General > General Technical Chat
Force multiplier
<< < (27/71) > >>
cbutlera:

--- Quote from: electrodacus on February 08, 2023, 08:36:56 pm ---
--- Quote from: cbutlera on February 08, 2023, 08:21:12 pm ---The total horizontal force exerted by the ground on the table would be the vector sum of F1 and F2.  The total horizontal force exerted by the table on the ground would be of equal magnitude to this vector sum, but directed in the opposite direction.  Those two combined forces do constitute a Newton’s third law force pair.

If F1 did somehow represent the total horizontal force exerted by the ground on the table, and F2 did somehow represent the total horizontal force exerted by the table on the ground, then the table would accelerate in the direction of F1.  F1 would be the only horizontal force acting on the table.  F2 would be acting on the ground (if you are going to treat it as the Newton's third law reaction force to F1) and so would not restrain this acceleration.  This is obviously wrong.

--- End quote ---
...
... whatever you wanted to say in the above comment makes no sense.

--- End quote ---

Do you perhaps regard F1 and F2 as bi-directional forces?  So in the case of F1 for instance, do you think of it as both pushing against the wheel in the direction shown by the arrow, and by implication also pushing back against the treadmill in the opposite direction with an equal magnitude?
electrodacus:

--- Quote from: cbutlera on February 08, 2023, 08:56:25 pm ---Do you perhaps regard F1 and F2 as bi-directional forces?  So in the case of F1 for instance, do think of it as both pushing against the wheel in the direction shown by the arrow, and by implication also pushing back against the treadmill in the opposite direction with an equal magnitude?

--- End quote ---

Yes there is of course the pair to F1 pushing against the treadmill but we are interested in the forces acting on the vehicle and not on the treadmill as we want to know if there is a net force on the "vehicle".

You can see the treadmill as a motor with the stator connected to ground and rotor connected to the vehicle right wheel but due to the way the vehicle is configured internally it will act as a solid block and so the left wheel will push against ground which is basically the same as the motor stator.
So the stator is basically rigidly connected to the rotor thus the motor can not move even if it can apply a force.
Force needs to be high enough for something to slip and depending on what slips will determine in what direction the vehicle will move.
Left wheel slips it moves in the direction of applied force.
Right wheel slips it moves less intuitively against the direction of the applied force and that can only happen with energy storage.
cbutlera:

--- Quote from: electrodacus on February 08, 2023, 09:14:24 pm ---
--- Quote from: cbutlera on February 08, 2023, 08:56:25 pm ---Do you perhaps regard F1 and F2 as bi-directional forces?  So in the case of F1 for instance, do think of it as both pushing against the wheel in the direction shown by the arrow, and by implication also pushing back against the treadmill in the opposite direction with an equal magnitude?

--- End quote ---

Yes there is of course the pair to F1 pushing against the treadmill but we are interested in the forces acting on the vehicle and not on the treadmill as we want to know if there is a net force on the "vehicle".
...

--- End quote ---

That pair to F1 is the Newton's third law reaction force to F1.
electrodacus:

--- Quote from: cbutlera on February 08, 2023, 09:19:40 pm ---
That pair to F1 is the Newton's third law reaction force to F1.

--- End quote ---

Yes that pair will also be Newton's third law reaction force. So will the F2 and the counterpart acting on the red box.
But to know if vehicle moves you will only look at the balance of forces acting on the vehicle and that will be F1 and F2

Reducing everything to just the electric motor powering the treadmill you can sat the the rotor is connected to the right wheel and the stator to the left wheel.
Since the way the belt connects the two wheels only allows the wheel to move in the same direction the motor will stall so no movement just a force proportional with the current applied to the motor.  All energy will be dissipated as heat and there will be no movement.

If you push on a wall with F1 using your hands the wall pushes back on you with an equal and opposite force but then at your feet there is also another pair of forces you pushing against ground and ground pusing on you and at the base of the wall there is also an say F2 both wall pushing against ground and ground against the wall.
In any case nothing will move the same as in the example (a) unless the applied force is large enough for the wheel to slip at the connection to ground.

I'm no longer sure what you are confused about. Is only that vehicle is a "locked gearbox" so thus can be treated as a single solid object or even if that was the case you still do not agree with the fact that F2 is equal and opposite to F1.
cbutlera:

--- Quote from: electrodacus on February 08, 2023, 09:45:06 pm ---
--- Quote from: cbutlera on February 08, 2023, 09:19:40 pm ---
That pair to F1 is the Newton's third law reaction force to F1.

--- End quote ---

Yes that pair will also be Newton's third law reaction force. So will the F2 and the counterpart acting on the red box.
But to know if vehicle moves you will only look at the balance of forces acting on the vehicle and that will be F1 and F2
...

--- End quote ---
This is true, F1 and its pair constitute a Newton's third law force pair, as do F2 and its pair.  Newton's third law has now been applied to both F1 and F2 independently.  It tells you nothing about the relationship between F1 and F2.


--- Quote ---If you push on a wall with F1 using your hands the wall pushes back on you with an equal and opposite force but then at your feet there is also another pair of forces you pushing against ground and ground pusing on you and at the base of the wall there is also an say F2 both wall pushing against ground and ground against the wall.

--- End quote ---
This is also true and again each of those pairs constitute a Newton's third law force pair.


--- Quote ---In any case nothing will move the same as in the example (a) unless the applied force is large enough for the wheel to slip at the connection to ground.
...

--- End quote ---
Whether or not anything moves has nothing further to do (beyond its application to these force pairs) with Newton's third law.  So you cannot keep using it as the reason why you think that everyone else is wrong.

Edit: added text in brackets for clarity.
Navigation
Message Index
Next page
Previous page
There was an error while thanking
Thanking...

Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod