[Yahya (OP)]

The equation :
F=ma won't apply to this situation.
Both mass M and mass m are at free fall , they will have the  acceleration A and a  according to pulley ratio since they start from speed zero meters per second and the speed ratio is always the same according to pulleys' ratios.

I think of the same system as follows :
The mass M has acceleration say 4 m/s.s after three intervals of time 1, 2,3 seconds its speed becomes 4 , 8, 12 m/s respectively think of any other system connected in such way , let say pulley ratio is 1:4
then when pulley #1 has speed 4m/s then P2 will have 16 m/s according to pulley ratio , when P1 has 8 m/s P2 has 32  m/s , P1 has 12 m/s , P2 has 12*4=48 m/s , let's calculate acceleration on P2:
time intervals are 1,2,3 speeds are 16,32,48 respectively , a=16 m/s.s which is =A*4
 At that case the equation F=ma won't satisfy for mass m,  i.e force on mass m won't equal to its mass multiplied by acceleration:
Let's have mass M is 4 times mass m. M=4m, let the ratio this time be 1:6
force on mass M is its weight minus gm/6=gM-gm/6
A=F/M=(gM-gm/6)/M=(gM-gM/24)/M= (1-1/24)g

force on mass m is 6gM-gm
a=F/m=(24gm-gm)/m=23g
A is not equal to 6 times a as it should be above.

“The reason is because the third law won't apply as well , objects do not exert the same force however a force according to the pulley ratio , another law should be presented for such situations where the action and reaction are not equal, objects M and m do not touch directly so that there is equal  action and reaction"

so will the second law apply? why not?

[guest39538]

The equation :
F=ma won't apply to this situation.
Both mass M and mass m are at free fall , they will have the  acceleration A and a  according to pulley ratio since they start from speed zero meters per second and the speed ratio is always the same according to pulleys' ratios.

I think of the same system as follows :
The mass M has acceleration say 4 m/s.s after three intervals of time 1, 2,3 seconds its speed becomes 4 , 8, 12 m/s respectively think of any other system connected in such way , let say pulley ratio is 1:4
then when pulley #1 has speed 4m/s then P2 will have 16 m/s according to pulley ratio , when P1 has 8 m/s P2 has 32  m/s , P1 has 12 m/s , P2 has 12*4=48 m/s , let's calculate acceleration on P2:
time intervals are 1,2,3 speeds are 16,32,48 respectively , a=16 m/s.s which is =A*4
 At that case the equation F=ma won't satisfy for mass m,  i.e force on mass m won't equal to its mass multiplied by acceleration:
Let's have mass M is 4 times mass m. M=4m, let the ratio this time be 1:6
force on mass M is its weight minus gm/6=gM-gm/6
A=F/M=(gM-gm/6)/M=(gM-gM/24)/M= (1-1/24)g

force on mass m is 6gM-gm
a=F/m=(24gm-gm)/m=23g
A is not equal to 6 times a as it should be above.

“The reason is because the third law won't apply as well , objects do not exert the same force however a force according to the pulley ratio , another law should be presented for such situations where the action and reaction are not equal, objects M and m do not touch directly so that there is equal  action and reaction"

so will the second law apply? why not?

Looks more like a friend type relationship to me, nice try at the drawing out of your ideas, shame they have no real weight .  Perhaps you should add something electrical to your ideas like charges.  Make up some charges even ,   why not hey if it suits .   But expect the charge to repel back or escape into space , the charge may not help you at all, electrons bond protons , protons are only repulsive to protons because of negative energy between them .

M is strong where m is weaker, M holds onto m and wherever M goes , m follows because of G . You can't see an invisible link unless shown the link , a strong bond does not need contact necessarily, but relationships can be var x and incrimination of x in cyber space is meaningless and invalid . 

The particle differential  is fascinating hey?   Disabled particles don't lie about their spatial position, under a lot of gravity and stress they are mainly fixed to a position ,  however in dealing with this  position , the lead particle M also becomes severe stressed to a tension level where it also becomes fixed, the amount of energy levels, helping with some freedom of the particles . 
Particles can't traverse without enough energy to traverse.

[guest39538]

Darn't it, I gone and got myself all upset again and you know what , science is just not worth me.  It is not me who should be begging science for a job , it should be them begging me to come work with them .  I put more effort into my science than most, for what ? 

Hope , dreams, God,   not even worthy of being words,  can you believe I was praying to God to win some money just so I could take my children away ......I must be insane for praying.  I wish I could back in time, I would have not had children, I would of been a full bore super criminal instead.   

Enough , had my moan...

[Bored chemist]

Both mass M and mass m are at free fall

No.
They are not in free fall.
They have things tied to them.

It would be sensible for you to learn physics before trying to claim that physics is wrong.

[Bored chemist]

will the second law apply?

What do you think "law" means?

[guest39538]

They have things tied to them.

Tied , can be such a loose term to use.   Subjective is subjective....

[Bored chemist]

They have things tied to them.

Tied can be such a loose term....

Shut up, you useless idiot.

[guest39538]

They have things tied to them.

Tied can be such a loose term....

Shut up, you useless idiot.

Yup that is me, useless idiot, what gives you the right to tell me to shut up? 

Are you drunk again?

I love your banter, don't run off. 

[Bored chemist]

Yup that is me, useless idiot, what gives you the right to tell me to shut up? 

Since you accept that your "contributions" are useless, you should stop wasting others' time + bandwidth.
There's nothing special about me telling you that- you should already know it. I'm just reminding you because I happen to be here.

[guest39538]

Yup that is me, useless idiot, what gives you the right to tell me to shut up? 

Since you accept that your "contributions" are useless, you should stop wasting others' time + bandwidth.
There's nothing special about me telling you that- you should already know it. I'm just reminding you because I happen to be here.

I know I am useless , but does that mean I cant have a conversation ?   I have nothing else in my life but my kids and conversation.

[Bored chemist]

Cluttering up people's threads isn't conversation, it's vandalism.

[guest39538]

Cluttering up people's threads isn't conversation, it's vandalism.

You started the vandalism....not me, well maybe it was me when I got to the feeling sorry for myself.  Stuff it my computer is going tomorrow , I am going have the internet turned off .  Happy now ?

[Bored chemist]

You started the vandalism....not me, well maybe it was me when I got to the feeling sorry for myself. 

Nope
This "Tied , can be such a loose term to use.   Subjective is subjective...." wasn't meaningful- it just cluttered the forum.

[guest39538]

You started the vandalism....not me, well maybe it was me when I got to the feeling sorry for myself.

Nope
This "Tied , can be such a loose term to use.   Subjective is subjective...." wasn't meaningful- it just cluttered the forum.

When objects are tied together by gravity , they can be viewed loosely,  consider any object on the ground, it is loose relative to ground,  do you even understand gravity ?

I think your scientific understanding is loose and not precise at all.   My science is more precise, I account for all the mass.   I do not think you account for all the mass, my theories is much tighter. 

[Yahya (OP)]

Both mass M and mass m are at free fall

No.
They are not in free fall.
They have things tied to them.

It would be sensible for you to learn physics before trying to claim that physics is wrong.

I know what free fall is I meant it in a deeper sense, the only external force affecting on the system they consist of is gravity and each other force affects on each one of them is due to gravity , and I didn't say physics is wrong here, I said the law does not apply to this situation , that does not mean it is wrong, that does not mean it should be replaced but extended as I mentioned
so your answer is Yes .Could you explain my results?

[alancalverd]

Objects in free fall towards a much larger mass (such as the earth) have exactly the same acceleration. If they don't, then one or other (or both) is not in free fall, by definiition.

Rather than get involved in complex arithmetic, consider Galileo's thought experiment. If rock A is heavier and falls  faster than rock B, what happens if we tie them together? Does B slow down A or does A speed up B? How do they choose?

Let's suppose A speeds up B. Obviously B is reluctant to travel as fast as A so it exerts some drag, so speed (A + B) < A.

Now instead of string, we use glue. Thus since mass (AB) > mass (A), speed (AB) > A.

So the speed of a falling object depends on whether it is held together by string or glue.

Bollocks (a loose translation of Dialogo sopra i due massimi sistemi del mondo (1632)).

[Bored chemist]

I know what free fall is I meant it in a deeper sense,

If you choose to use a phrase like "free fall"
https://en.wikipedia.org/wiki/Free_fall
which has a well defined meaning, for something else you are being foolish.

Either use "free fall" to mean what everyone understands it to mean, or use some other word or phrase.

Don't try to tell us that your use of it is somehow "better".

Try again, but this time  try to communicate your meanng properly.

[Bored chemist]

I said the law does not apply to this situation

Spoiler alert; it does.

[Yahya (OP)]

I said the law does not apply to this situation

Spoiler alert; it does.

it doesn't think of a gearbox for speed reduction with very low speed and very high torque , the output gear will have acceleration relative to the input gear according to gear ratio , but the output gear will have very large torque and very large force , in that case F=ma for the output gear won't satisfy. because when I increase gear ratio mass is constant , but acceleration decreases and force increase:
F=ma m constant   , acceleration decreases F should decrease as well ,but F increase according to gearbox ratio " torque " then the equation is invalid.

[Bored chemist]

If you define F as the number of rhododendrons in my local park, M as the number of gospel saints and A as the average body-weight of a duck then there's very little chance that F=MA.

However, if you use the conventional definitions where they represent the force acting on; the mass of, and the acceleration of THE SAME OBJECT then the equation holds.

By looking at the force on one cog, but the acceleration of another you are simply not using the equation correctly.

If you don't make that stupid mistake, then F=MA.