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Force multiplier
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cbutlera:

--- Quote from: electrodacus on February 07, 2023, 08:29:40 pm ---Yes that is correct for cylinder block attached to the ground.

If cylinder is not attached to ground the cylinder itself will move to the right and assuming cylinder has no mass F2 = F1 but if cylinder has some mass then F2 = F1 + m *a thus no longer function as a force multiplier.
So maybe (b) is not a great example as if piston F1 moves to much the cylinder will fall off (as it is drawn). If there is say a stop on the wide part so that large piston can not fall off the cylinder then cylinder will accelerate to that end then cylinder will push against the large piston F2 and so when at that point F2 = F1 as there is no longer than m*a factor.
In any case at no point unless by coincidence F2 ≠ F1 * A2/A1 instead is F2 = F1 + m*a until the stop at the end then F2 = F1

--- End quote ---

So to sum up.  You now agree that the cylinder block will accelerate to the right if it is not attached to the ground. You also agree that during the acceleration F1 will not be equal to F2.  As a consequence, do you also now agree that F1 and F2 cannot be a Newton’s third law force pair, because such a pair must necessarily be equal and opposite under all circumstances?
electrodacus:

--- Quote from: cbutlera on February 07, 2023, 08:46:02 pm ---
So to sum up.  You now agree that the cylinder block will accelerate to the right if it is not attached to the ground. You also agree that during the acceleration F1 will not be equal to F2.  As a consequence, do you also now agree that F1 and F2 cannot be a Newton’s third law force pair, because such a pair must necessarily be equal and opposite under all circumstances?

--- End quote ---

Yes but that will be true in any acceleration frame as Newton's 2'nd law is also involved on top of the 3'rd.

So to sum up for case (b) you have
F2 = F1 * A2/A1  when cylinder body is attached to ground thus force amplification or force multiplier.

F2 = F1 + m*a  when cylinder is floating and in an accelerating reference frame.
F2 = F1             when cylinder is floating but not moving due to a limit stop so cylinder will not fall off.


For case (a) is similar if you attached the vehicle/gearbox body to ground you just have D2/D1 as the gear ratio.
If body is floating then you have F2 = F1 unless left wheel slips the as long as it accelerates it will still be F2 = F1 * m*a but once it stops accelerating say moves at constant speed is F2 = F1 again.
And there is the more complex case where right wheel (input wheel) slips then there is energy storage and over that period F2 = F1 but right wheel rotates meaning energy is stored then the stored energy is converted to vehicle kinetic energy so again Newton's 2'nd law gets involved.

Do you agree with all this ?
cbutlera:

--- Quote from: electrodacus on February 07, 2023, 09:04:25 pm ---
--- Quote from: cbutlera on February 07, 2023, 08:46:02 pm ---
So to sum up.  You now agree that the cylinder block will accelerate to the right if it is not attached to the ground. You also agree that during the acceleration F1 will not be equal to F2.  As a consequence, do you also now agree that F1 and F2 cannot be a Newton’s third law force pair, because such a pair must necessarily be equal and opposite under all circumstances?

--- End quote ---

Yes but that will be true in any acceleration frame as Newton's 2'nd law is also involved on top of the 3'rd.

So to sum up for case (b) you have
F2 = F1 * A2/A1  when cylinder body is attached to ground thus force amplification or force multiplier.

F2 = F1 + m*a  when cylinder is floating and in an accelerating reference frame.
F2 = F1             when cylinder is floating but not moving due to a limit stop so cylinder will not fall off.


For case (a) is similar if you attached the vehicle/gearbox body to ground you just have D2/D1 as the gear ratio.
If body is floating then you have F2 = F1 unless left wheel slips the as long as it accelerates it will still be F2 = F1 * m*a but once it stops accelerating say moves at constant speed is F2 = F1 again.
And there is the more complex case where right wheel (input wheel) slips then there is energy storage and over that period F2 = F1 but right wheel rotates meaning energy is stored then the stored energy is converted to vehicle kinetic energy so again Newton's 2'nd law gets involved.

Do you agree with all this ?

--- End quote ---

So now that you have agreed that the hidden hand of Newton’s third law does not reach out and restrain the cylinder block in your example b), you can apply the same reasoning to example a).

In the past you have been arguing that the wheels will slip because Newton’s third law is preventing the vehicle from accelerating.  If you now accept that this hidden hand does not intervene in example b), why do you think it will still reach out and restrain the vehicle in example a)? As you seem to have hinted at above, the vehicle will accelerate with no slip required.  (Assuming that the treadmill itself accelerates steadily from a stationary start at time t = 0)
electrodacus:

--- Quote from: cbutlera on February 07, 2023, 09:22:33 pm ---
So now that you have agreed that the hidden hand of Newton’s third law does not reach out and restrain the cylinder block in your example b), you can apply the same reasoning to example a).

In the past you have been arguing that the wheels will slip because Newton’s third law is preventing the vehicle from accelerating.  If you now accept that this hidden hand does not intervene in example b), why do you think it will still reach out and restrain the vehicle in example a)? As you seem to have hinted at above, the vehicle will accelerate with no slip required.  (Assuming that the treadmill itself accelerates steadily from a stationary start at time t = 0)

--- End quote ---

Did you miss the Newton's 3'rd law ? It is present in all non accelerating frames and of course when whe are in an accelerating frame the 2'nd law also applies.
Have you missed the fact that in none of the case where body is floating there is no force multiplication ?
Yes the vehicle at (a) in order to move will require one of the wheels to slip as it is a locked gearbox.
The same is true in b) as neither the piston pushed by F1 nor the one pushed by F2 moves unless fluid is compressible (not normally the case). But if it is compressible it is no different from vehicle (a) where belt is elastic and so then piston F1 will move as energy is stored same as wheel on the right moves storing energy.
Unless one of the wheels slips vehicle (a) will not move. It is your choice if back wheel on the left slips or the one on the right and depending on that we can discuss what the forces are but I already mentioned what they are multiple times.


Edit: Also I asked but you did not answered if you agree with the equations I wrote.

Only for the case of body connected to ground you can have force multiplication F2 = F1 * A2/A1
When there is no force multiplication you will have either F2=F1 in non accelerating frame or F2 = F1 + m*a last part being the second law which deals with acceleration frames. None of this is force multiplication.
IanB:

--- Quote from: electrodacus on February 07, 2023, 10:01:47 pm ---Yes the vehicle at (a) in order to move will require one of the wheels to slip as it is a locked gearbox.
--- End quote ---
There is no locked gearbox.


--- Quote ---The same is true in b) as neither the piston pushed by F1 nor the one pushed by F2 moves unless fluid is compressible (not normally the case). But if it is compressible it is no different from vehicle (a) where belt is elastic and so then piston F1 will move as energy is stored same as wheel on the right moves storing energy.
--- End quote ---
No one ever said the fluid was compressible. In all the discussion above the results are true for an ideal incompressible fluid.

In case (b) it is impossible to apply a force F1 or F2 because as soon as you try the system will move to cancel out the force. So F1 and F2 are always exactly zero (or you have to introduce a non-ideal F = ma term). This situation is just the same as case (a) where F1 and F2 are also zero.
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