Calculating force: pulley and gear on the same axle / shaft

[VeronicaT]

Dear Forum,

Please see the attached file below.
I have trouble understanding what happens at the light green gear which drives the big brown one.
I read somewhere that the force that the belt applies on the surfaces of the two pulley's is the same, so 353, 68 Nm. The torque and rpm on both pulley's is different, as is the torque.

Is it correct to say that the torque on de small gear is the same as on the big pulley? Because they are both on the same axle?
Do I calculate the force that the small gear applies on the big one like this? 53,05 Nm / 0, 037m?

Thanks for your time!

 

[Merlin3189]

Hello and welcome to PF. I hope you find it interesting and helpful here.

I'm happy with your pulleys and the values of speed and torque. (I'll leave aside the tension in the belt for the moment.)

When you get to the small gear driving the big ring, I think I agree with what you write.
The small gear torque is the same as the big pulley torque, 53.05 Nm.

The forces are equal at the teeth of the two gears. So your method of dividing torque by radius to get the force is right. F=53.05 / 0.037m
Then the torque on the big gear is the force x the radius.

You could also have simply used the gear ratio, if you only want to know the final torque and not the force:
Torque Out = Torque In x Mechanical Advantage = Torque In x Output teeth / Input teeth (assuming perfect efficiency - which, of course, will not be true.)

=======================================
Tension in the belt:

I read somewhere that the force that the belt applies on the surfaces of the two pulley's is the same, so 353, 68 Nm. The torque and rpm on both pulley's is different, as is the torque.

I'm not sure what you are saying here.
Your 353.68 Nm is shown as 353.68 N on the diagram, a force rather than a torque.
I guess the figure 353.68 comes from the calculated force applied by the belt to the pulleys, which was 176.83 N. Then you've doubled that force.
The 176 N is the difference in tension between the two sides of the belt as it wraps around the pulley. If one side of the belt were slack, then the tension in the other side would be 176 N, but then the belt might slip a bit. So usually there will be a little tension in both sides - say 40N in the 'slack' side and 216 N in the tight side. The difference is the force applied to turning the pulley. (The 40N is just a wild guess, as its calculation is not too simple.)

Now those two tensions are the same on both pulleys, just trying to turn them in opposite directions. The total force on each pulley due to the belt will be the vector sum of the two tensions. The sides of the belt are not parallel. The angles of the belt depend on the radii of the pulleys and their separation.

 

[CWatters]

Is it correct to say that the torque on de small gear is the same as on the big pulley? Because they are both on the same axle?

Yes.

The motor produces 12.73Nm so the torque available at the big pulley is

12.73 * 0.3/0.072 = 53.05Nm

That's transmitted along the shaft to the small gear. I have no idea why it says 13.09 Nm next to the small gear.

Do I calculate the force that the small gear applies on the big one like this? 53,05 Nm / 0, 037m?

Do you want the force or the torque? I assume you mean torque...

The ring gear has 120 teeth, the small gear has 12 teeth so the torque available to turn the ring gear is 53.05 * 120/12 = 530Nm

or you can do it this way...

The ring gear has a radius of 0.374m, the small gear has a radius of 0.037m so the torque available to turn the ring gear is 53.05 * 0.374/0.037 = 530Nm

That works because the pitch of the teeth must be the same on both gears.

Cross posted with Merlin.

 

[VeronicaT]

Dear Forum,
Thank you for your constructive replies! With your help I have come to the following calculations. I have written it all out again.
I attached letters to each step for clarity: A=small pulley, B=big pulley, C=small gear, D=big gear and finally E=knife.
My final aim is to buy a second hand concrete mixer and be able to select the right one.
The more powerful the more expensive and maybe I can do with a small one. For that I need to take the shear strength of the material into account, but that is the next step :-)

A: torque electrical motor
torque(Nm) = power(N) x 9.549 /rpm
torque = 9.5488 x 1600(W) (delivered by electrical motor) / 1200 (rpm) = 12.7317 Nm

B: rpm
Pulley-B is 4,1666667 times bigger as pulley-A. The bigger the pulley, the slower it turns.
(Diameter B) 0.30/(diameter A) 0.072= 4.1666667
1200 tpm / 4.1666667 = 288 tpm

B: power and torque
torque(Nm) = power(N) * radius(m)
power = torque/radius

given: torque (12.73 Nm) and radius (0.036m) → power =353.6583 N
Power is constant through the belted drive, the torque and speed change to keep this constant.

Torque driven pulley B: 353.66 N x 0.15 m =53.04875 Nm

C: torque
torque-C = torque pulley-B while on the same axle: 53,05 Nm

C: power on gear/pulleys on same axle
Rpm pulley-B and gear-C are the same: 53.05 Nm
Power gear-C is 53.05Nm/0.037m= 1433.75 N

D power
Power on teeth of both gears is the same: 1433,75 N
Also: ratio pulley-B and gear-C is 1 : 4.054054 (4,054054 * 353,6583 N (pulley-B) = 1433,75 N gear-C)

D torque
torque(Nm) is power(N) x radius(m)
torque is 1433,75 N * 0.374 m = 536,2225 Nm

E torque
Torque knife-E = torque gear-D while on the same axle: 536,22 Nm

E: power
power = torque/radius
power is 536,22 Nm / 0.06m= 8937N

 

[CWatters]

A: torque electrical motor
torque(Nm) = power(N) x 9.549 /rpm
torque = 9.5488 x 1600(W) (delivered by electrical motor) / 1200 (rpm) = 12.7317 Nm

B: rpm
Pulley-B is 4,1666667 times bigger as pulley-A. The bigger the pulley, the slower it turns.
(Diameter B) 0.30/(diameter A) 0.072= 4.1666667
1200 tpm / 4.1666667 = 288 tpm

OK so far

B: power and torque

torque(Nm) = power(N) * radius(m)
power = torque/radius
given: torque (12.73 Nm) and radius (0.036m) → power =353.6583 N
Power is constant through the belted drive, the torque and speed change to keep this constant.

The bit in red is wrong. Power is measured in Watts (W). You mean Force (N). The correct equation is..
Torque(Nm) = Force(N) * radius(m)
However you don't really need to think about forces to analyse this system, see below.

Torque driven pulley B: 353.66 N x 0.15 m =53.04875 Nm

You can get the torque at B by multiplying the torque at A by the belt ratio eg..
12.73Nm * 4.166 = 53.04Nm

C: torque
torque-C = torque pulley-B while on the same axle: 53,05 Nm

Correct.

C: power on gear/pulleys on same axle

Rpm pulley-B and gear-C are the same: 53.05 Nm
Power gear-C is 53.05Nm/0.037m= 1433.75 N

Again you mean Force not Power.

D power
Power on teeth of both gears is the same: 1433,75 N
Also: ratio pulley-B and gear-C is 1 : 4.054054 (4,054054 * 353,6583 N (pulley-B) = 1433,75 N gear-C)

Again you mean Force not Power.

D torque

torque(Nm) is power(N) x radius(m)
torque is 1433,75 N * 0.374 m = 536,2225 Nm

Again you don't need to think about the force. The torque at D is the torque at C multiplied by the gear ratio which is 120/12
Torque D = 53.04 * 120/12 = 530.4N.

E torque
Torque knife-E = torque gear-D while on the same axle: 536,22 Nm

Correct reasoning (but it should be 530Nm)

E: power
power = torque/radius
power is 536,22 Nm / 0.06m= 8937N

Again you mean force not power.

If you want to know the power in Watts at E then that's easy to calculate. You don't mention any losses in any of the stages so the power at E is the same as the power at the motor A. In practice there will be some losses in each stage.

 

[VeronicaT]

Dear CWatters,
Thank you very much for your reply. OK, force, not power, I will not forget that now. I meant force. My native language is Dutch and the use of everyday words in physics sometimes confuses me. Besides, the highest grade I ever got for physics was a C+...
To take the gear ratio into account while calculating torque as you suggest, is easier and also more accurate in my case. I copied the outer and inner measurements of gear D from a spare part advertisement and didn't know were the radius should start: from the inside, outside or from the middle of the teeth? I chose the inside, which was wrong presumably, because the outcome is wrong.
Maybe interesting to hear: I also posted my question on a Dutch (physics) website where I was told gear E exerts the same force (in Nm) as pulley-A and pulley-B because "it adopts all energy of the motor". From your answer I take it he meant power...

 

[CWatters]

Dear CWatters,
Thank you very much for your reply. OK, force, not power, I will not forget that now. I meant force. My native language is Dutch and the use of everyday words in physics sometimes confuses me. Besides, the highest grade I ever got for physics was a C+...

No worries. I lived in Belgium for a few years and your English is way better than my Flemish.

To take the gear ratio into account while calculating torque as you suggest, is easier and also more accurate in my case. I copied the outer and inner measurements of gear D from a spare part advertisement and didn't know were the radius should start: from the inside, outside or from the middle of the teeth? I chose the inside, which was wrong presumably, because the outcome is wrong.

In such cases it's easier to use the ratio of the number of teeth on each gear.

Maybe interesting to hear: I also posted my question on a Dutch (physics) website where I was told gear E exerts the same force (in Nm) as pulley-A and pulley-B because "it adopts all energy of the motor". From your answer I take it he meant power...

Yes he must mean power. This comes from the law of conservation of energy. If there are no losses in the gears the power output must be the same as the power input.