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I'm designing a lifting device powered by a motor, i would like to know if the forces exerted by the load on pulley 1 and 2 are equal when the motor star running , or if in pulley 1 is exerted more force because the location of the load at the end of the lifting platform. The lifting platform only moves vertically, restricted by rails. What I want to know if the load will exert a perpendicular force that could create friction on the rails

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    $\begingroup$ as drawn there will be a rotation moment simply because the load is on one side and the pulleys on the other. Even if pulley 1 carried all the weight it will still rotate $\endgroup$ – ratchet freak Jul 5 '17 at 14:18
  • $\begingroup$ @ratchetfreak technically "attempt to rotate" given perfect restraining rails. But your point is well taken. $\endgroup$ – Carl Witthoft Jul 5 '17 at 15:31
  • $\begingroup$ voting to close as unclear and non-response to clarification questions. $\endgroup$ – agentp Nov 2 '17 at 17:41
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Of course the load will exert a moment, to study the behaviour of your system you can simply model it as a constrained beam in pure bending if the elasticity of your platform is not negligible or as a constrained rigid body. Every point of the rail is a constrained point, analyzing the strain diagrams you can compute how much perpendicular force you will have.

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  • $\begingroup$ But if the tension along the cord keeps constant once the motor start running, how can it be that forces exerted on pulley 1and 2 be different. $\endgroup$ – Jor Camero Jul 6 '17 at 21:02
  • $\begingroup$ Reactions are equal but different in directions. Between the 2 pulleys you will have 2 opposite bending moments. $\endgroup$ – Matteo Caruso Jul 7 '17 at 15:32
  • $\begingroup$ @JorCamero whoever claimed the pulley forces are different? The question really should be improved by showing in the graphic what keeps the platform from rotating. $\endgroup$ – agentp Sep 5 '17 at 13:53
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The drawing is lacking detail on how to the load and platform will not rotate. Despite this, you can assume accelerations are zero; therefore static. The sum of the forces, and the sum of the moments in the system must equal zero. Then a safety factor of should be computed per ASME B30.

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