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Linear reaction force on motors?

  1. Mar 15, 2015 #1
    I have two motors(m1 &m2) placed on a vertical position(on the ##y-axis##), that have gears connected to their shafts, and I placed the gears next to one another. Fixing them to move together. I now powered both motors allowing their torques to oppose one another and cancel out each other's torque like so:

    ltqFY65.jpg -


    I would like to know the possibility of having a linear force acting on the motors(in both cases of frictional surface and non-frictional surfaces).
    I understand that if the gears of the motors cancel out their torques the reaction force(newtons 3rd law) to it is the stator of the motor moving in opposite direction of the gears.

    However, is there a force that forces the motors to move forward/backward(##z-axis##) or right/left(##x-axis##)?

    As diagrammed here:


    J6sbYx7.jpg
    Using the coordinate system right/left forces(##F_L## & ##F_R##) are on the ##x##-axis and ##F_f## ##F_b## are the forward/back forces via the ##z-axis##, X being backward force(out of the page), while O is the forward force(into the page).

    Also, what if I attached the motors fixed to the ground/table via these cylindrical objects as diagrammed here:

    zaQVnEI.jpg

    I'm confused as to what forms of force(reaction) will there be(aside from the stators rotating)when the gears are fixed after canceling out their forces). I thought about an actuator it was easy to solve, however, never thought about torque and motors and their reaction forces. I posted this earlier in the EE section, but I want to understand the mechanics and theory much more in depth. Will there be any form of linear forces acting on the motors or the objects the motors are attached/placed onto? Or there will only be a torque acting on the stators as a reaction?
     
    Last edited: Mar 15, 2015
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  3. Mar 15, 2015 #2

    OldEngr63

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    There is no reason to expect a net force on the system. There will, however, be a separating force between the two motors due to the gear action. The force between the two motors occurs along the line of contact, which at an angle to the line of centers. For example, for gears made to a nominal 20 deg pressure angle and mounted at their nominal center distance, the force will be at an angle of 20 deg from the normal to the line of centers. Changing the center distance will make a small change, and using a different nominal pressure angle will also change the angle value.
     
  4. Mar 16, 2015 #3
    @OldEngr63 Then I am mistaken to assume any form of linear forces as I diagrammed them above. Only the force of separation I believe is the involute gear?
    A member also explained the separation force yet not something I completely understand yet. What direction will it be? How might one calculate this form of reaction from the action of the gears?
     
  5. Mar 16, 2015 #4

    OldEngr63

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    Involute gear teeth have the form of an involute of a circle. That circle is called the "base circle," and it is not visible (there is no physical manifestation of the base circle). Each gear has a base circle, and the action of the gear pair is exactly like winding a cord off one base circle and onto the other. The path of that cord is called the "line of action." As with a cord, the force transferred from one gear to the other is along the line of action.

    Standard gear teeth today are based on either 20 deg or 25 deg pressure angles, where the pressure angle is the angle between the line of action and the normal to the line of centers. Thus if you know the torque transferred and the nominal pressure angle and assuming the nominal center distance, you know then both the direction of the force between the gears and the magnitude of the force. The separating force is the component of that force that acts parallel to the line of centers.

    I strongly recommend that you draw some pictures to get a handle on this problem. If you are unfamiliar with gear terminology, etc. you might want to try the AGMA web site or a theory of machines textbook.
     
  6. Mar 18, 2015 #5
    I found some great resources to explain such ideas to beginners like me.
    In general, to apply newtons 3rd law to such applications(opposing gears/motors) the action would be the the torque applying from the motors(each one) the reaction would be the separation force? Or the cancellation of motion or... a torque somehow applied to the motors? Bit confused about that point.
     
  7. Mar 18, 2015 #6

    OldEngr63

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    Again, I say, draw some pictures!

    Draw a FBD for each rotor and the associated gears. Show the bearing reactions on the rotor and the interaction forces between the gears. Show also the E/M torque on the rotors. You will get more out of this than all the words we can write.
     
  8. Mar 26, 2015 #7
    I drew multiple FBD and came to the conclusion that there shouldn't be any net-force in the system aside from the separation force as you stated earlier. I worried a lot about the reaction forces, however, the motors would transmit a reaction torque that is equal and opposite to the one it applied as it's action torque. There isn't any form of linear force, and since the motor's stator is fixed to the cylinder objects it's reaction torque does nothing. What do you think @OldEngr63 of this conclusion?
     
  9. Mar 27, 2015 #8

    OldEngr63

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    If I understand correctly your original situation, the two motors are stalled against each other, so that the torque of one is opposing the torque of the other. In this case, because they are geared together (with no idler gear between) the torque on each of the rotors is in the same sense. Thus the E/M reaction torque on the stators is also in the same direction for each of the motors (the direction opposite the torque on the rotors). Thus there will be a net reaction torque on the support common to the two stators.
     
  10. Mar 27, 2015 #9
    @OldEngr63 Yes, and I agree. There is reaction torque on the support that is holding the motor opposite to the torque the motors apply on each other. However, the torque will just force the support to rotate on the same axis that the as the rotor however, opposite to it.

    I figured it out thanks!
     
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