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Force required to move a 30000lb cart

  1. Jan 30, 2009 #1
    I have a 30000 lb cart with 8" steel wheels on a steel surface, that I need to pull and push out of a large oven. How would you go about determining the force required to pull the cart out of the oven?
     
  2. jcsd
  3. Jan 30, 2009 #2

    mgb_phys

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    In reality - try it!
    The wheels are rolling, so assuming that they are fairly smooth and free from bits of whatever is in the oven you have more resistance from the wheel bearings.
    Modelling the rolling resistance inside the bearings is tricky.

    Are they running on rails? You can (slowly) pull a 25t empty railway car with one (strong) person.

    And finally there is how fast do you need to go, in theory (ignoring friction) any tiny force will produce a very slow movement.
     
  4. Jan 30, 2009 #3
    I am trying to determine the force so that I can put together a motor and gear train to pull it. Currently I am using a forklift to tow it.
     
  5. Jan 30, 2009 #4

    stewartcs

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    Hook a load cell up to the cable you're towing it with.

    CS
     
  6. Jan 30, 2009 #5
    Cant say that I have one. Are there cheap ones available? I guess there are two forces to look at? 1.The torque required to overcome friction and start the wheels rolling any suggestions on this calculation? The 2nd would be the force required to accelerate from rest to the constant velocity speed it will be pulled at (F= MA)?
     
  7. Jan 30, 2009 #6

    mgb_phys

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    It's all a matter of the friction, unfortunately it's difficult to calcualte for a real world situation where it's dominated by a bit of rust on the track or a sticky bearing.

    Assuming you only need the thing to move slowly the F=ma becomes pretty negligible.
    Say you wanted to move it 5m in 30seconds and it's stopped by just hitting a buffer (so you don't need any force to slow it down)
    The speed equation is distance = 1/2 acceleration * time^2 and F = ma

    So force = mass * 2 * distance/time^2 = 6800kg * 2 *5 /30^30 = 75N or 15 lbf
     
  8. Jan 30, 2009 #7
    Ok, so overcoming friction is going to be the max required force. What equation would you use to determine this? Neglecting bearing friction for the time being and only dealing with the friction between the wheels and the ground what equation would you use to determine the force required to get the cart rolling?
     
  9. Jan 30, 2009 #8

    mgb_phys

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    The equation is just force = coeff_friction * weight
    The problem is guessing the coeff, for steel wheels on a rail it could be anything from 0.0001 to 0.005
     
  10. Jan 30, 2009 #9
    That is the equation that I was using, but i was using the static coeff for steel on steel of .74. What are the coeff that you listed above?
     
  11. Jan 30, 2009 #10

    mgb_phys

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    The rolling resistance, 0.74 sounds like the sliding friction.

    Sorry to be obtuse on this but static friction metal-metal is tricky. The metal actualy bonds together so you are breaking microscopic welds - in very clean applications with very high precision parts it's a real problem, if you put two peices of metal together you will never get them apart.
     
  12. Jan 30, 2009 #11
    I think you have enlightened me. So if the four wheels were locked up and sliding on the steel deck I would use the coef of sliding friction .74. But since they ideally wont be sliding I would use the rolling resistance.
     
  13. Jan 30, 2009 #12

    mgb_phys

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  14. Feb 4, 2009 #13
    Would the value calculated for rolling friction be the force to keep it moving at a constant velocity? What about the force required to start the wheels rolling.

    Is there a higher friction value to use from the stationary position? Similar to static and kinetic sliding friction?
     
  15. Feb 4, 2009 #14

    minger

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    Typically yes, these are called static and kinetic friction coefficients. It's just like anything really. Try to move a couch across your floor. Once you get it moving you don't want to let it stop because its hardest to "get it going".

    When looking at a force applied vs friction plot, (see http://www.roymech.co.uk/images11/friction.gif) the friction force will increase linearly with the applied force until the object "breaks loose". At this point, the frictional force will be constant regardless of force applied. This frictional force is then used to compute the kinetic coefficient of friction.

    The peak, or point 'just' before the object breaks loose is the highest friction force. It's used for the static coefficient. Typically the values can vary quite a bit (think back to your couch).
     
  16. Feb 4, 2009 #15
    Can any one point me to so tabulated friction coefficients for rolling resistance both when moving and from a stopped position? I'd like to see the varience between the two.
     
  17. Feb 26, 2009 #16
    This is a pretty tricky calculation. If the load is low enough - you can substitute a hanging scale for the load cell (Google). You might also look for a compact mover/pusher off the shelf like the one shown on PowerHandling.com. Hope that helps.
     
  18. Apr 20, 2009 #17
    F = M A!!! i.e. force = mass x acceleration. I recommend converting to metric but thats probably because I'm a Canadian :) Anyways in theory any amount of force would get it out of the oven even if the force was quite small but in real life you have friction so you need to take that into account. I'd say for a crude calculation, determine the acceleration you want and plug it into the equation then find the force required for that acceleration then test it on the object but measure the actual acceleration and either figure that its close enough (which it prob wont be) or allow that number to allow you to calculate the friction in the system then take that into consideration and make another more accurate calculation :D... just makes sense to me but I'm not an engineer (yet) thou he never said anything about it being super accurate... i mean... move something big in and out of something else gives you a lot of room to play with lol.
     
  19. Apr 21, 2009 #18

    minger

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    Unfortunately that coefficient of friction is far too important here; there are huge masses involved. Think about the difference in force needed if your coefficient of friction is off only 10%.
     
  20. Apr 21, 2009 #19
    For this case, you need to be very sure the surface is level; if the gradient is 0.01 foot per foot for example, you will need an extra 300 pounds just to push the cart uphill.
     
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