The human body special mass

[Yahya A.Sharif]

If you do an experiment in which you lie on a block or beam, so that no part of your body other than your belly is in contact with any support, then you will feel a lot of pressure on it and could indeed injure yourself.

So you agree that lying on my belly on concrete block by my weight 59 kg must damage my belly severely just like putting the 59 kg rock on my belly.
In my experiment I did so. No contact on the chest or so, just the belly. And I bear it for 1 minute. Not even slight injury.

Also my body is balanced.

 

[exchemist]

So you agree that lying on my belly on concrete block by my weight 59 kg must damage my belly severely just like putting the 59 kg rock on my belly.
In my experiment I did so. No contact on the chest or so, just the belly. And I bear it for 1 minute. Not even slight injury.

Also my body is balanced.

OK, I think you are taking quite a lot on your lower ribs, though. I didn't say putting a 59kg weight just on your belly would damage it severely. I said might injure yourself, that's all. If you were to place the object you are draped across on your belly and put a weight on it, you would be able to manage it without too much trouble, so long as you braced your abdominal muscles appropriately. At the rowing club we sometimes would get someone to stand on our abdomens for fun, just to show how strong those muscles were.

 

[Michael 345]

If you do an experiment in which you lie on a block or beam, so that no part of your body other than your belly is in contact with any support, then you will feel a lot of pressure on it and could indeed injure yourself.

I took it to be

• lying on his belly on the block his belly is not crushed
• but
• the block on his belly would (yes it would) crush his belly

Yes that is so so silly but it is how I read what he posted

 

[James R]

Newton theories do not work for human body mass I have an experiment not just a theory.

By your own admission, you're too afraid to do the experiment in which you put the rock on your belly. So, you have half an experiment, so far. You need to do the other half to complete the experiment.

Why if the forces are the same the pressure on the belly by the rock is much bigger than the pressure by my body ?

It isn't. The pressure would be the same, as long as the contact area is the same, like I said previously.

Do you want me to prove that Newton is wrong?

Yes please.

 

[Yahya A.Sharif]

Why if the forces are the same the pressure on the belly by the rock is much bigger than the pressure by my body ?

It isn't. The pressure would be the same, as long as the contact area is the same, like I said previously.

It is. For the same area " The belly area" a smaller force" The bock" gives greater pressure" crushes the belly" while a greater force 59 kg do not crush the belly" lower pressure"

Do you want me to prove that Newton is wrong?

Yes please.

In the experiment What is the force f in the scale when the human lifts his body? what its value and why?

A human stands on a scale. The scale reads his mass 60 kg. If the human lifts his body up like someone trying to pick a fruit from a tree, he exerts a force "f" on the scale. The scale will exert the same force "f" upwards on the human, this force "f" by the scale is the force that lifts the body.
When the human lifts his body up like someone tries to pick a fruit from a tree with constant speed the scale will increase by f N in which the total measurement displayed on the scale will be 588+f N (the mass of the human displayed is 60 kg, the force of weight is 588 N, g=9.8 m/s/s) The scale will show the "f" force; which is the total read of the scale (588+f) minus the weight 588 or f N. This force f N which lifts the human is always less than 588 N or less than the human weight. In fact the force turns out to be relatively very small and this implies that the constant for any equation is a small fraction.

 

[James R]

It is.

How do you know? You haven't measured it. By your own admission, you have only done half the experiment. You're too scared to try the other half.

In the experiment What is the force f in the scale when the human lifts his body? what its value and why?

This is just another simple application of Newton's second law of motion.

When a person stands on a set of scales, he pushes down on the scales with a certain force, N. The scales apply on equal and opposite upwards forward on the person.

The forces on the person, then, are the person's weight, W, downwards, and the force from the scales, N, upwards. As an equation, the upwards acceleration of the person is then given by

$$a=\frac{N-W}{m}$$

where $$m$$ is the person's mass.

When the person is standing still on the scales, he is not accelerating, so $$a=0$$ and then we must have $$N=W$$. Since the reading on the scales is essentially the $$N$$ value, the scales show the person's weight in this case.

If the person stretches upwards (e.g. as if trying to pick fruit from a tree), then his centre of mass will rise. For a short time, during the stretch, the person will be accelerating upwards, which means $$a$$ in the above equation has a positive value (as opposed to $$a=0$$ when the person is just standing on the scales). It follows that the force $$N$$ must, for this short time, now be larger than the person's weight. Of course, when the person stops moving again, his acceleration goes back to zero and the scales again read his usual weight.

If the person holds his arms above his head and then lets them drop, as they are falling the person's centre of mass will be accelerating downwards for a short time. So, while this is happening the scales will momentarily read a value smaller than the person's weight.

There is no mystery in any of this. There's nothing special about a human body.

You can test the same equation by taking your scales into an elevator and placing a rock on the scales. Observe how the reading changes when the elevator starts or stops moving up or down (i.e. when the elevator is accelerating). Note that, when the elevator is moving at a constant speed, or is stationary, the scales just read the weight of the object on them.

Human bodies on scales in elevators do not behave differently to rocks on scales in elevators, in case you're wondering.

 

[Yahya A.Sharif]

In the experiment What is the force f in the scale when the human lifts his body? what its value and why?

This is just another simple application of Newton's second law of motion.

When a person stands on a set of scales, he pushes down on the scales with a certain force, N. The scales apply on equal and opposite upwards forward on the person.

The forces on the person, then, are the person's weight, W, downwards, and the force from the scales, N, upwards. As an equation, the upwards acceleration of the person is then given by

$$a=\frac{N-W}{m}$$

where $$m$$ is the person's mass.

When the person is standing still on the scales, he is not accelerating, so $$a=0$$ and then we must have $$N=W$$. Since the reading on the scales is essentially the $$N$$ value, the scales show the person's weight in this case.

If the person stretches upwards (e.g. as if trying to pick fruit from a tree), then his centre of mass will rise. For a short time, during the stretch, the person will be accelerating upwards, which means $$a$$ in the above equation has a positive value (as opposed to $$a=0$$ when the person is just standing on the scales). It follows that the force $$N$$ must, for this short time, now be larger than the person's weight. Of course, when the person stops moving again, his acceleration goes back to zero and the scales again read his usual weight.

If the person holds his arms above his head and then lets them drop, as they are falling the person's centre of mass will be accelerating downwards for a short time. So, while this is happening the scales will momentarily read a value smaller than the person's weight.

There is no mystery in any of this. There's nothing special about a human body.

You can test the same equation by taking your scales into an elevator and placing a rock on the scales. Observe how the reading changes when the elevator starts or stops moving up or down (i.e. when the elevator is accelerating). Note that, when the elevator is moving at a constant speed, or is stationary, the scales just read the weight of the object on them.

Human bodies on scales in elevators do not behave differently to rocks on scales in elevators, in case you're wondering.

All this is true for objects, not for a human body. I do not argue about that.
You did not answer my question. In the experiment. The scale shows 60 kg for the human body weight, when the human lifts his body the scale reading increases by 60+f kg, f is the increment in the weight, what this f kg represents?

A human stands on a scale. The scale reads his mass 60 kg. If the human lifts his body up like someone trying to pick a fruit from a tree, he exerts a force "f" on the scale. The scale will exert the same force "f" upwards on the human, this force "f" by the scale is the force that lifts the body.
When the human lifts his body up like someone tries to pick a fruit from a tree with constant speed the scale will increase by f N in which the total measurement displayed on the scale will be 588+f N (the mass of the human displayed is 60 kg, the force of weight is 588 N, g=9.8 m/s/s) The scale will show the "f" force; which is the total read of the scale (588+f) minus the weight 588 or f N. This force f N which lifts the human is always less than 588 N or less than the human weight. In fact the force turns out to be relatively very small and this implies that the constant for any equation is a small fraction.

 

[Yahya A.Sharif]

Calves' mucsles and feet muscles are weak so they lift the massive human 60 kg many times with force f of only several kilograms.

 

[Michael 345]

Calves' mucsles and feet muscles are weak so they lift the massive human 60 kg many times with force f of only several kilograms.

How it’s possible for an ordinary person to lift a car

https://www.bbc.com/future/article/20160501-how-its-possible-for-an-ordinary-person-to-lift-a-car

Makes interesting reading and perhaps something to chew the cud over

 

[Randwolf]

Yahya A.Sharif,

Your posts about "The human body special mass" are very, well, fascinating...


You can perform a simple two-part experiment to prove your theory that:

The human uses force far less than the maximum 60 kgF or far less than the human weight 60 kgF to lift his body.

You will need a volunteer from the audience registering 60 kg (or more) on a scale.



Part one relates to belly muscles:

Why when I lie on my belly on a concrete in the video I could bear my weight 59 kg for 1 minute, while if I put a 20 kg concrete block on my belly it will damage it severely?

Have your volunteer stand on your stomach for a minute. Extra points if your volunteer shows 80+ kg on the scale or if s/he jumps up and down a few times. Record the experiment on video.



Part two illustrates how a human

will not be able to even move a rock of 60 kg with his all body muscles

but

A human of 60 kg can lift his body up holding a bar many times with his arm muscles

Just record a video of you lifting your volunteer up many times with your arm muscles.



When you post the videos here everyone will understand.

Very simple...

 

[James R]

All this is true for objects, not for a human body. I do not argue about that.

1. The human body is not special. In terms of forces, it obeys the same laws of physics as any other object.
2. Clearly you believe the human body has special supernatural powers. Where do you think these powers come from? Are they magical? Does some god give them to you? Or what?

You did not answer my question.

Yes I did. The 'increment in weight' that you refer to will vary, depending on the acceleration of the centre of mass of the human body. I even supplied the relevant equations for you.

Calves' mucsles and feet muscles are weak so they lift the massive human 60 kg many times with force f of only several kilograms.

Those muscles aren't weak.

How have you measured the forces the muscles provide, seeing as you claim to know this?

 

[Yahya A.Sharif]

Those muscles aren't weak.

How have you measured the forces the muscles provide, seeing as you claim to know this?

In the video I measure my maximum calves and feet muscles force which turns out to be approximately 8 kg-force which is relatively weaker compared to other human body muscles.
I must exert the maximum 8 kg-force or less to lift my body as in the video. This force of 8 kg-force I lift my body with is far less than my weight 59 kg-force.
The movement of pushing the scale is equivalent to the movement to lift my body.

This 8 kg-fore or less is equivalent to the force "f " I lift my body with in the experiment:

A human stands on a scale. The scale reads his mass 60 kg. If the human lifts his body up like someone trying to pick a fruit from a tree, he exerts a force "f" on the scale. The scale will exert the same force "f" upwards on the human, this force "f" by the scale is the force that lifts the body.
When the human lifts his body up like someone tries to pick a fruit from a tree with constant speed the scale will increase by f N in which the total measurement displayed on the scale will be 588+f N (the mass of the human displayed is 60 kg, the force of weight is 588 N, g=9.8 m/s/s) The scale will show the "f" force; which is the total read of the scale (588+f) minus the weight 588 or f N. This force f N which lifts the human is always less than 588 N or less than the human weight. In fact the force turns out to be relatively very small and this implies that the constant for any equation is a small fraction.

How it’s possible for an ordinary person to lift a car

https://www.bbc.com/future/article/20160501-how-its-possible-for-an-ordinary-person-to-lift-a-car

Makes interesting reading and perhaps something to chew the cud over

This is not an ordinary person.

 

[Yahya A.Sharif]

You did not answer my question.

Yes I did. The 'increment in weight' that you refer to will vary, depending on the acceleration of the centre of mass of the human body. I even supplied the relevant equations for you.

The human lifts his body with constant speed " a=0" the force "f" is a constant . This constant f is the force a human lifts his body with.

A human stands on a scale. The scale reads his mass 60 kg. If the human lifts his body up like someone trying to pick a fruit from a tree, he exerts a force "f" on the scale. The scale will exert the same force "f" upwards on the human, this force "f" by the scale is the force that lifts the body.
When the human lifts his body up like someone tries to pick a fruit from a tree with constant speed the scale will increase by f N in which the total measurement displayed on the scale will be 588+f N (the mass of the human displayed is 60 kg, the force of weight is 588 N, g=9.8 m/s/s) The scale will show the "f" force; which is the total read of the scale (588+f) minus the weight 588 or f N. This force f N which lifts the human is always less than 588 N or less than the human weight. In fact the force turns out to be relatively very small and this implies that the constant for any equation is a small fraction.

 

[Yahya A.Sharif]

Yahya A.Sharif,

Your posts about "The human body special mass" are very, well, fascinating...

You can perform a simple two-part experiment to prove your theory that:


You will need a volunteer from the audience registering 60 kg (or more) on a scale.



Part one relates to belly muscles:



Have your volunteer stand on your stomach for a minute. Extra points if your volunteer shows 80+ kg on the scale or if s/he jumps up and down a few times. Record the experiment on video.



Part two illustrates how a human

but



Just record a video of you lifting your volunteer up many times with your arm muscles.



When you post the videos here everyone will understand.

Very simple...

Thank you. I will think about it.

 

[origin]

In the video I measure my maximum calves and feet muscles force which turns out to be approximately 8 kg-force which is relatively weaker compared to other human body muscles.

All I can tell you is that you have done a poor job on the experiment because the average person can easily produce a force of 59 kgf with one foot using their calf muscles.

 

[Yahya A.Sharif]

All I can tell you is that you have done a poor job on the experiment because the average person can easily produce a force of 59 kgf with one foot using their calf muscles.

This means if I am lying on the ground and put a 59 kg rock on one foot and exert force with my calf muscles of one leg then I can lift the rock that short distance with one foot ? An average human cannot do that. But an average human can lift his body like someone tries to pick a fruit from a tree short distance several times

 

[Dywyddyr]

This means if I am lying on the ground and put a 59 kg rock on one foot and exert force with my calf muscles of one leg then I can lift the rock that short distance with one foot ? An average human cannot do that. But an average human can lift his body like someone tries to pick a fruit from a tree short distance several times

That's because the leverage is different in each of those cases. Ergo, apples and oranges.

 

[Yahya A.Sharif]

That's because the leverage is different in each of those cases. Ergo, apples and oranges.

When I lie down on the ground and my legs are up and then putting a rock of 59 kg on foot sole and then tries to lift it not by my legs but by my foot " one foot" short distance equivalent to moving my body up short distance like someone tries to pick a fruit from a tree with one foot. The two movements are equivalent, are not they? I cannot lift the rock 59 kg but I can lift my body 59 kg several times.

Let's say for the experiment the foot is 20 cm or 0.2 meters long , Now let's calculate for a 0.2 m lever. First the lever will be class 2 :
The weight for 59 kg will be 59 kgf. Class 2 is the fulcrum at the toes , and in this case both the weight of my body and the force of my calves muscles I lift my body with will be at the ankle(The body can be balanced to be exactly vertical) The feet muscles specifically will be distributed throughout the feet but let’s assume we get their maximum strength in which the feet muscles are concentrated on the ankle:

F: the force of my weight

f: the force of my muscles.

L: the distance of the weight from the ankle to the toes.

l: distance of the muscles force from the ankle to the toes.

f * l=FL.

F=59 kgf and L=l =0.2 meters

f*0.2=59*0.2

f=59 kgf

The force needed to lift my body is 59 kgf. The foot as a lever does not change the force needed to lift the human when the human lifts his body like someone tries to pick a fruit from a tree. I must exert a force equals to my weight 59 kgf to lift my body(or the smallest force that is greater than my weight 59 kgf) In other words the foot as a lever does not increase or decrease the force to lift my body, the force to lift my body must be greater than my weight 59 kgf.

 

[origin]

This means if I am lying on the ground and put a 59 kg rock on one foot and exert force with my calf muscles of one leg then I can lift the rock that short distance with one foot ? An average human cannot do that.

Of course they can.

 

[origin]

I cannot lift the rock 59 kg but I can lift my body 59 kg several times.

Yes, you can raise a 59 kg weight if you can raise your body.
In the gym where I work out there is an apparatus that allows you to lift weights in an inclined position. I can easily lift a 60 kg weight with 1 foot using my calf muscles.

This thread is terribly silly. The issue is that you cannot or will not perform a simple experiment to prove it to yourself.