Sorry, F1 cannot be bigger than M1g, because then the latter would fly up into the sky.
Are you familiar with the equation F = ma, Heiwa?
What does it represent?
Now, suppose you have a mass falling downwards (a = g, therefore F = mg), and after falling a certain distance (and thereby gaining speed), it impacts a body which provides a resistance force of R (which, being a resistance, is opposite in direction to F).
To re-iterate, you have a mass (m) falling under gravity (g) onto another object. By the time the falling mass hits the object, it has gained some speed (V, or else it wouldn't be falling, would it?).
So it is moving, and moving downwards. Even you can agree with the assumption that a mass must move downwards to strike an object below it, requiring said mass to be in motion.
Now, you have a mass with speed V. It impacts an object which provide a resistance force of R.
You state that if R is greater than mg, the mass will move upwards.
That
acceleration vector will be upwards. That I grant.
But the
motion of the object will still be
downwards. It's downward acceleration will be slowed, but it will
not "fly up into the sky"
Follow me here (and remember, forces and acceleration are
vectors)
mg (force of falling mass) = F.
Well, we know that R will impart an acceleration on the mass. R is a force, therefore we can write R = ma.
Let's look at three cases (and note that I take upwards as the "positive" direction):
1) R = -F (forces R and F are equal in magnitude and opposite in direction),
2) R < -F (force R is less in magnitude and opposite in direction to force F),
3) R > -F (force R is greater in magnitude and opposite in direction to force F)
Starting with scenario 1):
R = -F
Therefore ma = -mg.
We cancel out the like term (m), and we wind up with a = -g.
Thus, we see that the acceleration imparted by the resistance force is equal to the acceleration of gravity. There is no net acceleration.
But the mass was already moving! Therefore, in this case, the moving mass continues to move downwards at a constant velocity! Exactly as Dave Rogers said!
Now on to scenario 2):
R < -F
Therefore ma < -mg.
We cancel out the like term (m), and we wind up with a < -g.
Thus, the acceleration imparted by the resistance force is less than the acceleration of the mass under gravity. There is a net acceleration downwards.
Therefore, in this case, the moving mass continues to accelerate downward! This, by the way, is what was observed on September 11, 2001.
Finally, scenario 3):
R > -F
Therefore ma > -mg.
We cancel out the like term (m), and we wind up with a > -g.
Thus, the acceleration imparted by the resistance force is greater than the acceleration of the mass under gravity. There is a net acceleration upwards.
Therefore, in this case, the moving mass slows it's descent! It will slow down to a complete stop. It cannot shoot into the air, because that would remove it from the object providing the resistance force, and thus the upwards acceleration would no longer be imparted. This is what you seem to think should have happened. Your errors in this assumption, however, numerous though they are, have been pointed out to you before, and in great detail, and so I'm not going to cover them here.
But surely you can no longer deny that for the upper block to slow down and stop moving requires a resistance force greater than the force of the falling block, not equal to it.
Dave Rogers is correct in what he said to you..
