Physics of exercising and forces felt

[bo reddude]

TL;DR

How does momentum affect the forces our body feels?

Let's imagine an ideal scenario where you're lifting your own weight in its entirety. Let's say a woman weighing 100 lbs. Suppose she's doing an idealized handstand and pushup from that position. So she's lifting 100 lbs. Let's say ideally all of the forces are on her arms only. Do these forces on her arm increase or decrease due to momentum? So as she moves downward, the speed and the mass of the woman change the force she feels?

She would just feel F=ma she was only doing the handstand. But how about while she's moving downward and then up? How do I figure this out? Thanks for any help.

 

[kuruman]

HI @bo reddude and welcome to PF.


I am not sure what you mean by these forces increasing or decreasing due to momentum. Are you perhaps asking if these forces will be different in magnitude if the person is moving up as opposed to down?

 

[DaveC426913]

I suspect the OP is asking whether, at the bottom of a pull-up, is the subject feeling only 100lbs weight on her arms.

No. The subject's weight and her decel/acceleration upwards are cumulative.

This is why its not uncommon for people who swing on a rope to either let go too soon (because they can't resist the extra force) or have the rope break.

Am I about right, OP?

 

[kuruman]

I suspect the OP is asking whether, at the bottom of a pull-up, is the subject feeling only 100lbs weight on her arms.

If that's the case, then one can actually do the experiment at home without the necessity of a headstand, just a foot-stand. Stand on a bathroom scale and bend your knees up and down at different speeds keeping an eye on the display.

 

[DaveC426913]

If that's the case, then one can actually do the experiment at home without the necessity of a headstand, just a foot-stand. Stand on a bathroom scale and bend your knees up and down at different speeds keeping an eye on the display.

Yup. Exactly what I was thinking. Walking down a flight of stairs can put as much as 400 pounds of weight on one's feet.

 

[bo reddude]

HI @bo reddude and welcome to PF.


I am not sure what you mean by these forces increasing or decreasing due to momentum. Are you perhaps asking if these forces will be different in magnitude if the person is moving up as opposed to down?

Yes, that's exactly what I'm asking. What are the equations to be used in figuring that out?

 

[bo reddude]

I suspect the OP is asking whether, at the bottom of a pull-up, is the subject feeling only 100lbs weight on her arms.

No. The subject's weight and her decel/acceleration upwards are cumulative.

This is why its not uncommon for people who swing on a rope to either let go too soon (because they can't resist the extra force) or have the rope break.

View attachment 279378

Am I about right, OP?

Yes. I see where you're coming from. but the rope does break on occasion. why doesn't it break immediately at the start of the swing? because there is more force towards the end of the pendulum?

 

[bo reddude]

If that's the case, then one can actually do the experiment at home without the necessity of a headstand, just a foot-stand. Stand on a bathroom scale and bend your knees up and down at different speeds keeping an eye on the display.

Ok. That does make sense. It WILL change the weight. so the momentum does change the force felt.

 

[kuruman]

Ok. That does make sense. It WILL change the weight. so the momentum does change the force felt.

Not the momentum itself but the change in momentum. So if you are moving up and down at constant speed your muscles will exert the same force(s) as if you were at rest. It's only when you change your speed that your muscles exert more or less force than when you are at rest. Newton's second law says that force is the rate of change of momentum over time.

 

[A.T.]

She would just feel F=ma she was only doing the handstand. But how about while she's moving downward and then up? How do I figure this out?

With F=ma, where "a" is the acceleration relative to free fall (proper acceleration that an accelerometer measures).

 

[hmmm27]

Momentum doesn't play a huge role. Imagine lying horizontally on one of those mechanic's dollies for sliding underneath cars. Now, push against the wall : the effort it takes to get rolling up to speed is the change in "momentum".

Push at the same speed as a handstand-pushup extension. Not much effort required. In an actual handstand, almost all the sweat is from combatting gravity (and keeping balance).

 

[bo reddude]

can anyone actually write out the equations for this?

let's say, hand stand. 50 kg woman.

F_rest = 50 kg * 9.81m/s^2 = 490.5 kg*m/s^2

at rest with arm stretched out.

if she were to move down at 2 seconds interval and if her arm length is 14 inch or 0.3556 m. then at the bottom she's traveling at 0.3556m/s

then the p = 50 kg * 0.3556m/s = 17.78 kg*m/s

then how do i convert that to additional force? is it divded by the time it takes for that travel, ie., 2 seconds?

17.78kg*m/s * (1/2s) = 8.89 kg*m/s^2

then the final force acting on her body is the sum of those two numbers?

490.5 kg*m/s^2 + 8.89 kg*m/s^2 = 499.39 kg*m/s^2

Is that right?

 

[A.T.]

Momentum doesn't play a huge role. Imagine lying horizontally on one of those mechanic's dollies for sliding underneath cars. Now, push against the wall : the effort it takes to get rolling up to speed is the change in "momentum".

Push at the same speed as a handstand-pushup extension. Not much effort required. In an actual handstand, almost all the sweat is from combatting gravity (and keeping balance).

The required accelerations depend on how quickly you do those experiences.

Also, talking about "all the sweat" confuses the external physical force that the body produces with the physiological effort. Muscles are not perfectly efficient, and their ability to produce force is highly dependent on their current contraction/extension velocity.

 

[A.T.]

if she were to move down at 2 seconds interval and if her arm length is 14 inch of 0.3556 m. then at the bottom she's traveling at 0.3556m/s

At the bottom she needs to have 0m/s. The interval alone doesn't fully specify the accelerations during that interval. But for starters you could assume a sinusoidal movement and get the accelerations from that. For a more accurate estimate of the acceleration you would have to track the movement, or much simpler: put her on a scale.

 

[bo reddude]

At the bottom she needs to have 0m/s. The interval alone doesn't fully specify the accelerations during that interval. But for starters you could assume a sinusoidal movement and get the accelerations from that. For a more accurate estimate of the acceleration you would have to track the movement, or much simpler: put her on a scale.

but you still need the equations to explain it right?

 

[A.T.]

but you still need the equations to explain it right?

The support force is:
F = m * (a + g)
where a is the upwards acceleration of the center of mass relative to the ground.

But practically it's much simpler to measure F than a accurately.

 

[sophiecentaur]

but you still need the equations to explain it right?

Yes - @A.T. gives one in the above post.
But you also have to know some realistic values and you can only know the value of acceleration you are causing by using an acclerometer. Lucky that you can almost certainly get one for you cell phone and it will tell you the max, min and mean etc. of the acceleration.

A relevant point is that when you are doing an 'exercise' you may not want to be very efficient - just to build muscle or CV fitness, whereas, when you are aiming at stamina in running, you will develop running techniques that are efficient and make best use of what you've got. Hence, the answer to your question will probably be 'it depends' and measurement will be the only true indication of what's going on. There is no universal formula for the way your body works.

 

[A.T.]

Yes - @A.T. gives one in the above post.
But you also have to know some realistic values and you can only know the value of acceleration you are causing by using an acclerometer. Lucky that you can almost certainly get one for you cell phone and it will tell you the max, min and mean etc. of the acceleration.

@bo reddude Note that the accelerometer will already give you the (a + g) value, so you don't need add g to the accelerometer reading. You just multiply the accelerometers vector magnitude by mass to get force.

 

[bo reddude]

ok thanks everyone.

The support force is:
F = m * (a + g)
where a is the upwards acceleration of the center of mass relative to the ground.

But practically it's much simpler to measure F than a accurately.

so in the play example i gave above:

50kg woman in hand stand.

F_rest = 50 kg * 9.81m/s^2 = 490.5 kg*m/s^2

then the p = 50 kg * 0.3556m/s = 17.78 kg*m/s

17.78kg*m/s * (1/2s) = 8.89 kg*m/s^2

the final force acting on her body when moving up

490.5 kg*m/s^2 + 8.89 kg*m/s^2 = 499.39 kg*m/s^2

and the final force moving down

490.5 kg*m/s^2 - 8.89 kg*m/s^2 = 481.61 kg*m/s^2

that should be equivalent to the value from the accelerometer as below?

490.5 kg*m/s^2 +/- accelerometer value kg*m/s^2 =final force kg*m/s^2

does that seem reasonable?

 

[A.T.]

so in the play example i gave above:

As already said, your approach above doesn't make much sense:

if she were to move down at 2 seconds interval and if her arm length is 14 inch or 0.3556 m. then at the bottom she's traveling at 0.3556m/s

No, at the bottom her velocity is 0. Those 0.3556m/s is the average velocity for the whole descent.

then how do i convert that to additional force? is it divded by the time it takes for that travel, ie., 2 seconds?

No, you cannot use the whole 2 seconds, because only part of the decent is slowing down. And if you assume constant acceleration in that phase (as you do here), then you have to divide the maximal (not the average velocity) by the duration. But assuming constant acceleration is not reasonable in a case of dynamic biomechanics in the first place.

A better estimate would be to assume sinusoidal motion. Maximal acceleration can be found easily for this:
https://www.spaceagecontrol.com/calcsinm.htm

However, 'better' doesn't mean 'good'. I suspect the actual acceleration is quite different from such idealized models. This can be easily tested using a scale or a smart phone accelerometer.

 

[sophiecentaur]

A better estimate would be to assume sinusoidal motion.

That would be one step closer to the right model (if 'right' is the word for it).
If the OP feels the need to quantify things, an accelerometer or recording force meter would be of more use than trying to include an idealised model - which would need some measured values to be inserted in the first place. Parameters such as Maximum Acceleration and Mean Force would perhaps be appropriate but is there a need to go further? Perhaps including the measurements of individual's limbs (lengths and muscle cross section, maybe) could also be relevant.

 

[jbriggs444]

but the rope does break on occasion. why doesn't it break immediately at the start of the swing? because there is more force towards the end of the pendulum?

It is very helpful to have played on rope swings and chain swings as a child. That trains the intuition well. One can most definitely feel the increase in rope tension at the bottom of the swing's path and the easing of tension at the ends.

There are two effects going on. [Or at least one can account for things that way]. Both are most obvious when one is swinging very high. Both work to increase rope tension at the bottom of the swing path and decrease it at the ends.

The first effect has to do with components. The rope tension acts radially. Gravity acts vertically. The rope only has to resists that component of gravity which acts in a radial direction. If the swing is in a very high path so that the rope gets horizontal, the rope goes completely slack -- the component of gravity acting in the radial direction is zero. At the bottom of the path, gravity acts exactly in the radial direction and its contribution to rope tension is at a maximum.

The second effect has to do with centrifugal force (or with centripetal acceleration as your physics teacher may prefer to have you think). At the bottom of the swing's path when the rider is going fastest, the centripetal acceleration is the greatest. The rope tension supplies this centripetal acceleration.

When "pumping" very high swings (think circus trapeze rather than playground swings) at high amplitudes, the proper approach is to lift one's body at the bottom of the stroke and extend it back out at the ends -- doing work and injecting energy into the system when the resting force is greatest. If you can pump up to near horizontal, you'll be facing about 2 g's of tension at the bottom of the stroke and 0 g's at the ends (remembered, not calculated).

It is not easy to deal with 2 g's with hand strength (rope swings). I had the great good fortune to also have access to a 15 foot chain swing with a wooden "stand-up" bench. Getting that thing up to horizontal was not difficult. Getting it much past horizontal was very challenging. You lose a lot of energy and significant control when the chains go slack and then snap back into tension. I suspect that skate boarders face much the same challenge for identical reasons, feathering out their foot pressure above the top of a half-pipe.

 

[sophiecentaur]

Max G pulled on a swing: If you start with the rope horizontal you pull 2G at the bottom, however long the rope is (assuming it doesn't go slack). 3G Max on a loop if you only just make it to the top without falling down.

 

[JT Smith]

Similar with a spring. If you start from a neutral position of no extension and then let a rigid mass slump onto the spring, at maximum extension the force is twice the weight of the mass.

I wondered if someone doing a pushup (or squats) might decelerate more linearly in order to minimize the force. But apparently I was wrong:

https://iopscience.iop.org/article/10.1088/1757-899X/245/2/022053/pdf

 

[bo reddude]

It is very helpful to have played on rope swings and chain swings as a child. That trains the intuition well. One can most definitely feel the increase in rope tension at the bottom of the swing's path and the easing of tension at the ends.

There are two effects going on. [Or at least one can account for things that way]. Both are most obvious when one is swinging very high. Both work to increase rope tension at the bottom of the swing path and decrease it at the ends.

The first effect has to do with components. The rope tension acts radially. Gravity acts vertically. The rope only has to resists that component of gravity which acts in a radial direction. If the swing is in a very high path so that the rope gets horizontal, the rope goes completely slack -- the component of gravity acting in the radial direction is zero. At the bottom of the path, gravity acts exactly in the radial direction and its contribution to rope tension is at a maximum.

The second effect has to do with centrifugal force (or with centripetal acceleration as your physics teacher may prefer to have you think). At the bottom of the swing's path when the rider is going fastest, the centripetal acceleration is the greatest. The rope tension supplies this centripetal acceleration.

When "pumping" very high swings (think circus trapeze rather than playground swings) at high amplitudes, the proper approach is to lift one's body at the bottom of the stroke and extend it back out at the ends -- doing work and injecting energy into the system when the resting force is greatest. If you can pump up to near horizontal, you'll be facing about 2 g's of tension at the bottom of the stroke and 0 g's at the ends (remembered, not calculated).

It is not easy to deal with 2 g's with hand strength (rope swings). I had the great good fortune to also have access to a 15 foot chain swing with a wooden "stand-up" bench. Getting that thing up to horizontal was not difficult. Getting it much past horizontal was very challenging. You lose a lot of energy and significant control when the chains go slack and then snap back into tension. I suspect that skate boarders face much the same challenge for identical reasons, feathering out their foot pressure above the top of a half-pipe.

thanks for the insight. very well explained.

 

[bo reddude]

Similar with a spring. If you start from a neutral position of no extension and then let a rigid mass slump onto the spring, at maximum extension the force is twice the weight of the mass.

I wondered if someone doing a pushup (or squats) might decelerate more linearly in order to minimize the force. But apparently I was wrong:

View attachment 279543

https://iopscience.iop.org/article/10.1088/1757-899X/245/2/022053/pdf

that paper was very helpful in understanding this was a way more difficult question to ask than i had previously thought. thank you

 

[sophiecentaur]

But apparently I was wrong:

You could be right when dealing with someone who works efficiently. People doing gym exercises are probably not like that

 

[A.T.]

I wondered if someone doing a pushup (or squats) might decelerate more linearly in order to minimize the force. But apparently I was wrong:

You could be right when dealing with someone who works efficiently.

Minimizing the maximal force is not most efficient in terms of energy per squat. The muscles have to support your weight against gravity, so they waste a lot of energy, if you do the squat slower to minimize the accelerations and forces.

 

[JT Smith]

Minimizing the maximal force is not most efficient in terms of energy per squat. The muscles have to support your weight against gravity, so they waste a lot of energy, if you do the squat slower to minimize the accelerations and forces.

Good point. As the owner of a human body that should have been intuitively obvious to me. E.G., lowering slowly on a pull-up bar is very noticeably more energy consumptive than lowering quickly. The clock is always ticking unless you're hanging on your bones.

But with more than body weight I think the balance is different. Too fast and you may not be able to hold on.