B Explaining forces acting on gears

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In a simple gear system, when force is applied to Cog A, it transmits that force to Cog B, which pushes back due to Newton's Third Law. This interaction means that to rotate Cog A, a greater force must be applied than the reaction force from Cog B, especially when starting from a stationary position. Introducing a third cog complicates the system, potentially increasing the required force to rotate Cog A. The forces at the contact points between the gears are indeed Third Law pairs, but the driving force on the input shaft is distinct from these contact forces. Understanding the dynamics of these forces is crucial for analyzing gear systems effectively.
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Hi all
In respect to a simple gear system, two cogs are linked together, Cog A and Cog B. You apply a force to cog A, and cog A transmits this force to cog B through contact of the teeth on each cog (third law pair?)

If cog A pushes on cog B, does cog B not push back on cog A? Does this not therefore make it harder (if I am applying a force) to rotate cog A. Is it better to say that that resultant force on cog A is transmitted to Cog B?

If a third cog is introduced, does this also make it subsequently harder to rotate Cog A?
Finally how can the alternating rotation of the cogs be explained in terms of forces (correct me if I am wrong but it is not due to the third law pair that was mentioned earlier?)
Additionaly how does the attached image lend to an explanation that links to the above?
Thanks
 

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If any two bodies are in contact (gears or other bodies), if the first pushes on the second, the second must necessarily push back on the first. This is Newton's 3rd Law.

The picture you show (from a textbook?) Is poorly drawn, both in regard to the location and direction of the contact forces. The forces are always normal to the surfaces at the point of contact. If the gears are involutes (the common form), the line of action of the forces will alway act through a fixed location called the "pitch point." Your figure fails to show the forces as colinear as they must in fact be.

Do an internet search on "involute gear teeth" and I'm sure you will find a better picture that will explain what happens.
 
Thanks, so if I am rotating cog A must I apply a larger force to cog A than the reaction force from cog B in order to accelerate the system from a stationary position?
 
If not how does the force that is labelled in the diagram on the input cause the rotation that is shown on the input?
 
http://www.engineeringexpert.net/Engineering-Expert-Witness-Blog/tag/spur-gear

This site talks about opposing forces f1 and f2 and says " To get a stationary locomotive to move, mechanical energy must be transmitted from the driving gear that’s attached to its traction motor, then on to the driven gear attached to its axle. At their point of contact, the driving force of the motor, F1, is met by the resisting force of inertia, F2. In order for the train to move, F1 must be greater than F2. If F1 is less than or equal to F2, then the train won’t leave the station. "

This must mean F1 and F2 are not Newton 3rd law pair forces, I'm mm head they are the better version of the textbook image I attached and so should be 3rd law pairs! Help!
 
Mr Balmond said:
the resisting force of inertia
There is no such thing (or at least certainly not as a 3rd law partner force). Toss that explanation straight to the trash heap.
 
Thank you Briggs, can you provide a clear link between the forces acting in a simple gear system? Dr D says that the two forces are third law pairs, which I had originally taken to be the case when I asked the question and although that website says incorrectly "resisting force or inertia" it does show two forces acting at the point of contact between the gears and the model shows the forces being equal in size and inferring that they acting on different bodies i.e. each gear?
My original question can be summed up as: is the driving force on the input shaft a completely different force to the forces acting in the contact between the two gears?
 
Mr Balmond said:
Thanks, so if I am rotating cog A must I apply a larger force to cog A than the reaction force from cog B in order to accelerate the system from a stationary position?

Another way to write that would be to say that in order to accelerate a gear the net torque on that gear must not be zero. Perhaps think about it from the perspective of drawing a free body diagram for that gear.
 
Mr Balmond said:
My original question can be summed up as: is the driving force on the input shaft a completely different force to the forces acting in the contact between the two gears?

Yes. But note that the diagrams show only one force acting on the driving gear at the contact point between the gears. The other force shown in the diagrams is its Third-Law partner, and it acts on the other gear.
 
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