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Figuring out rotational force?

  1. Mar 27, 2006 #1
    I have a winch to build. This winch has 5 drums spaced equally over a 40' length. They are connected by a common shaft of 2" solid bar. In the center of the shaft/drum assembly is a gear motor turning the shaft.

    Attached to the wire rope coming down from the 8" dia. grooved drums is a 40' aluminum I-beam. This ibeam will be supporting trolley that will roll along the ibeam.

    The weight to be lifted is 3500lbs (after a 5:1 safety factor is included) and needs to travel 17' in about 4 seconds.

    My question is how to figure out the horsepower needed to lift this much weight, assuming a gear reducer of 30:1 after the motor. Motor is turning max rpm of 3500. Also need to figure the amount of torque needed to hold the load in the air safely.

    Is it safe to assume all 5 drums will be doing the same amount of work, regardless of the placement of the rolling trolley? (I would think not.)

    Is it also safe to assume that the amount of torque needed to LIFT the weight is the same as the amount of torque needed to sustain the load in a static state? (Hope I said that right.)

    I guess what I'm looking for are the formulas to figure these out, not neccassarily the answers to the questions.
  2. jcsd
  3. Mar 28, 2006 #2


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    A few questions to clear up some possible misinterpretations:

    Is the Ibeam vertical?

    I assume there is one cable that winds from one drum to the other and finally connects to the trolley?

    Is the cable guided by a pulley (system) such that it always aligns with the beam when winding or unwiding - that is the cable always pull inline with the Ibeam?

    Does your safety factor mean that the system will actually have to lift about 700 lbs?
  4. Mar 28, 2006 #3
    Thanks so much for the reply and hopefully I can explain this better.

    This Line Winch is essentially what I'm talking about, except with 4 drums and 40' length.


    The beam is on a horizontal plane, as are the drums. One cable comes down from each drum and the drums are equally spaced along the 40 foot of beam. the drums and motor are mounted in the "ceiling" and the I-beam hangs from the cables, wrapping up on the drums to raise the I-beam. So each cable is picking up 1/4 of the weight of the beam. All drums are on a common shaft so they all turn at the same rate. The motor and gear reducer will be mounted in the center of the common shaft, not at either end.

    The cables are mounted directly above the beam and have no other sheaves involved, just straight down from the cable drums to the beam. There is adequate clearence for the wire to "walk" across the 8" wide grooved drum.

    The trolley is only mentioned in that it will be rolling along the length of the beam, independent of the cables, etc. Like an overhead crane in a factory.

    Yes, the safety factor is in there. The actual weight lifted is only 700lbs.

    Hope that clears up everything. I'm not an engineer so I'm not sure I got all the terms right but I'll be happy to explain anything furthur.
  5. Mar 29, 2006 #4


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    The motor has to do all the work irrespective of the gear reduction which just transmits it. The rate at which it need to do work [itex]P[/itex] in order to lift the 3500 lbs of weight through a vertical distance of 17 feet in 4 seconds comes to
    [tex]P=\frac{\Delta W}{\Delta t}[/tex]
    where [itex]\Delta W[/itex] is the work it has to do in the 4 seconds
    [tex]P=\frac{F\times d}{4}[/tex]
    which comes to
    [tex]P=\frac{3500\times 17}{550\times 4}[/tex]
    horse-power. This reduces to
    [tex]\frac{35\times 17}{12}=27\ hp[/tex]
    does that seems to be a realistic value?

    A source of great confusion for us who are used to the metric system is whether a pounds quantity needs to be multiplied by [itex]32\ ft/s^2[/itex] in order to get the actual weight (gravitational attractive force on) of the object. Your value seems to be a pounds-force , [itex]lbf[/itex], value and not pounds weight (mass), [itex]lb[/itex], quantity?
    Last edited: Mar 29, 2006
  6. Mar 30, 2006 #5


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    The static torque caused by the weight on the cable (when the weight is not being hoisted) will be
    [tex]\tau\ =r\times \ W[/tex]
    where [itex]W[/itex] is the total weight being supported by all the cables and [itex]r[/itex] is the radius of the drum - that is the force is applied tangentially to the drum at this distance from the centre of the shaft.
    [tex]\tau\ = \frac{3500}{3}=\ 1170\ lbf.ft[/tex]
    This torque will be reduced by a factor [itex]30[/itex] on the motor's side to approx [itex]40\ lbf.ft[/itex] since the pinion is on its side.
    The dynamic torque needs a bit more work, but it is doable.
    Last edited: Mar 30, 2006
  7. Mar 31, 2006 #6


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    The dynamic torque will only be different from the static value during the initial phase of lifting (accelerating) the weight. One can make an estimate of this if more is known about the properties of he cable.
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