. Move the pulley under the light to the top of the middle section. Note that the sections won't move evenly. The freest one will move first. $\endgroup$
    – Transistor
    Mar 5 at 17:25
  • $\begingroup$ Will this arrangement work if I replace the pulleys to screw eye hooks? If not, I can implement a another pulley in the second stage will the cable that goes over it $\endgroup$
    – AK RAJ
    Mar 5 at 17:43
  • $\begingroup$ Will this arrangement work if I replace the pulleys to screw eye hooks? If not, I can implement another pulley in the second stage will the cable that goes over it. This arrangement should be work I think. Even then I don't how to calculate the torque, are there any resources that I can use to understand this concept. $\endgroup$
    – AK RAJ
    Mar 5 at 17:49
  • $\begingroup$ Add a second diagram. $\endgroup$
    – Transistor
    Mar 5 at 17:53
  • $\begingroup$ @Transistor I have added the second cable arrangement in the question. Do you think this will work. I will also give the link to a video where I got this idea from too. youtube.com/watch?v=gPZPs-l5LiU $\endgroup$
    – AK RAJ
    Mar 5 at 18:03
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You can workout the tension in the rope using energy/work principle. In stage one, the mases will move a distance of h1, the work done equals to W1 = (m1+m2+m3)gh1 = T1r. Assume your design can stop mass 1, then the additional work for mass 2 and mass 3 to reach the final position is W2 = (m2+m3)g(h2-h1) = T2r. Thus at the end of the journey, the total work done W = W1 + W2, and T = T1 + T2 = W/r.

enter image description here

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If we assume no friction and ignore the mass of your jack, because of the moving pulleys T tension will be 1/4 of the weight at the step motor:

let's assume your weight to be lifted is

  • w =100kg= 980N

  • T tension in cord= 980/4=245N

  • r the radius of the step motor pulley is 2cm

then the torque required is t=Fr

$$\tau_s= 245*2/100=490N/100= 4.9Nm$$

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  • $\begingroup$ How do I find the tension in the cable? $\endgroup$
    – AK RAJ
    Mar 5 at 16:24
  • $\begingroup$ any pulley moving along with the cord will half the tension. stationary pullies just redirect the cord. It appears that you have two moving pulleys, or correct me if I misinterpret your figure. The rest are just changing the direction of the cord. $\endgroup$
    – kamran
    Mar 5 at 17:20
  • $\begingroup$ Do I not have three moving pulleys; two on the third stage and one on the second stage. $\endgroup$
    – AK RAJ
    Mar 5 at 17:30
  • $\begingroup$ yes, i assumed 2 moving pulleys. but i assumed the right bottom puley is moving in a slot not showing. if no the jack will lock up as @Transistor says. $\endgroup$
    – kamran
    Mar 5 at 17:39
  • $\begingroup$ What if I to replace the pulleys to screw eye hooks, will this cable arrangement work then? $\endgroup$
    – AK RAJ
    Mar 5 at 17:45
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kamran has already done the maths for you. You seem to be confused about the cable geometry.

enter image description here

Figure 1. Note that when taut the short sections of cable should be as close to vertical as possible.

The cable will have a certain tension, T. The amount of lift that tension can generate is T cos θ where θ is the angle between the string and vertical. When θ is zero, cosine is 1 and that gives maximum lift.

Failure to keep the short sections vertical means that the tension must increase by a factor of 1/cos(θ) the higher the load is lifted.

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  • $\begingroup$ Thank you for the response, " Failure to keep the short sections vertical means that the...." what were trying to say here please? $\endgroup$
    – AK RAJ
    Mar 9 at 13:30
  • $\begingroup$ Sorry I started to write that and then went back and worded the previous paragraph to cover it. I've completed the sentence but it's saying the same thing. Do you understand your original errors and why you need to keep those sections vertical? $\endgroup$
    – Transistor
    Mar 9 at 14:10
  • $\begingroup$ yh thank you. I have modified the design according to what you have taught me . Could you have look at that design for me plz? $\endgroup$
    – AK RAJ
    Mar 9 at 22:00

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