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Force multiplier

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electrodacus:

--- Quote from: IanB on February 07, 2023, 05:59:19 am ---
I know it works and I know how it works. I have shown you it working with an animation. Animations are ideal, they cannot have any sticking or slipping or any other non-ideal behavior. Animations are perfect, and they work perfectly.

Please stop creating these pointless threads.

--- End quote ---

Anything you want can work in an animation without a physics model. Same is true with math.
You will be surprised when you will try the real model. You will observe the same thing I showed in my video.
You had no explanation for why the gearbox is locked in those 15 seconds while the vehicle is dragged in the direction of applied force. That is because you do not understand the physics involved and specifically Newton's 3'rd law.

cbutlera:

--- Quote from: electrodacus on February 07, 2023, 01:53:09 am ---...
The internal mechanism in a) and b) can not do force amplification / multiplication in the way they are set up.
For that to be true the body of the vehicle in a) (blue part) will need to be connected rigidly to ground and same in case b) the cylinder will need to be connected to ground.
Yes fluid pressure is the same also F2 equal and opposite to F1.
...

--- End quote ---

OK, let’s examine your example b) more closely.

Assume that your diagram b) represents the state of the system at time t = 0, and at that time, all of the parts of the system are at rest.

Also assume that a force is applied to the lever in the direction toward the cylinder starting at time t=0 such that:

F1 = 0 for t < 0
F1 = X for t >= 0, where X is a constant.

What will happen to F2?  How do you think that the system will evolve?

electrodacus:

--- Quote from: cbutlera on February 07, 2023, 06:11:32 pm ---OK, let’s examine your example b) more closely.

Assume that your diagram b) represents the state of the system at time t = 0, and at that time, all of the parts of the system are at rest.

Also assume that a force is applied to the lever in the direction toward the cylinder starting at time t=0 such that:

F1 = 0 for t < 0
F1 = X for t >= 0, where X is a constant.

What will happen to F2?  How do you think that the system will evolve?

--- End quote ---

If X is less than needed to move the red box then F2 will have the same value and opposite direction to F1

If the body of the cylinder was to be connected to ground then F1 will be applied between the small piston and the cylinder body that is rigidly connected to ground and then F2 can be say 2 * F1 as it will be pushing relative to the cylinder body that is now referenced to ground and not relative to F1

As long as cylinder body is floating as in diagram (b) F2 can not be anything other than equal and opposite of F1.

cbutlera:

--- Quote from: electrodacus on February 07, 2023, 06:57:24 pm ---
--- Quote from: cbutlera on February 07, 2023, 06:11:32 pm ---OK, let’s examine your example b) more closely.

Assume that your diagram b) represents the state of the system at time t = 0, and at that time, all of the parts of the system are at rest.

Also assume that a force is applied to the lever in the direction toward the cylinder starting at time t=0 such that:

F1 = 0 for t < 0
F1 = X for t >= 0, where X is a constant.

What will happen to F2?  How do you think that the system will evolve?

--- End quote ---

If X is less than needed to move the red box then F2 will have the same value and opposite direction to F1

If the body of the cylinder was to be connected to ground then F1 will be applied between the small piston and the cylinder body that is rigidly connected to ground and then F2 can be say 2 * F1 as it will be pushing relative to the cylinder body that is now referenced to ground and not relative to F1

As long as cylinder body is floating as in diagram (b) F2 can not be anything other than equal and opposite of F1.

--- End quote ---

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?

electrodacus:

--- 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.

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