Jump to content

Recommended Posts

Posted

I'm pretty sure that the OP has posted this idea on other fora and failed to listen to the replies he got.

I can't see why it's going to work better here.
(and the same goes for his other ideas based on magic.)

Posted
8 hours ago, awaterpon said:


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

 

By experiment, you mean you’ve done this?

 

8 hours ago, awaterpon said:


2) The force gravity exerts downwards to be stored in the spring which turns out to be  the weight 570 N

Which is it? 80 N or 570 N? It won’t read both.

 

8 hours ago, awaterpon said:


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"

Sure it does.

8 hours ago, awaterpon said:

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" 

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

 

8 hours ago, awaterpon 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
I hope this is more clear

No, this makes no sense.

Posted
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

Posted
9 minutes ago, awaterpon said:

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

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

Posted
37 minutes ago, awaterpon said:

 

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"

This is fiction

 

37 minutes ago, awaterpon said:

The upwards normal force of gravity does not exist as I stated earlier"

I think this is something we can all agree on, as gravity and the normal force are two very different things.

37 minutes ago, awaterpon said:

 

**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

Again, this is fiction. This does not become true just because you say it.

 

Earlier you mentioned an experiment. Have you done an actual experiment? Or are you just making this up?

 

 

Posted (edited)
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 mentionimage2.thumb.png.7743bd4e892ddbc52c2bb0670cc30d34.pngimage1.thumb.png.b565f14a0ca6038d42e31a5c0ef9a1ee.png

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

Edited by awaterpon
Posted

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.

Posted
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

Posted
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

Posted
5 minutes ago, awaterpon said:

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

That's useless, it explains nothing.  How could you use a scale to measure this alternative weight?

Posted
1 hour ago, awaterpon said:

The right term is alternative mass

!

Moderator Note

I was hoping for a rigorous explanation, rather than a superficial hand-wave. 

Answer Bufofrog’s question (How could you use a scale to measure this alternative weight?) or this is finished.

 
Posted (edited)
1 hour ago, awaterpon said:

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

That seems like an attempt at describing something*. It is not a definition. 
 

*) It looks like a description of a phenomenon that simply does not exist. 

Edited by Ghideon
Grammar, clarity
Posted
On 3/12/2021 at 10:51 PM, Ken Fabian said:

Levers, gears and pulleys are all ways for less force to lift objects that weigh more (have greater gravity force) than that.

I felt the urge to quote this. Fun fact: A Greek influencer named Archimedes became famous for his obsession for lifting heavy objects with little force by using a lever.

Posted (edited)

 

1 hour ago, timo said:

Fun fact: A Greek influencer named Archimedes became famous for his obsession for lifting heavy objects with little force by using a lever.

That got me thinking; maybe this helps OP: In the video below the lecturer finds the force exerted on the Achilles tendon when a person is standing on tip-toes. The video will guide you through a simple model of the foot and lower leg and apply basic mathematics and physics.   

 

 Talking of archimedes; maybe OP can try standing on toe partially submerged in water and report back....

Edited by Ghideon
clarity
Posted

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.


 

Posted
41 minutes ago, awaterpon said:

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. 

!

Moderator Note

No, physics doesn’t work this way. Only external forces can cause an acceleration, and it has no effect on mass.

You need experimental evidence for your claims. You’ve been asked for it, and haven’t provided any, despite multiple references to your experiment.

Do you have an experiment? If you don’t answer this question, as your next action in this thread, it will be locked

 
Posted (edited)

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.

Edited by awaterpon
Posted (edited)
5 hours ago, awaterpon said:

Human body effortless walking, running  jumping, standing, dancing and other movements by human force on its own mass.

Let's try another point: "effortless" walking is a psychological effect, not physics, in this case. An experience of "effortless" walking is not an indicator that mass magically is reduced. An example of your flawed logic: Assume a well trained human A runs "effortlessly" at the same speed as a not so well trained human B. A and B has the same mass. The fact that B struggles to keep up with A does not mean that B have some unspecified "alternative mass", A and B have the same mass. B's struggle is more likely due to being less fit than A. 

Another example: After 10km of running I do not run effortless anymore. How much "alternative weight" have I gained according to your idea?

Have you investigated the possibility that human experience "less effort" because human benefits from not having to feel like walking requires a lot of effort?

Edited by Ghideon
added example
Posted
2 hours ago, awaterpon said:

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.

This is kind of irrelevant. Your scale reads a value, and we can compare those values, since g isn’t changing 

2 hours ago, awaterpon said:


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 .

How do you measure just your calves? 

 

 

Posted
3 hours ago, awaterpon said:

I used the scale to measure my maximum force of my calves' muscles I can exert it turns out to be  8 kg .

You screwed up your experiment, if you could only lift 8 kg with your calves you wouldn't be able to walk.  I can easily calf press 120 kg.

Posted (edited)
11 hours ago, Bufofrog said:

You screwed up your experiment, if you could only lift 8 kg with your calves you wouldn't be able to walk.  I can easily calf press 120 kg.

Lifting and walking both applies to my concept , if I lift my body with 8 kg I also use small force " 8 kg or less " to walk effortlessly.
 

12 hours ago, Ghideon said:
17 hours ago, awaterpon said:

Human body effortless walking, running  jumping, standing, dancing and other movements by human force on its own mass.

 

 

12 hours ago, Ghideon said:

Let's try another point: "effortless" walking is a psychological effect, not physics, in this case. An experience of "effortless" walking is not an indicator that mass magically is reduced. An example of your flawed logic: Assume a well trained human A runs "effortlessly" at the same speed as a not so well trained human B. A and B has the same mass. The fact that B struggles to keep up with A does not mean that B have some unspecified "alternative mass", A and B have the same mass. B's struggle is more likely due to being less fit than A. 

 

If both A and B have the same mass then  both will have the same alternative mass" alternative mass is constant " if both runs with the same speed then B needs to exert the same force that he can't bear " not trained"
 

 

12 hours ago, Ghideon said:

Another example: After 10km of running I do not run effortless anymore. How much "alternative weight" have I gained according to your idea?

 

You do not run effortless any more because you are  pushing with tiny force.
You do not gain more mass. The alternative mass is constant. Let's for instance a human of 60 kg has  10 kg alternative mass" alternative mass is always smaller than the actual mass" The Newtonian equation is separate whether the human  pushes himself or another person pushes him.
m=F/a
I can use the equation to calculate the two cases .In the first case when I lift myself the body will be lighter and will have greater acceleration substitute this acceleration will give smaller mass"alternative mass"
The second case the body will be heavier "the normal movement of the actual mass "  will have smaller acceleration and bigger mass this mass is the actual mass

 

11 hours ago, swansont said:

How do you measure just your calves? 

The whole lower leg's muscles are involved in lifting myself. greater percentage for my calves muscles', feet, front legs muscles are also involved.Lifting myself when trying to pick a fruit from a tree "in the experiment" and pushing the scale, both these movements are identical, I do them  with the same leg's muscles' force. I push the scale with 8 kg and this 8 kg lift my body 57 kg regardless what exactly muscles are involved.
 

Edited by awaterpon
Posted
1 hour ago, awaterpon said:

 The whole lower leg's muscles are involved in lifting myself. greater percentage for my calves muscles', feet, front legs muscles are also involved.Lifting myself when trying to pick a fruit from a tree "in the experiment" and pushing the scale, both these movements are identical, I do them  with the same leg's muscles' force. I push the scale with 8 kg and this 8 kg lift my body 57 kg regardless what exactly muscles are involved.

 

I'm not asking for your explanation. I'm asking how you are actually doing this measurement; i.e. I want an explanation as if I were going to try and replicate it myself. You stand on the scale and it reads 57 kg. When you reach up like you're picking fruit it drops to 8 kg?

 

Posted (edited)
1 hour ago, swansont said:

I'm not asking for your explanation. I'm asking how you are actually doing this measurement; i.e. I want an explanation as if I were going to try and replicate it myself. You stand on the scale and it reads 57 kg. When you reach up like you're picking fruit it drops to 8 kg?

 

If I stand on scale surface the scale will read my weight 57 kg when I press the surface to pick a fruit from a tree I will push by small force 8 kg or less to raise my body*

While I lifting myself the scale will start reading small forces x in which the scale total read will read 57+x kg . I push the scale with small forces maximum 8 kg and the total force on the scale will be 57+8 kg or less when I stop upwards the scale will drop to 57 kg which is my actual  weight"no forces acting"

*These details are above.

Edited by awaterpon

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.