awaterpon

I didn't speak about my experiment in details:
The scale I use uses force in kilograms which is the gravity force for each kilogram so the scale when I use to measure my weight it instantly gives me my mass instead of my weight and I need  to multiply the scale quantity  times the gravity acceleration to get  my weight.
If I use the scale to measure my calves muscles force and it reads 8 kg then this can be equivalent to force of gravity on mass of 8 kg so I can convert from mass to weight and from weight to mass.
I used the scale to measure my maximum force of my calves' muscles I can exert it turns out to be  8 kg .
When I stand on my feet in my experiment I did  several movements one of them is equivalent to the movement a person does when trying to pick a fruit from a tree , which is short feet distance upwards.Both this movement and the movement I did to measure my calves muscles force are exactly equivalent both uses the same muscles forces.
So if my maximum force of calves muscles I can exert is 8 kg then the maximum force I can exert to lift my body as I described is only 8 kg .
If I use the  lever concept my fulcrum will be at the toes and both my weight and the force I exert are on the heel " lever class 3"
F*S=f*s
F is the force of my calves' muscles 
S is the distance of my foot
s=S
f is force of my weight
By simple calculations force of calves' muscles F must be equal to my weight f
Now let's see what I did:
My calves' muscles force was 8 kg "F"
My weight is 57 kg "f"
F must be greater than my weight to lift it " for force to lift a mass the force must be slightly greater than mass weight"
So the maximum force I exert to lift my body "F" is only 8 kg and the force needed to lift my mass  must be greater than  57 kg 

This satisfy my observations in the real world I already presented in which human can move or lift itself with small forces

The concept of the alternative mass which based on this is above.

I would like to extend my posts so that my idea is more understandable:

The internal force of human on its mass is the force a human exerts on his own body, body force on body mass. 

The external force is the force of human exerted on any other mass but not the body. internal force of human to lift  its own body is very smaller than the external force of human to lift another object  both the human mass and the object mass are equal. 
 

The observations of this phenomenon are: 
Human body effortless walking, running  jumping, standing, dancing and other movements by human force on its own mass.
 

Force can be exerted on a human body with two different ways one by the human force on his own body "internal force" and the other way by using an external force

If I push myself by a force of my own then the body is lighter, it will gain a greater acceleration "internal force" however if someone pushes me with the same force "external force" my body will be heavier and  it will gain less acceleration. 

So same force, different acceleration will give different masses:
F=ma, m=F/a

The different masses are the actual mass of the body and the other mass is the alternative mass

The alternative mass  is substituted for physical equations when the force acting is of the human on the human"internal force"

F: my body force on my body 
m: my alternative mass
a: the acceleration I gain

In my experiment the maximum force of my calves muscles is approximately 8 kg or 80 N this force can lift my 57 kg mass as in the video. This proves that I can lift my body 57 kg with only 8 kg or 80 N force which satisfy my observations of how l can lift or move my body with force very smaller than my weight " in my experiment the small force is the maximum force of calves muscles 8 kg and the mass I lift is my body 57. 

To lift a mass with some force the force must be slightly greater than the mass weight. 

Then if I lift my body with 8 kg then I am actually lifting 8 kg " alternative mass" because the alternative mass is all about lifting or moving my body then I lift 8 kg with 8 kg. 

My body weight force downwards on the spring has no relation to body lifting or moving itself it is direct external  force in the spring "gravity force" 

Because this all about body lifting itself then I will push the ground  with only 8 kg as in my experiment the ground will push me with the force 80 N the two forces give slight pressure  on my soles the slight pressure is  because both are small . Then I have my body"57 kg " which presses the spring and my alternative mass  which press the ground floor" pressing the floor is force of action and reaction on human body internal force " then the floor will be pushed by alternative mass " "8 kg" if the mass is in vacuum the only force the mass falls with its weight "57 kg" when the mass presses the ground it pushes the ground with smaller force 8 kg.

2 minutes ago, Ghideon said:

How is “alternative mass” defined?

The two masses are alternative because they work in different situations and both are the mass of the body

7 minutes ago, swansont said:

You need to establish the validity of this “equivalent mass” nonsense. Your other speculation is based on this speculation. The rules don’t permit you to bootstrap like this.

The right term is alternative mass

18 hours ago, Ghideon said:

If I understand your claim correctly:
1: A person with a mass of 57kg could stand on their toes on a surface that would collapse under a force grater that 80N.  
2: It follows that when I carry a person with mass 57kg and the person is standing on toe on my shoulders then I would be using 80N of force to support that persons weight and not 570N*. 

As the above is obviously not how physics works, can you provide a picture showing what you mean so misunderstandings can be sorted out?

 

*) Approximating g=10 in this discussion

 

These are two pictures one is for the forces involved the other is for the forces in the example you mention

:The block doesn't have equivalent mass but you can consider it as a human body

49 minutes ago, swansont said:

 

When a body moves or lifts itself , body mass will be equivalent to a smaller mass, so inertia as well will be small"effortless walking and running" however force of gravity will be  for the actual mass. We can compare two equal masses on a two-legged scale one is an object 70 kg and the other is a body 70 k both will have the same gravity force  even though the body will have some equivalent mass

49 minutes ago, swansont said:

No, this makes no sense.

All this is intuitive interpretation of how if human is lighter then it is equivalent to a smaller mass.This mass is represented by body moving or lifting itself, the actual mass is what the scale reads.So we can treat body as small mass when human moves or lifts itself and we can treat it as the actual mass in case of force of gravity on it" Newton gravity equation"

58 minutes ago, swansont said:

Quote

Second I will press the scale surface with my body mass but the mass here not equal to but equivalent to a smaller mass 8 kg so actually the surface pushes me with only 8 kg "80 N" and the force I lift my body with also small "the 80 N" 

 

1 hour ago, swansont said:

You can’t do this without also being supported at some other point, like one foot on the scale and one on the floor.

 

 

The upwards normal force of gravity does not exist as I stated earlier" and it doesn't because the small pressure on toes shows small forces "80 N" but  the weight is big 570 N" Then the only force upwards is  the equivalent mass weight* , equivalent mass is 8 kg** and it presses ground with its weight 80 N and this because the force I use to lift my body is 80 N so the surface will push with same force of 80 N

*I choose equivalent mass weight here because it is all about body lifting itself.

**equivalent mass is 8 kg because if I am able to lift my body with 80 N then the weight must be slightly less than 80 N , if the weight is slightly less than 80 N the equivalent mass will be slightly less than 8 kg

Suppose I have another scale with a surface and a spring I stand on the surface to do the measurements by pressing the spring.

I refer to three forces :
1) The force I press on the spring surface to lift my body and it turns out to be 80 N as my scale reads in my experiment
2) The force gravity exerts downwards to be stored in the spring which turns out to be  the weight 570 N
3) The normal force the surface of the scale pushes upwards which equals the force I lift my body with " two forces in opposite directions"

If I stand on the scale I will do two things:
One I compress the spring with my 57 kg mass  by gravity force 570 N and the spring will store my weight 
The supposed normal force by the surface upwards equal to my weight doesn't exists in this case"for humans"

Second I will press the scale surface with my body mass but the mass here not equal to but equivalent to a smaller mass 8 kg so actually the surface pushes me with only 8 kg "80 N" and the force I lift my body with also small "the 80 N" 

When a body moves or lifts itself , body mass will be equivalent to a smaller mass, so inertia as well will be small"effortless walking and running" however force of gravity will be  for the actual mass. We can compare two equal masses on a two-legged scale one is an object 70 kg and the other is a body 70 k both will have the same gravity force  even though the body will have some equivalent mass
I hope this is more clear

9 hours ago, Ghideon said:

I think we have had this discussion already*. The forces are as predicted by physics. Please draw a simple diagram with the forces you are uncertain about so that we can clarify the situation and correct misunderstandings.

I have assumed you mean the action/reaction forces floor<->person.If you wish to discuss forces in the foot we can probably do that using a simple model.

The model is not a proof. So could we discuss how what I propose violates physics predictions ?

1 hour ago, Ghideon said:

Your are affecting the ground with the force of your weight* and ground will push back with an equal and opposite force. Stating on your toes does not change that. Basic physics and something you can verify with a bathroom scale or a seesaw. 

*) m*g

 

If the two forces are weight and ground force that will appear in the toes pressure , the toes pressure is slight and 60 kg is massive to the toes or body to bear.Yet the scale reads the actual mass 60 kg and the gravity force is for a mass of 60 kg

3 hours ago, swansont said:

These are not equivalent motions, so this comparison is not particularly useful

 

These are close:

I carry 20 kg rock and I jump compared to jumping with only my 60 kg weight 

Running while carrying the 20 kg rock vs running with only my 60 kg weight

20+60 kg will not give that significant difference 

21 minutes ago, Ghideon said:

If your maximum force in a toe press exercise would be 80N then your would not be able to stand on your toes if your weight is 57kg. 

In case I lift my body with small force the ground will push with the same small force the two forces will give slight pressure on the toes 

I thought it was convincing anyway I started the discussion again in the main Thread

These are other observations:

An average human can jump high raising his 60 kg weight while he barely moves a rock of 60 kg

A human can run fast with his massive 60 kg weight while he barely moves the same rock of 60 kg

A human can lift his body of 60 kg holding a bar many times but he will barely move a rock of 60 kg

This is an experiment of how a human can lift or moves his body mass with force less than the force needed to lift or move an equivalent mass:

The scale reads approximately 8 kg"80 N" which is the maximum force my calves' muscles can exert .So in the video for me to lift my 57 kg weight I should only use the maximum 8 kg"80 N" or less to lift my weight of 57 kg. This means in the video I only use 8 kg "80 N "or less to lift my massive body of 57 kg .

Video

Should I wait for my Thread in speculations  if it is good ? Or I need a next step?

Thanks,

1 hour ago, swansont said:

2. Spacetime is a coordinate system, not a substance

Mass curve space-time .What if  a piece "m" is inside M,  what  does it  do  to space-time? does it curve the space-time inside M? or the piece m doesn't affect space-time and the curvature is done by the whole M?

2 hours ago, joigus said:

Why do you say all space-time to be curved is inside the mass? (My emphasis.)

If I put a piece of sponge  in water the sponge  can't displace water , instead water will fill all the sponge .This is an example of space-time inside mass, the mass can't curve space-time because it is filled with it. No displace in water by the sponge no curvature.by mass.

on the other hand , If I put an empty bucket in water , this empty bucket represent mass with no space-time inside it, this bucket will displace water "curvature" , but in this case there will be no space-time points in the universe.
 

How space-time curvature works ?

If space-time exists everywhere including the mass itself, in this case a mass can't curve space-time because all space-time to be curved is inside it.Also if space time exists inside mass  then  existence or non-existence of mass are the same.

If space-time doesn't exists inside mass or mass displace space-time, then when a mass move it will leave a broken space-time.

I'm glad this can be tested that easily.

weight of 60 kg is 60*9.8 or 588 Newton which is a huge number.

Try to pull toes with your hands and compare it with lifting a 60 kg mass you will see how these muscle are weak and won't bear even far less  that mass

46 minutes ago, Ghideon said:

If forces are not balanced then the body is accelerating down through the earth. Does that match your observations?

The force which represent weight is one direction downwards opposed with small force upwards.Theses forces are not equal .The body will be at rest and no acceleration since it actually opposed by earth surface . in the case for instance the object accelerating into water , this is applied according to the calculation of what I suggested. Taking in consideration acceleration as well as water resistance

13 minutes ago, Bufofrog said:

I find it quite believable and quite unremarkable.

Thanks you.

45 minutes ago, Strange said:

Also, the total force on the ground will be exactly equal to the weight. And the force of the ground on the person will also be equal to that. So Newton is redeemed once again.

Thanks

46 minutes ago, pzkpfw said:

For one thing, a body is made of parts.

If you lift a 60 kg weight with your arm, the arm has to cope will all 60 kg.

A 60 kg person lifting themselves, does it in parts. Their neck lifts their head. Their core lifts their torso plus neck plus head. Their legs lift all the above.

It's not an apples to apples comparison. A 60 kg person doing a one-handed pull up or press up, would be a fairer comparison.

Thanks.

According to classical mechanics for a force to  lift a mass it should be slightly greater than its weight .

My hypothesis  is that a human body can lift itself  by a force far less than its weight .

It is obvious phenomenon that when lifting an object  of 60 kg up , it would be extremely hard than lifting one's body " 60 kg" .while standing.

This applied to many phenomenon  .A body will seem to have inertia far less than its actual mass inertia , moving and walking effortlessly , standing effortlessly , lifting one's body parts easily.

In this special case the Newtonian equations doesn't apply , however we could measure the ratio between the force lifting a body and the force lifting an object both body and the object have the same mass.

My first thread about this topic:

F=GmM/r²
This equation is not valid and wrong.
What is wrong with this equation?
Mathematics equations are a away to measure quantities related to other quantities .
Equations works for finite values, I measure r = 4 meters and find out F to be x Newtons
Equations won't work for infinite values , that is equations themselves  works finitely but not infinitely .

Let say F decreases without bound while r increases without bound, then the equation itself as a functional tool will approach infinity and the equation limit as r approaches ∞ is ∞, the equation is undefined and wrong.
If the equation definition is the tool that finds values of F for each value of r, then I can't find all values of F for all values of r , then the equation won't work for all values of F and r and it is useless and invalid.That means the equation itself approaches ∞ in measuring the quantities and undefined

I can say at F=0 , r doesn't exist and equals ∞, so the equation won't work  and undefined in such case and invalid, but some scientists might say F won't reach 0 ever , so I presented the explanation above.

If space-time is infinite ,how gravity extends to infinity? we know infinity is unreachable because it continues forever and no-one reach a finite point.How gravity extends to infinite distances while  infinity is unreachable? how gravity bends and curve space-time everywhere while space time end is unreachable? for gravity to bend space-time everywhere it should reach its end , how gravity bends space-time end while this end is unreachable?