[Yahya (OP)]

Let say we have a gearbox of total ratio 1:4 , and we have input acceleration of a and output acceleration of A , then :
A=0.25a
F=4f

So why the acceleration  decreased but the force increased ?

When mass m on the input gear exerts force on mass M on the output gear through a gearbox for speed reduction of ratio say 1:4 , then the reaction of mass M on mass m is smaller than if they exert force on each other by a normal contact at a normal contact force exerted by mass m on mass M equals reaction of mass M on mass m, but in the case of gearbox the reaction is smaller than the force exerted .

Because inertia and resistance of mass M is small mass m will have the opportunity to exert more and more force making F bigger than f, if the ratio is 1:4 as above then F=4f .

And because the inertia and resistance of mass M is small , in gearbox contact mass M tends to absorbs motion than to reflect it , when absorbing motion , mass M tends to not to move from its position making motion little and acceleration small , but when reflecting it in case of normal contact it will again reflect and hit mass M making it move faster and with larger acceleration.

[Colin2B]

It really has to do with the difference in torque on each gearwheel, hence force at meshing point, and the diffence in angular rotation of each gear caused by the same distance being travelled along the circumference of each wheel when meshed.
It really is quite simple.

[Bored chemist]

Let us say I have a table in front of me here in my house.- it has a mass of 10 Kg
On the table are two rocks - one is 5 Kg and the other is 3 Kg.

Gravity acts on the two rocks. The local value for the acceleration due to gravity is near to 10 m/s/s

So the forces acting on the table are about 50N and 30N both acting downwards.
So the force due to the weights of the rocks produce a total of 80 N on the table.
Well, if the force is 80N and the mass is 10Kg, do you think the table will accelerate downwards at F/M i.e. 80/10 = 8 metres per second per second?

Or do you realise that you have to take other forces into account?

[Yahya (OP)]

Let us say I have a table in front of me here in my house.- it has a mass of 10 Kg
On the table are two rocks - one is 5 Kg and the other is 3 Kg.

Gravity acts on the two rocks. The local value for the acceleration due to gravity is near to 10 m/s/s

So the forces acting on the table are about 50N and 30N both acting downwards.
So the force due to the weights of the rocks produce a total of 80 N on the table.
Well, if the force is 80N and the mass is 10Kg, do you think the table will accelerate downwards at F/M i.e. 80/10 = 8 metres per second per second?

Or do you realise that you have to take other forces into account?

I have to take the whole mass including the table as mass and the whole weight including the table as force
I have to take in account the weight of the table as a force

[Bored chemist]

OK, the total mass is 18 Kg (3+5+10)
and the additional force- the weight of the table is 100 N
So the total force (downward) is 180N, and the total mass is 18Kg.
So, do you really think my table is accelerating downwards at  180/10 i.e. 10 m/s/s?

Or are you still missing something?

[Yahya (OP)]

OK, the total mass is 18 Kg (3+5+10)
and the additional force- the weight of the table is 100 N
So the total force (downward) is 180N, and the total mass is 18Kg.
So, do you really think my table is accelerating downwards at  180/10 i.e. 10 m/s/s?

Or are you still missing something?

It has acceleration =0, the floor pushes back against the total force.

[Bored chemist]

OK, the total mass is 18 Kg (3+5+10)
and the additional force- the weight of the table is 100 N
So the total force (downward) is 180N, and the total mass is 18Kg.
So, do you really think my table is accelerating downwards at  180/10 i.e. 10 m/s/s?

Or are you still missing something?

It has acceleration =0, the floor pushes back against the total force.

So, you understand that you need to add up all the forces to calculate the acceleration.

Now do that for the first cog in your gearbox.