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A mass can be be lifted with force less than its weight


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1 hour ago, Prof Reza Sanaye said:

As an almost direct result , intelligent people like awaterpon come to fall in doubt as about the feasibility of

 

 

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.

You might want to use the quote function. The above looks like a support for awaterpon's claims, was that the intention?

 

Can you elaborate; how does what you describe result in awaterpon's claims flesh (and bones) and physics?

On 3/23/2021 at 10:04 PM, awaterpon said:

Flesh.Perhaps bones as well

 

 

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Lol considering I am only  5 foot seven and 155 lbs my legs can press 250 kg for ten reps on a Universal machine. 60 kg is nothing.

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 ext

You talking to me?   yes, I have done this. When I raise myself up on my toes, the reading on my scale increases above my weight by a few pounds. If I am settling back down, it decreases bel

Posted Images

 

I found this article and measurement of the reaction force from the ground on a person performing a standing jump.

https://www.researchgate.net/figure/Dynamic-and-kinematic-curves-for-the-countermovement-jump-sample-output_fig1_233734813

 

It includes this force plot I have highlighted in pink.

Note the researchers recorded a dip, as I did.

Quote

Dynamic-and-kinematic-curves-for-the-countermovement-jump-sample-output.jpg.2d1c87a13951c38376236389e47dd8b0.jpg

 

9 minutes ago, Ghideon said:

In the context of this thread I believe that Newton mechanics and specifically F=mg will predict what happens when a mass m is put on a typical household scale. Flesh or wood or bones or quick silver or gravestone or whatever will not have an effect*, Newtonian physics is applicable in this case. OP seems to argue that Newtonian physics fails to predict the force if the mass m consists of flesh. I fail to find any evidence in this thread or in any mainstream physics supporting that opinion. 

 

(Disclaimer: Of course not counting engineering limitations, using the scale outside of limits, not keeping the mass stable or other out of context reasons.)

 

I agree, Newton rules OK.

 

How about this.

You stand on the scale
You are quite still.

So the vertical momentum is zero.

You lunge upwards, acquiring vertical momentum.
Then come to a halt and are again still.

Before and after the vertical momentum is zero.

During the upward lunge this must still be true ie there must be a counterbalancing downward momentum somewhere.
In the scale mechanism.

So the mechanism reads a lowering of weight.

The faster the upward lunger the greater the rate of change of vertical momentum and the the greater the corresponding reduction in scale reading.

 

I am far from an expert in human body mechanics so I would welcome comments from those who are as to how this might work.

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@studiot Thanks for adding science to the thread. Letters a-h are from your image:

a: standing with straight legs without movement. Reading of scale is constant since body is at rest.
b: Bending knees. While doing this the centre of mass of body is initially accelerating down; any force from the scale acting on the body is less than the weight of the body. Scale is reading less than mass of body.
b-c: tightening of leg muscles stops acceleration down to have the body stationary with bent legs. While the body decelerating the scale reads more than the mass m.
c starting the upwards push. There is no flat segment of the curve so the body is not at rest with bent legs for any extended amount of time.
d: beginning to push/straighten legs to jump.
e: (approximately) heels leaves the scale since legs are straight, pushing with calves muscles; force is lower than while using upper leg muscles.
f: the body is airborne.
h: landing

There is a brief moment e-f where the scale would read less than the body mass while the body is standing on toes. The body is not at rest relative the scale during that period of time.

 I am not an expert, my interpretation may be completely incorrect.

Edited by Ghideon
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2 hours ago, studiot said:

 

During the upward lunge this must still be true ie there must be a counterbalancing downward momentum somewhere.
In the scale mechanism.

So the mechanism reads a lowering of weight.

 

So if you’re 60 kg, and the scale is 1 kg, you’re saying the scale mechanism is moving 60x faster than you are. Does that seem reasonable?

You bend your knees and then we start the exercise: you straighten up, moving ~1m, in 2 seconds. How can the scale compress at 30 m/s for 2 seconds?

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34 minutes ago, swansont said:

So if you’re 60 kg, and the scale is 1 kg, you’re saying the scale mechanism is moving 60x faster than you are. Does that seem reasonable?

You bend your knees and then we start the exercise: you straighten up, moving ~1m, in 2 seconds. How can the scale compress at 30 m/s for 2 seconds?

I don't believe I said any of this.

Clearly the countermomentum must act over the same timespan so nothing is moving 60 times faster.
But the scale spring or whatever must have stiffness so that the movement of its scale over that same timespan is only some fraction of the moving mass of the body.

I'm also not saying that when straightening up the bulk of the body mass moves anything like 1 metre.
If you are refering to the video it involved starting flatfooted on the scale and raising the body slowly to tiptoe, about 0.1m or 1/10 of that figure.

I also asked for some help posting a short video of the scale acting as I described.
Photographs I can do but they are not useful in this case.

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As you can somewhat tell from the force plot, from rest to final rest at the same point the downward force on the scale will average exactly the weight over time. (assuming an idealized constant gravitational field, and in a vacuum) 

Not all scales will show that exactly, but that is exactly what would happen.

4 hours ago, studiot said:

 

 

I am far from an expert in human body mechanics so I would welcome comments from those who are as to how this might work.

Same also....but the expert can only add incite into what the different parts of the plot might look like...they are still constrained by Newton's laws (which I think we all agree apply)

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11 hours ago, studiot said:

I don't believe I said any of this.

Clearly the countermomentum must act over the same timespan so nothing is moving 60 times faster.
But the scale spring or whatever must have stiffness so that the movement of its scale over that same timespan is only some fraction of the moving mass of the body.

I'm also not saying that when straightening up the bulk of the body mass moves anything like 1 metre.
If you are refering to the video it involved starting flatfooted on the scale and raising the body slowly to tiptoe, about 0.1m or 1/10 of that figure.

I also asked for some help posting a short video of the scale acting as I described.
Photographs I can do but they are not useful in this case.

I’m giving a scenario which maximizes the CoM motion, to show how your explanation can’t be correct. The momentum is taken up by the earth, not the scale (to first order). The scale does compress, and more than if you were just standing on it - that part is correct. But if it compresses more, the reading has to go up.

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1 hour ago, swansont said:

I’m giving a scenario which maximizes the CoM motion, to show how your explanation can’t be correct. The momentum is taken up by the earth, not the scale (to first order). The scale does compress, and more than if you were just standing on it - that part is correct. But if it compresses more, the reading has to go up.

So why does it go down ?

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34 minutes ago, studiot said:

So why does it go down ?

Please have a look at my attempt at explaining the researchers recording. My interpretation is that the scale reading is low while the body is moving down ("A" in picture). The individual in the measurement bends their knees fast and that results in a low force pressing the scale.

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1 hour ago, studiot said:

So why does it go down ?

Because you push less against it.

 

16 hours ago, Ghideon said:


b: Bending knees. While doing this the centre of mass of body is initially accelerating down; any force from the scale acting on the body is less than the weight of the body. Scale is reading less than mass of body.

This

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2 hours ago, studiot said:

So why does it go down ?

It doesn’t, if the scale works by measuring the normal force. 

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15 hours ago, swansont said:

It doesn’t, if the scale works by measuring the normal force. 

I assumed Studiot was referring to the deweighting phase, in preparation for the jump, when the normal force on the scale would be reduced below that of the weight.

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On 3/19/2021 at 3:13 PM, studiot said:

Note again that the reading on my bathroom scale drops when I stand on tiptoe.

 

On 3/19/2021 at 3:13 PM, studiot said:

At no time does it exceed 70kg.   -   the static value.

 

On 3/27/2021 at 8:38 PM, studiot said:

Note the researchers recorded a dip, as I did.

My point is: when during the attempt did they record dip? 

 

1 hour ago, J.C.MacSwell said:

I assumed Studiot was referring to the deweighting phase, in preparation for the jump, when the normal force on the scale would be reduced below that of the weight.

I did as well, thanks for your comment that made me look again. I now draw the conclusion that the scale is reduced below that of the weight a second time possibly matching @studiot's observation(?). First time is at approximately from a to c, while bending knees to prepare as mentioned before. Second time is near point f just before leaving the scale on the way up. Assume the individual would not actually jump but instead reduce their force just so that they end up standing on tiptoe. Then the flat segment would show the static value (the mass of the individual) at some point slightly later than f. 

 image.png.fa07ffc998017589eb235b792a269b4c.png 

 

I tried myself and unfortunately my digital bathroom scale locks the value once a static value is reached. My scale is not affected by movement once it has performed a measurement, I need to step off and wait for it to "reset". This has no impact on others observations, I'm just noting that bathroom scales seems to be constructed to behave rather differently under the dynamic conditions described in this topic.

Edited by Ghideon
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It’s true that if you did an “aborted jump” and ended on your toes the scale reading would go down, the description is not consistent with saying the force never exceeds the static value - it should be larger than that during the earlier time that the CoM is rising.

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8 hours ago, swansont said:

It’s true that if you did an “aborted jump” and ended on your toes the scale reading would go down, the description is not consistent with saying the force never exceeds the static value - it should be larger than that during the earlier time that the CoM is rising.

You are correct. I did not intend to argue that the force never exceeds the static value, sorry if my post was not clear on that. 

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10 hours ago, swansont said:

It’s true that if you did an “aborted jump” and ended on your toes the scale reading would go down, the description is not consistent with saying the force never exceeds the static value - it should be larger than that during the earlier time that the CoM is rising.

Yes but also prior to your jump. In preparation for it as you bend your knees and your center of gravity is allowed to dip below your initial position.

 

On 3/27/2021 at 6:05 PM, Ghideon said:

@studiot
b: Bending knees. While doing this the centre of mass of body is initially accelerating down; any force from the scale acting on the body is less than the weight of the body. Scale is reading less than mass of body.

 

Of course this (stage) doesn't happen if you don't "wind up" and instead simply roll up on your toes compressing the scale, but that is what is shown in the graph.

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  • 2 weeks later...
Posted (edited)
On 3/26/2021 at 7:53 PM, Prof Reza Sanaye said:

 

As an almost direct result , intelligent people like awaterpon come to fall in doubt as about the feasibility of

 

 

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.

The ratio between the alternative mass and the actual mass is constant. I can use Newton's laws separately one for a person pushing himself and the other is another person pushing him. Newtonian equation F=ma deals with any mass ,acceleration and force ,I have separate equation one has actual mass ,force and actual acceleration, the other is for alternative mass,force and alternative acceleration.

So:

F1=m1a1

This when a person pushes himself

F=ma 

This is when another external force pushes him.

 

 

 

Edited by awaterpon
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6 minutes ago, awaterpon said:

The ratio between the alternative mass and the actual mass is constant. I can use Newton's laws separately one for a person pushing himself and the other is another person pushing him however if I use the Newton's equation not separately the equation will not apply,.Newtonian equation F=ma deals with any mass ,acceleration and force ,I have separate equation one has actual mass ,force and actual acceleration, the other is for alternative mass,force and alternative acceleration.

So:

F1=m1a1

This when a person pushes himself

F=ma 

This is when another external force pushes him.

"Using Newton's Laws" and "using Newton's laws correctly" are two different claims.  There is no "alternative mass" and "alternative acceleration" This is just an excuse to do the analysis incorrectly.

 

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28 minutes ago, awaterpon said:

The ratio between the alternative mass and the actual mass is constant. I can use Newton's laws separately one for a person pushing himself and the other is another person pushing him. Newtonian equation F=ma deals with any mass ,acceleration and force ,I have separate equation one has actual mass ,force and actual acceleration, the other is for alternative mass,force and alternative acceleration.

So:

F1=m1a1

This when a person pushes himself

F=ma 

This is when another external force pushes him.

Apply* Newton's laws F=ma (2nd law) and when an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A (3rd law). You will find that the mass is always m, or in other words, that the ratio m/m1=1. Your definition will result in F1=F, m1=m and a1=a or an unphysical situation according to Newton.

I may get some time to post an example later if you wish.

*) correctly applying them that is.

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Trying to figure this out.  90 + posts essentially because someone doesn't understand the difference between force, pressure and work?

Edited by Ken Fabian
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I have managed to upload a 14 second video to youtube.

The person standing on the scale starts quietly at rest to give a still reading.
Then raises their heels slowly up a couple of inches and slowly returns back down to rest.
The repeats quickly.

The scale reading can be seen to drop in both cases.

https://www.youtube.com/watch?v=NCwgvQf05SU

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5 hours ago, awaterpon said:

The ratio between the alternative mass and the actual mass is constant. I can use Newton's laws separately one for a person pushing himself and the other is another person pushing him. Newtonian equation F=ma deals with any mass ,acceleration and force ,I have separate equation one has actual mass ,force and actual acceleration, the other is for alternative mass,force and alternative acceleration.

So:

F1=m1a1

This when a person pushes himself

F=ma 

This is when another external force pushes him.

 

 

 

Excuse me Awaterpon ; 

How can I be made to understand an alternative for mass and yet another alternative for acceleration  ??  !!  

 

+++++++++++++++++++++++++++++++++++++++++++++++++++ 

 

@ everybody ; 

I am real open to plough back the whole physics from the very beginning 

I am NOT talking rhetorically or metaphorically 

I am , rather , talking REALLY 

In case there  IS  anybody here offering any alternative brand of classical and/or modern physics , then I shall certainly be more than glad to read and learn ....

Edited by Prof Reza Sanaye
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On 4/13/2021 at 2:50 PM, studiot said:

I have managed to upload a 14 second video to youtube.

Unfortunately the video doesn't seem to be public, was that intentional? When I click the link YouTube responds with: 

"The video is unavailable"
"This video is private."

 

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37 minutes ago, Ghideon said:

Unfortunately the video doesn't seem to be public, was that intentional? When I click the link YouTube responds with: 

"The video is unavailable"
"This video is private."

 

+1

Thanks for the response I was hoping someone would try it as I have never put anything on youtube before.
It did say something about private and then something about waiting a day whilst it 'processed' when I asked to make it public. That dfay has now passed.

But I really don't know what I am doing so welcome all the help I can get.

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+1 for the effort but any discrepancy in the scale readings from what they would be due to simple calculations of Newton's laws will no doubt be due to the idiosyncrasies of the scale, assuming the test was otherwise accurately performed.

In both cases described it should drop...after it rises first. You simply cannot elevate from rest without some force acting with greater force than your weight, nor lower from rest without the net force being lower.

Either would violate conservation of momentum.

 

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