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Torque needed to start rolling motion on a inclined surface

  1. Mar 29, 2017 #1

    I need to choose a hydraulic hub motor for a vehicle which weighs 25 000 N when fully loaded. The vehicle must be able to start moving (fully loaded) on a inclined surface (30 degrees angle). So I need to determine the torque needed to get vehicle moving. The vehicle has three wheels from which only one (diam. 450 mm) will be driving (ohter 2 are driven). This has to be an approximate calculation so I have only taken into account the weight of the vehicle and the force needed to overcome rolling friction. The maximum speed to be achieved is 4 km/h but as a first step I just need to know how much torque is needed to get the thing moving at all.
    Is my solution valid or am I doing it wrong, because i get a pretty big torque value? Any help and opinion is welcome !
  2. jcsd
  3. Mar 29, 2017 #2


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    Staff: Mentor

    Welcome to the PF.

    Your calculation of F1 looks correct. But what do you mean by "rolling friction"? Unless your drive wheel slips, the friction between the tire and the ramp surface doesn't matter. You need to know the "rolling resistance" instead to know how much extra torque it takes to get the vehicle moving. The rolling resistance will consist of the axle friction(s) and whatever resistance you get from the deformation of the tires.

    Does that make sense?
  4. Mar 30, 2017 #3
    Hello berkeman,

    Thank you for your answer!
    By rolling friction I meant indeed the resistance which is caused by the deformation of the tires (my physics vocabulary in english is poor :) ). The formula I used, is from a mechanical engineering handbook. And I would think that the friction in the bearings is rather small (compared to other factors) or is it not?
    My gut feeling tells me that the most of the torque is needed due to weight (F1) but I may be wrong on that one.

    And the main question is still, how to get the needed torque value? Did I got this one right?
  5. Mar 30, 2017 #4
  6. Mar 30, 2017 #5

    Thanks for answering!
    The motor you are refering, has a max torque of 4650 in-lbs. Is in-lbs the same as in-lbf?
    Because if it is, then it`s equal to approx. 500 Nm, which is about 6 times less than needed according to my calculation..
  7. Mar 30, 2017 #6
    Sorry I thought that was ft lbs
  8. Mar 30, 2017 #7
    No problem and thank you anyway.
  9. Mar 30, 2017 #8
    It looks like putting all the torque in one wheel is to much the biggest I can find is almost enough if you use all three wheels
  10. Mar 31, 2017 #9
    Your calculation is correct, the value you got is large because of the weight of the car. 25000N is equivalent to 2,5 tons, that means your car is very heavy and even if there is no rolling resistance, your car must have to overcome about 2600Nm due to incline. Major part of the torque arises from the weight.
  11. Mar 31, 2017 #10


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    Science Advisor
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    Gold Member

    Will you gave enough grip? Just a ball park but google said the coefficients of friction for rubber tyre on asphalt road is about 0.7 (dry) or 0.4 (wet). So the value of the friction force Fr is in the range..

    25000 Cos (30) * 0.4 < Fr < 25000 Cos (30) * 0.7

    8,660N < Fr < 15,155N

    The force required to go uphill you calculated at 12,500N so it's just ok in the dry but the wheels will spin in the wet.
  12. Apr 1, 2017 #11
    Hello CWatters,

    Indeed, if we take real-life conditions into account, then things get more complicated. But with this calculation I just wanted to get the ballpark torque value (in almost ideal conditions).
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