The human body special mass

[Yahya A.Sharif]

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.
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Here is a picture of how a human lifts his body up short distance like someone tries to pick a fruit from a tree. https://ibb.co/s9c7R3g

You press the 60 kg weight by an equivalent movement in which you do not move your leg just your foot to lift it up short distance.

 

[origin]

You press the 60 kg weight by an equivalent movement in which you do not move your leg just your foot to lift it up short distance.

Yes, it is called a calf press. Any average person can easily press 60 kg using a machine or free weights.

 

[Yahya A.Sharif]

Any average person can easily press 60 kg using a machine or free weights.

An average human of weight 60 kgf cannot lift a rock of 60 kg with one foot. I am not sure about your weight. Try to lift a load equals your weight. I am also not sure about your muscles strength. Try do repeat lifting the weight and repeat lifting your body and see how many times you are able to do for each case.

 

[origin]

An average human of weight 60 kgf cannot lift a rock of 60 kg with one foot.

You are wrong, it is just that simple.

Try to lift a load equals your weight.

I've done it, no problem.

Here is the point, if you can raise yourself with one foot then you can raise 60 kgf with one foot. The thought that your body is somehow magically lighter than your weight is somewhat insane. You have been making this crazy claim for years and it is sad that you still believe it.

 

[exchemist]

An average human of weight 60 kgf cannot lift a rock of 60 kg with one foot. I am not sure about your weight. Try to lift a load equals your weight. I am also not sure about your muscles strength. Try do repeat lifting the weight and repeat lifting your body and see how many times you are able to do for each case.

Go to a gym and try it. There are machines there that allow you to raise weights by pushing with one or both legs. I think you will find you can do it without too much trouble.

 

[Yahya A.Sharif]

You have been making this crazy claim for years and it is sad that you still believe it.

Because no-one could give a logical scientific explanation for this:
https://www.youtube.com/shorts/KieNlBBgXYc
My weight is 57.7 kgf, when I lift my body the measurement rose to a maximum of 59.2 kgf. If I lift my heel 5 cm up then my body can return back to the ground with a weight of 60 kgf and distance 0.05 that is a work of 60*9.8*0.05= 29.4 joules, so I should exert the same work I got for energy conservation or I should exert 60 kgf by my calf muscles to lift my body this 5 cm distance. But the scale measures increment in force of only 1.5 kgf, 59.2-57.7=1.5 kgf. If you can give me a logical scientific explanation for this according to current physics then you solved my problem.

 

[DaveC426913]

If you can give me a logical scientific explanation for this according to current physics then you solved my problem

You are reading way too much into the scale's display. It is not capable of providing an accurate weight in real-time. That's why it goes blank for a moment when you first step on it - to hide these garbage readings.

The readings while you are in motion are nonsense. You cannot count them as data.

You won't like this answer, but it is - as you requested - the logical scientific explanation according to current physics, and it has - as you requested - solved your problem.

 

[Yahya A.Sharif]

You are reading way too much into the scale's display. It is not capable of providing an accurate weight in real-time. That's why it goes blank for a moment when you first step on it - to hide these garbage readings.

The readings while you are in motion are nonsense. You cannot count them as data.

You won't like this answer, but it is - as you requested - the logical scientific explanation according to current physics, and it has - as you requested - solved your problem.

Are you saying the scale should read 60+60 kg but it does not because it is not accurate? whether the scale is accurate or not what measurement should the scale read when someone lifts his body while standing on it? at the time I lift my body by what force I press the scale?

 

[DaveC426913]

The only readings that can be considered reliable on that scale are readings that are stable for at least, say, two seconds. Any readings that are stable for less than that time are junk and must be discarded.

You will find that the ONLY readings that are valid will give you 57.7kg.

 

[origin]

Are you saying the scale should read 60+60 kg but it does not because it is not accurate?

No, your scale will not read 120 kgf.

what measurement should the scale read when someone lifts his body while standing on it? at the time I lift my body by what force I press the scale?

Let's answer that question.

What is the force on the scale while standing still? We would use the equation F = ma.
F = 60kg * 9.8 m/s^2 = 588N or 60 kgf.

Now if you stand on your toes how much extra force are you exerting on the scale? You simply use the F = ma equation again. You have stated the distance you move up is 5 cm, it is closer to 10 cm when I measure myself but we will use your number.

So we need to calculate the acceleration when standing on your toes. We can use the following equation: $$ a = 2 \times \frac{(\Delta d - (v_i \times \Delta t)}{\Delta t^2}$$.
The initial velocity is zero, $$v_i \times \Delta t = 0$$

If we assume it takes 0.5 seconds to raise up on your toes, we have:
$$ a = 2 \times \frac{.05 m}{{0.25 s}^2} = 0.4 \frac{m}{s^2}$$

Now we can use F = ma. $$60 kg \times 0.4 \frac{m}{s^2} = 24 N $$ or 2.4 kgf.

So your claim is that when you stand on your toes the scale reading increases 1.5 kgf. My calculation comes out to an increase of 2.4 kgf. The differences in the numbers is due to several factors such as the actual height you raise yourself and the time it takes to raise to your toes.

At any rate you can clearly see that there is no way that the scale should read 120 kgf for simply standing on your toes.

Unless you can find an error in this analysis you must agree that there is nothing special about the weight of the human body.

I have the feeling you will find no error and yet you will not give up you silly claim, but I hope I am wrong.

 

[Yahya A.Sharif]

No, your scale will not read 120 kgf.

Let's answer that question.

What is the force on the scale while standing still? We would use the equation F = ma.
F = 60kg * 9.8 m/s^2 = 588N or 60 kgf.

Now if you stand on your toes how much extra force are you exerting on the scale? You simply use the F = ma equation again. You have stated the distance you move up is 5 cm, it is closer to 10 cm when I measure myself but we will use your number.

So we need to calculate the acceleration when standing on your toes. We can use the following equation: $$ a = 2 \times \frac{(\Delta d - (v_i \times \Delta t)}{\Delta t^2}$$.
The initial velocity is zero, $$v_i \times \Delta t = 0$$

If we assume it takes 0.5 seconds to raise up on your toes, we have:
$$ a = 2 \times \frac{.05 m}{{0.25 s}^2} = 0.4 \frac{m}{s^2}$$

Now we can use F = ma. $$60 kg \times 0.4 \frac{m}{s^2} = 24 N $$ or 2.4 kgf.

So your claim is that when you stand on your toes the scale reading increases 1.5 kgf. My calculation comes out to an increase of 2.4 kgf. The differences in the numbers is due to several factors such as the actual height you raise yourself and the time it takes to raise to your toes.

At any rate you can clearly see that there is no way that the scale should read 120 kgf for simply standing on your toes.

Unless you can find an error in this analysis you must agree that there is nothing special about the weight of the human body.

I have the feeling you will find no error and yet you will not give up you silly claim, but I hope I am wrong

How it is possible that I exert 2.4 kgf for 0.05 m, a work of 2.4*9.8*0.05 = 1.17 joules but when my body returns to the ground it exerts 60*9.8*0.05=29.4 joules that means I get more work than I exert? That why a mass is lift at least with force equals its weight so that the minimum work done is 60*9.8*0.05=29.4 joules equals the work that obtained when the human feet return to the ground 60*9.8*0.05. And because the scale measures the force exerted when lifting then it should measure at least the minimum force which is 60+60 kgf.

 

[DaveC426913]

Stop pretending a 20 dollar bathroom scale can produce real-time data in the subsecond range.

 

[sideshowbob]

My father used to use a $5 thermometer graduated in 2-degree increments to tell the Weather Office they were wrong.

In a Physical Chemistry lab, we calibrated $150 thermometers that were off by 1.5 degrees.

 

[origin]

And because the scale measures the force exerted when lifting then it should measure at least the minimum force which is 60+60 kgf.

I just showed you why that is not true! Is there an error in the analysis I did? You must think there is an error for you to continue to claim the scale should read 120kgf so please point out the error.

 

[exchemist]

How it is possible that I exert 2.4 kgf for 0.05 m, a work of 2.4*9.8*0.05 = 1.17 joules but when my body returns to the ground it exerts 60*9.8*0.05=29.4 joules that means I get more work than I exert? That why a mass is lift at least with force equals its weight so that the minimum work done is 60*9.8*0.05=29.4 joules equals the work that obtained when the human feet return to the ground 60*9.8*0.05. And because the scale measures the force exerted when lifting then it should measure at least the minimum force which is 60+60 kgf.

No. You are not thinking clearly.

The 2.4kgf is the increase in the scale reading. The scale didn't read zero when you were standing still on it. It read 60kgf. So now, when you push with your toes, it reads 62.4kgf instead of 60. So the work done is enough to raise your mass, plus an extra (small) contribution due to the upward kinetic energy your body gains as a result of moving.

The 60 + 60 is just totally wrong. I have no idea where you get that from.

 

[Michael 345]

How it is possible that I exert 2.4 kgf for 0.05 m, a work of 2.4*9.8*0.05 = 1.17 joules but when my body returns to the ground it exerts 60*9.8*0.05=29.4 joules that means I get more work than I exert? That why a mass is lift at least with force equals its weight so that the minimum work done is 60*9.8*0.05=29.4 joules equals the work that obtained when the human feet return to the ground 60*9.8*0.05. And because the scale measures the force exerted when lifting then it should measure at least the minimum force which is 60+60 kgf.

One aspect of this tin foil hat crazy thought is WHY

Why would the mass of a body (if I am reading correctly) have a different force than a mass of a different material?

 

[origin]

Why would the mass of a body (if I am reading correctly) have a different force than a mass of a different material?

Yahya thinks human mass is magical. Why he thinks that is because he read some physics that he didn't understand and tried to apply that misunderstanding to his bathroom scale.

 

[Yahya A.Sharif]

No. You are not thinking clearly.

The 2.4kgf is the increase in the scale reading. The scale didn't read zero when you were standing still on it. It read 60kgf. So now, when you push with your toes, it reads 62.4kgf instead of 60. So the work done is enough to raise your mass, plus an extra (small) contribution due to the upward kinetic energy your body gains as a result of moving.
The 60 + 60 is just totally wrong. I have no idea where you get that from.

The human standing has weight 60 kgf downwards and normal force of this weight 60 kgf upwards the two forces are equal and in opposite directions no work can be done by these two forces. The normal force 60 kgf upwards does not count. What counts is the force I exert with calf muscles which will do work not the normal force 60 kgf. So if I am standing still I need to exert force equals to my weight to lift it upwards. The two forces, the weight 60 kgf and the normal force 60 kgf are conserved they do not generate work upwards. What generates work is an added force by calf muscles this force is measured by the scale to be f kgf, because before the motion the scale reads 60 kgf, and after lifting the scale reads 60+f kgf.

 

[exchemist]

The human standing has weight 60 kgf downwards and normal force of this weight 60 kgf upwards the two forces are equal and in opposite directions no work can be done by these two forces. The normal force 60 kgf upwards does not count. What counts is the force I exert with calf muscles which will do work not the normal force 60 kgf. So if I am standing still I need to exert force equals to my weight to lift it upwards. The two forces, the weight 60 kgf and the normal force 60 kgf are conserved they do not generate work upwards. What generates work is an added force by calf muscles this force is measured by the scale to be f kgf, because before the motion the scale reads 60 kgf, and after lifting the scale reads 60+f kgf.

That's wrong. When you lift a 60kg mass through a metre, the work done is 60xg x 1, i.e. ~600N-metres, or 600J, of work.

 

[origin]

The human standing has weight 60 kgf downwards and normal force of this weight 60 kgf upwards the two forces are equal and in opposite directions no work can be done by these two forces. The normal force 60 kgf upwards does not count. What counts is the force I exert with calf muscles which will do work not the normal force 60 kgf. So if I am standing still I need to exert force equals to my weight to lift it upwards. The two forces, the weight 60 kgf and the normal force 60 kgf are conserved they do not generate work upwards. What generates work is an added force by calf muscles this force is measured by the scale to be f kgf, because before the motion the scale reads 60 kgf, and after lifting the scale reads 60+f kgf.

You are going off into the weeds. Did you understand the analysis I did on your scenario? Do you understand that physics says that the scale should only read a couple more kgf when you stand on your toes instead of an additional 60 kgf?