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Net torque Problem in gear system

  1. Apr 11, 2010 #1
    If I have a gear system that consist of 2 gears that are meshed together
    the first has 16 teeth which is the driver and the 2nd has 36 teeths which is the follower,
    power transmitted by the driver equal to 5 kw and the driver is rotating at 30 rps and has a known diameter, the usual way to find the tangential force that is acting on both gears is :

    Ft=Power/Vdriver , Vdriver = 3.14xDiameter of the driver x 30 rps.

    what confuse me is that the driver as well as the follower are rotating at cosntant velocity but at the same time they have a net torque on each one :
    Net torque on the driver = FtxRadius of driver
    Net torqe on the follower =FtxRadius of follower
    Though we know that when a disk is rotating at constant speed the net torque exerted should be equal to 0 , but it is not the case in Gear systems , please explain .

    p.s : is there any other forces that should be taken into consideration in the following gears that if multiplied by their radius will cancel the torque and the net torque of each one of the gears will become "0" ????

    Thank you
  2. jcsd
  3. Apr 11, 2010 #2
    It dosent make sense !?!
  4. Apr 11, 2010 #3


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    Welcome to PF!

    Hi valkyrie016! Welcome to PF! :smile:
    But what is the rest of the question? :confused:

    What is the follower connected to (gears are usually connected to something, that's the whole point of a mechanical system)?

    Whatever the follower is connected to is exerting a torque on the follower, which in turn is exerting the same torque on the driver, which is balancing the applied torque. :wink:
  5. Apr 11, 2010 #4
    but if the follower is connected to a load lets say a fan , the load is no more connected to anything else , so only 1 torque will apply to it coming from the tangential force between the follower and the load , so again we are lost , the load would rotate at constant speed or accelerating ????
    if 1 torque is being applied to the load it should accelerate and not rotate at constant speed because "sum of torques = I . alpha" but in reality it is rotating at constant speed . there is something wrong we are unable to understand.
  6. Apr 11, 2010 #5


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    yes it is … it's connected to the air, which is resisting its rotation with a torque, equal to the torque applied.
  7. Apr 11, 2010 #6
    what if we put our system in a vaccum enviroment ?? ,I really respect your help , but I am not convinced how the air will affect the load by a torque that is equal to the torque affecting the driver which is in this case the fan?!
    and again what if we put it in a vaccum enviroment "like in space " what will happen thn.?
  8. Apr 11, 2010 #7
    Please somebdy help us in that matter :uhh:
  9. Apr 11, 2010 #8


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    If the system was in a vacuum, then the angular velocity of each component would increase if there is a torque input with no equal and opposing torque.
  10. Apr 11, 2010 #9

    jack action

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    The forces will be in equilibrium. The tangential force is the same (but opposite) on both gear teeth. Each gear has a reaction force at its center equal to the tangential force. What determines the torque applied on each gear is this couple times the distance between them (hence the radius of the gear). Since the radius of the gears are different, the torque will be different for each gear.

    The same thing applies to rotational speed. The tangential velocity is the same on both gears. But the rotational speed [tex]\omega[/tex] = Vt / R, so it is also different for each gear.

    What is the same throughout the system, it is the energy transmitted or, since everything happens during the same period of time, the power transmitted.

    Like you said, P = Ft*Vt, but it will also be the same for both gears, i.e. P = Td*[tex]\omega_d[/tex] = Tf*[tex]\omega_f[/tex] = Ft*Vt.

    This power will either serve to do some mechanical work to another system which the follower is connected to or to accelerate the gears' rotational speed or both.
  11. Apr 12, 2010 #10
    If we take each gear alone to analyse , lets take the driver for instance, the forces that act on it are : the tangential force that is common between the driver and the follower , the reaction of this force is not necessary acting on the driver it must be acting on of the bearings that support the ax of the drive , anyway the reaction force is not what we need to know , what we need to know is how the net torque on the driver is equal to zero in order to rotate at constant velocity , if only one force is acting on the driver whch is Ft, if we apply the net torque rule it is equal to "sum of Torques = F1xR1 + f2XR2 ETC..

    all we have here is just one froce not a couple acting on the driver which is Ft so the net torque = FtxRadius of the driver which in its turn is equal to IxAlpha , so the driver must be accelerating and not rotating at cosntant speed , but it is nt the case in gear systems , cause driver and follower do rotate at constant speed , so there is a missing force on the driver that balance the tangential force which is common between the driver and the follower to cancel the torque and to reach a net torque = 0 , what is this force and wherer it is coming ??
  12. Apr 12, 2010 #11

    jack action

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    The reaction force is ALWAYS acting on the bearing.

    From http://en.wikipedia.org/wiki/Couple_(mechanics)" [Broken]: A Couple is a system of forces with a resultant (a.k.a. net, or sum) moment but no resultant force. The resultant moment of a couple is called a torque. The simplest kind of couple consists of two equal and opposite forces whose lines of action do not coincide.

    If we have a torque, we necessarily have a couple. Force #1 is the one acting on the gear tooth, force #2 is the one equal and opposite reacting at the bearing and both are separated by the radius of the gear.

    Like I said:

    If the follower is connected to nothing and the rotational speed is constant, then all the power serves to fight the friction forces of the bearings. This energy loss will be converted into heat.

    The reacting torque is then the friction force times the radius of the bearing. Multiply this torque with the rotational speed of the bearing and you have the power loss of that bearing. Add up all the power losses from each bearing and you have the total power loss which will be equal to the power input if the gear system is at constant speed.
    Last edited by a moderator: May 4, 2017
  13. Apr 12, 2010 #12


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    Yup! :biggrin: (ie: I agree entirely with jack action's last post)
  14. Apr 12, 2010 #13
    Thank you :)
  15. Apr 13, 2010 #14
    The air can and does impart a force on the fan blades, which results in a torque. Don't underestimate what air can do. Sure at low speeds it doesn't seem like much, but as you increase the speed, the air puts out exponentially more force.

    Look at what a hurricane or a tornado can do. It can rip out trees and lift roofs....and in some cases cars and houses. Airplanes and helicopters routinely use thousands of horsepower to drive nothing more than fans. Some use tens of thousands to do nothing more than fight against the air.
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