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Force multiplier
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cbutlera:

--- Quote from: electrodacus on February 07, 2023, 07:13:13 pm ---
--- Quote from: cbutlera on February 07, 2023, 07:11:15 pm ---
Yes, lets assume that the red box is fixed to the ground and unable to move.

So are you saying that the system will reach a static equilibrium with F1 = F2 = X and with the cylinder block remaining stationary?

--- End quote ---

Yes that is exactly what I'm saying.

--- End quote ---

If the cylinder block remains stationary, then it is behaving exactly as it would if were attached rigidly to the ground.  There appears to be no consequence to detaching from the ground.  So by what mechanism can whether or not the cylinder block is attached to the ground alter F2?
electrodacus:

--- Quote from: cbutlera on February 07, 2023, 07:22:31 pm ---
If the cylinder block remains stationary, then it is behaving exactly as it would if were attached rigidly to the ground.  There appears to be no consequence to detaching from the ground.  So by what mechanism can whether or not the cylinder block is attached to the ground alter F2?

--- End quote ---

As it is in diagram (b) cylinder block floating F2 is a consequence of F1 equal and opposite.  Just imagine redbox had no mass and no friction with ground then you can not have an F2 and thus you can also not have an F1 and the displacement will be the same as they will move the same amount.

But if the cylinder body is connected to ground then F1 piston can move relative to the cylinder case and the other piston F2 will move say half the distance due to larger say double section area.
F2 will push against F1 with say force X but also against the cylinder body with force X so F2 = 2X but this is only possible if cylinder body is connected to ground so there is something to push against.
If cylinder body is floating everything will move as one object the small piston the large one and the cylinder case will travel the same distance and so F2 will just be equal and opposite to F1.
nctnico:

--- Quote from: Nominal Animal on February 02, 2023, 06:27:25 pm ---I forget who first described having done it in that other thread, but they tried basically the same thing with a heavy spool of soldering wire, with the wire coming off the spool at the bottom.  If the spool has sufficient traction (doesn't slip), pulling on the wire causes the spool to roll faster towards you than you pull the wire, re-spooling the wire back onto the spool!  Funky, eh?  But quite expected, physically speaking.

--- End quote ---
Basic physics  8) You can also have very interesting discussions with car enthousiasts that keep claiming torque is the main engine parameter. You can't make them understand that power is what makes a car go forward.  Even when you show this with simple math |O
cbutlera:

--- Quote from: electrodacus on February 07, 2023, 07:42:21 pm ---
--- Quote from: cbutlera on February 07, 2023, 07:22:31 pm ---
If the cylinder block remains stationary, then it is behaving exactly as it would if were attached rigidly to the ground.  There appears to be no consequence to detaching from the ground.  So by what mechanism can whether or not the cylinder block is attached to the ground alter F2?

--- End quote ---

As it is in diagram (b) cylinder block floating F2 is a consequence of F1 equal and opposite.  Just imagine redbox had no mass and no friction with ground then you can not have an F2 and thus you can also not have an F1 and the displacement will be the same as they will move the same amount.

But if the cylinder body is connected to ground then F1 piston can move relative to the cylinder case and the other piston F2 will move say half the distance due to larger say double section area.
F2 will push against F1 with say force X but also against the cylinder body with force X so F2 = 2X but this is only possible if cylinder body is connected to ground so there is something to push against.
If cylinder body is floating everything will move as one object the small piston the large one and the cylinder case will travel the same distance and so F2 will just be equal and opposite to F1.

--- End quote ---

In the case where the cylinder block is rigidly attached to the ground and the system has reached a static equilibrium, do you agree with the following?

If the fluid pressure (force per unit area) is P, and the working surface area of the F1 piston is A1, then

F1 = P * A1

Similarly, if the working surface area of the F2 piston is A2, then

F2 = P * A2

Therefore F2 = F1 * A2 / A1
electrodacus:

--- Quote from: cbutlera on February 07, 2023, 07:57:10 pm ---
In the case where the cylinder block is rigidly attached to the ground and the system has reached a static equilibrium, do you agree with the following?

If the fluid pressure (force per unit area) is P, and the working surface area of the F1 piston is A1, then

F1 = P * A1

Similarly, if the working surface area of the F2 piston is A2, then

F2 = P * A2

Therefore F2 = F1 * A2 / A1

--- End quote ---

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   
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