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**New Theories / Re: Will Newton's second law apply to this situation ?**

« **on:**20/07/2018 23:34:22 »

The equation :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 .

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?

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.

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