I am designing a scissor lifting table and I have an issue with the linear actuator. table This table should be lifting 100 kg. I studied the problem in two ways : a static study and a power study. My table is a 3-level table. Here is a schema of a one level table : lifting table sketch (1) and (4) are V-rails that guide wheels on points A and F. (5) is the linear actuator. (2) and (3) are scissors, and all the mechanical connections between the different parts are pivots (revolute).

In this static study, I found (page 3) that the two linear actuators have to lift a load of minimum 2158.2 N. So each linear actuator has to be lifting 1079.1N.

In this power study, I found (page 2) that the two linear actuators have to lift 12988 N (6494 N without the safety factor). So each linear actuator has to be lifting 6494 N (3247 N without the safety factor).

Between the two results, I have a ratio of 3. I guess the second study is false, but I'm struggling finding why, and I still have no solution...

I hope you can help me, and I thank you for your time :)


  • $\begingroup$ 3 is the number of scissor links, which is likely related $\endgroup$ Aug 1 '18 at 14:41
  • $\begingroup$ You should really typeset your calculations, or better yet calculate in a software and have it typeset your stuff. $\endgroup$
    – joojaa
    Aug 1 '18 at 15:37

Basically it works like a lever with the force or weight arm length as AC and resisting arm which is connected to the actuator as BC.

The ratio of mechanical advantage remains constant as the jack lifts the load because the ratio of the horizontal projection of the arms which is proportional to their respective lengths multiplied by the cosine of angle Alpha is the same.

Hence the demand on actuator is $ \space =\frac {( F\times AC)}{BC}$

By roughly scaling your sketch the force on actuator is $ = 100\times 9.8 \times 5/2 = 2450 N $

This is for the two actuators, thus we dived by 2 and get 1225 N per actuator. Now we have to multiply this by the safety factor.


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