3 gears and the torque on the support

[Cri85]

Homework Statement

I need to find if there is a torque on the support with 3 gears:

http://imageshack.com/a/img901/8954/mFzBas.png [Broken]

The Gear1 is forced to turn at 3w by an external device. Gear2 and Gear3 must follow.

Homework Equations

None

The Attempt at a Solution

I drawn all forces to the support F1, F2, F3, F4 (I don't drawn all forces on gears) :

http://imageshack.com/a/img538/4788/LhpPWd.png [Broken]

The support don't receive a torque only if d1=d2 ?

 

[OldEngr63]

When you ask if there is any torque, are you asking about torque due to shaft reactions on the support, or are you asking if there is a torque transfer from any (or all) of the shafts directly to the support?

If it is the first question, what is the reference point for the torque you are asking about?
If it is the second question, only friction torque transfers directly from shaft to the support, and this is usually very small.

 

[Cri85]

It is for the first question, the reference point is needed for a torque ? If necessary place it in the center of Gear1.

 

[Cri85]

When I said "only if d1=d2" I would like to be sure for d1=d2 there is no torque on the support and it is the unique solution.

 

[Cri85]

I guess the main center of rotation in the center of the Gear1. |F1|=|F2|=|F3|=|F4|=|F|, the torque on the support is +d1*F2+d1*F3-(d1+d2)*F4=2d1F-2d2F, if d1=d2, the torque is 0, is it correct ? The forces I drawn are correct ?

 

[haruspex]

I need to find if there is a torque on the support with 3 gears:

the reference point is needed for a torque ? If necessary place it in the center of Gear1.

That doesn't seem like a reasonable thing to do unless told to do so. What is the exact wording of the question?

 

[Cri85]

Without friction. An external torque (not drawn) is applied to the Gear1. The Gear3 applies a torque on an external device (not drawn). I would like to know if the support receives a torque from 3 gears.

1) My forces on the support are correct (F1, F2, F3, F4) ?
2) If d1=d2, the torque on the support is 0 ?

http://imageshack.com/a/img661/1192/vTEwet.png [Broken]

Note the rapport of reduction is 1/3.3333 not 1/3.

 

[billy_joule]

You free body diagrams are wrong. They are not in static equilibrium, the first and third gears would accelerate upwards.

I agree with haruspex, I think you have misinterpreted the question. Post the exact wording and we may be able to help.

 

[Cri85]

Sorry, each gear turn around itself because there is an axis of rotation:

http://imageshack.com/a/img908/7112/EncuiV.png [Broken]

 

[billy_joule]

Post the exact wording and we may be able to help.

 

[Cri85]

"Find a mechanical device that can reduce the rotationnal velocity by a factor K. The device shall not give a torque on the support."

I find this device but I'm not sure.

 

[haruspex]

"Find a mechanical device that can reduce the rotationnal velocity by a factor K. The device shall not give a torque on the support."

Then the torque should be assessed with the point of support as the axis.

 

[Cri85]

With:

http://imageshack.com/a/img673/6916/bDyNTC.png [Broken]

The torque on the suppport is -dF+2(d+d1)F-(d+d1+d2)F, if d1=d2, then -dF+2d1F-dF+d1F-d1F-d2F = 0

1) I don't know if my forces are correct or not
2) Like I said before the torque don't depend of 'd'

This is correct ?

 

[haruspex]

The torque on the suppport is -dF+2(d+d1)F-(d+d1+d2)F, if d1=d2, then -dF+2d1F-dF+d1F-d1F-d2F = 0

1) I don't know if my forces are correct or not
2) Like I said before the torque don't depend of 'd'

This is correct ?

I find the question a little strange, so I'm not really sure I've understood what's wanted.
But if the system is not accelerating and there's no axle friction then it follows immediately that there's no torque on the axle of the centre discs.

 

[Cri85]

I find the question a little strange

The question at start is : "Find a mechanical device that can reduce the rotationnal velocity by a factor K. The device shall not give a torque on the support." So I tried to find a solution.

there's no axle friction

No friction.

if the system is not accelerating

, the acceleration around the red axis or the acceleration of gears ? The rotationnal velocity around the red axis is supposed constant.

You know another solution with an efficiency 100% if there is no friction ?

 

[Cri85]

I was wrong when I drawn my forces, it's not possible to have the same forces in value in the Gear2, if the force at right is correct, then the force at left must be divided bt 2, no ?

 

[haruspex]

I was wrong when I drawn my forces, it's not possible to have the same forces in value in the Gear2, if the force at right is correct, then the force at left must be divided bt 2, no ?

I don't see why. The torques have to be different, but not necessarily the forces.
But I would draw to your attention that you have not satisfied the requirement about the 3:1 ratio of rotation rates. I suspect that throughout the thread you have posted images which show a combination of the original question and features you have added as part of a proposed solution. If so, I cannot be sure what the original question was. In particular, this d1=d2. That won't necessarily give you the desired ratio. However, it's hard to tell because there are no dimensions provided for gears 1 and 3, or the separation between them.
Another puzzle for me is the red 'main centre of rotation' added in later images. Perhaps this is the 'point of support' about which torque is to be avoided? If so, your proposal does not achieve that (but it's not obvious to me that it can be done).

 

[Cri85]

Perhaps this is the 'point of support' about which torque is to be avoided?

No I don't forgot it but I placed it because OldEngr63 seems to need it.

But I would draw to your attention that you have not satisfied the requirement about the 3:1 ratio of rotation rates.

I drawn with radius:

http://imageshack.com/a/img673/2359/iDvXea.png [Broken]

The ratio is not R1/R2a*R2b/R3 = 1/(4/3)*(2/3)/(5/3)=6/20 = 0.3 ?

In particular, this d1=d2.

It was just because I don't want a torque on the support. I don't have an image with the question (it's an old book).

I don't see why. The torques have to be different, but not necessarily the forces.

The Gear1 gives the torque on Gear2 and Gear2 gives to Gear3, so the Gear3 has a counterclockwise on it (a charge), I though it was not possible.

 

[haruspex]

I placed it because OldEngr63 seems to need it.

OK, so it is not part of the original problem.

The ratio is not R1/R2a*R2b/R3 = 1/(4/3)*(2/3)/(5/3)=6/20 = 0.3 ?

With those dimensions, yes you get 0.3. I assume R1, R3 and R2a+R2b are given.

It was just because I don't want a torque on the support.

As discussed, you would only have a net torque about the axle of gear 2 if there were some friction there. It has nothing to do with d1 and d2. If you vary d1 and d2 then the forces will adjust appropriately.

 

[Cri85]

It has nothing to do with d1 and d2.

You don't take in account F1, F2, F3 and F4 in the center of each gear ?

 

[haruspex]

You don't take in account F1, F2, F3 and F4 in the center of each gear ?

In your original attempt you considered torque about the axle of gear 1. That was where you calculated that d1=d2, right?

On the one hand, there was no reason to suppose that was the 'point of support' mentioned in the problem statement. If you think you still need d1=d2 in order to have no net torque about gear 2 axle, please post your working for that.

On the other hand, your calculation was wrong anyway. The axle at gear 1 cannot know anything directly about F2 etc. The only forces apparent there are F1 (having no torque about gear 1 axle) and the F you drew acting at the edge of gear 1. Thus there will be a torque F*R1 about gear 1 axle. This is completely unavoidable. There is an externally applied load torque at gear 3, so there must be a driving torque supplied at gear 1.