Linear force, acting on a motor?

[PhiowPhi]

I had a thought that's been puzzling me this whole morning, its more related to classical mechanics, however I am dealing with motors so... I thought it would be appropriate to post the question here.

Lets say we have to identical motors(with identical gears) placed vertically on a table. Those motors are free to move, they aren't fixed. We placed the gears together so that when rotating the first gear the second will rotate etc... and now, we turned on the motors. Since their torques are in opposition and equal in magnitude they cancel out, the gears will be at rest. Now I can't imagine what is the result to the motors themselves, will they fall? Will they just stay there at rest?

Is there any form of linear force acting on the motors(Aside from gravity)?
Since the motor is placed on the $y$-axis and gravity acts on that same axis, could there be a linear force acting on the motors on the $x$ ,$z$ axis?

Remember, the motors are free to move they are not fixed in any way to the table, and they are vertically placed.

 

[billy_joule]

The motors will fall. I don't think a gear profile can exists that does not have a radial component to the force. The most common profile, the involute gear (developed by Euler) has a line of action of some 20 degrees from tangential if I recall correctly.

 

[donpacino]

Assuming the motors won't fall over, and no friction on the tables, and the gears hold togther, and I am understanding you correctly, the rotor (connected to the gear) will stay still and the stator will spin like a top

 

[PhiowPhi]

The motors will fall. I don't think a gear profile can exists that does not have a radial component to the force. The most common profile, the involute gear (developed by Euler) has a line of action of some 20 degrees from tangential if I recall correctly.

Why would they fail? Looking at the radial force and the torques they should cancel out?

Assuming the motors won't fall over, and no friction on the tables, and the gears hold togther, and I am understanding you correctly, the rotor (connected to the gear) will stay still and the stator will spin like a top

Why would they fall in general? Or why might we consider that?
It makes sense to me that the stator would spin. However, is there any form of forward/backward force ($z$-axis) or right/left one($x$-axis) since the motor is vertical on the $y$-axis

 

[billy_joule]

Why would they fail? Looking at the radial force and the torques they should cancel out?

Draw a free body diagram. The radial forces do not cancel. This is sometimes descriptively called the 'separating force', or more often the radial load.
I took your statement: "Those motors are free to move, they aren't fixed." to mean the table has some friction. If the table is truly frictionless they will not fall, just move apart from each other, spinning on their axis as they do.

 

[PhiowPhi]

I'm confused to my understanding of what radial force is and need to review that.
However, any form of force that are coming from the motors are all rotational somewhat, there is no force pushing it forward/backward (z-axis) or a force moving it left/right (x-axis).

I'm trying to simulate this using a bottle of water, trying to apply torque and rotating it there shouldn't be any linear force forward/backward or left/right unless i apply one. The same thing comes to my mind when thinking about the motors.

 

[billy_joule]

The contact force (blue line) has both radial (passing through axis) and tangential components (perpendicular to radial)

http://en.wikipedia.org/wiki/Involute_gear

 

[PhiowPhi]

Im having trouble imagining another form of force besides torque acting on the motor(if the table we're frictionless) when the two gears cancel out their torques.
And this radial force is new to me, which I've been studying last night and still is puzzling.

Here is my understanding and please help me here @billy_joule & @donpacino, when the two motors have their rotor's torques cancel out, and they are placed to the table what other force is acting on the motor, and what is it equal to?

Lets assume instead of a table we placed the motors on separate moveable carts/discs any form of moveable object, (say two discs) will there be a force acting on the motors that will force the disc to rotate in anyway? Or is there a force due to the motor's that will move the cart forward/backward or left/right?

Finally, I still don't understand why the two motors would fall...? The contact force that I'm imagining here is what keeping the gears locked in at rest.

 

[billy_joule]

Im having trouble imagining another form of force besides torque acting on the motor(if the table we're frictionless) when the two gears cancel out their torques.
And this radial force is new to me, which I've been studying last night and still is puzzling.

Here is my understanding and please help me here @billy_joule & @donpacino, when the two motors have their rotor's torques cancel out, and they are placed to the table what other force is acting on the motor, and what is it equal to?

The radial force (due to geometric constraints in gear profile design).
The torque applied to the motor casing by the gear shaft.

Imagine holding a powered motor, what happens? the shaft will rotate.
Now imagine holding the shaft, now what happens? The motor rotates...
They are two sides of the same coin. The second applies to your situation.

Draw some free body diagrams and insert them in a post. It's the quickest way to see exactly where the problem is.

 

[PhiowPhi]

Yes it makes sense now, thanks @billy_joule and if the motor was also fixed there is no motion on the system at all.
My concern was a linear force acting on the motors, there isn't only torque acting on the motor forcing it to rotate, however no linear force moving it right/left or forward/backwards correct?

 

[PhiowPhi]

I know this a bit old... but if the stator stays still due to the opposing action forces(or torques), is the motor's rotation or the torque that forces it to move opposite to the stator considered the equal and opposite reaction?

Really need your clarification on that confusing point @billy_joule @donpacino.