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Gear help

  1. Jan 7, 2010 #1
    I have a picture of two gears, but I need some help with mechanical advantage. Lets say the larger gear is providing the work, and if there was no load supplied to the smaller gear, the large gear be spinning at 1,000 rpm, which would make the small gear spin at 3,000 rpm since the gear ratio is 1:3. Now lets say we add a load of 50lbs resistance to the small gear. Since the small gear is spinning 3 times as fast do we multiply that by 3? To 150lbs. Is that enough data to calculate what would happen to the rpm of the large gear?

    Even though the small gear is spinning faster, there is 3 times the torque on it because of the gear ratio. So would that cancel out the excess torque and equal increase in speed without the large gear slowing down?..confusing. I know the system would obviously slow down as soon as you add the 50lb load, but the added speed and torque relationship is whats confusing.

    Attached Files:

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    Last edited: Jan 7, 2010
  2. jcsd
  3. Jan 7, 2010 #2


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    In any system forces and moments are conserved. This means that the torque applied by the gear to the pinion is the same. Force however, will be multiplied because it's being applied at a smaller distance.
  4. Jan 7, 2010 #3
    Ok so the force of the pinion would be multiplied, I am guessing by a factor of 3 in this scenario. But it would take more force to move the pinon because it has the move the load 3 times as fast...right? So what happens?
  5. Jan 7, 2010 #4
    Huh? Mass and energy are conserved, not forces or moments.
  6. Jan 7, 2010 #5


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    No. You are directly applying load to the small gear, so whatever the load you apply is the load.

    What do you mean? What is driving the large gear? The additional load to the pinion means that the driver of the larger gear (usually called the bull gear) has to also pick up the reflected load or else the speed will decrease. It all depends on what is driving the large gear.

    Think of it this way...in terms of power. You have a torque and a speed. If one goes up, the other has to go down to maintain the same power (forgetting about some losses). So if the speed of the pinion is 3X the bull gear, that means it has to have less torque at the pinion to maintain the constant input power to the bull gear.

    So you have:
    [tex]\tau * \omega = P[/tex]

    [tex]F * d * \omega = P[/tex]
  7. Jan 8, 2010 #6


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    Conserved may not have been the word I was going for. Since there is no acceleration, forces and moments sum to zero.

    The important thing in introductory gearing is to draw a nice free body diagram. Once you understand that gears transmit torque equally from one to another, then determining forces on the gears and the shafts become apparent.
  8. Jan 8, 2010 #7
    I just felt like giving you a hard time. :approve:

    Thats what she said
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