# Impulse/force in pounds for the time frame

by waynexk8
Tags: frame, impulse or force, pounds, time
 P: 399 A question please in three parts, need the numbers for the first before I can asked the next. A Machine lowers from rest, 100 pounds under control, at 2m/s, for 1000mm. Then immediately stops the weight, and lifts it back up at 2m/s. At the transition from negative to positive, what impulse force in pounds, would be the maximum on the Machines components/parts, and for how long, untill the normal acceleration forces that would be on the components/parts if it lifted the weight from rest. Just in case I did not explain right. The force on the components/parts, lifting from rest would keep getting higher, lets go for every 10th of a second. Say from the lift at rest, the first 10th of a second would have at that vilocity ??? 105 pounds of force needed to lift the weight at that vilocity, then the next 10th of a second it may need 110 pounds as the vilocity went up, and more and more force would be needed to keep the vilocity going up, untill the decceleration. But what would be the impulse/force on/from the components/parts if the weight was lowered at the above vilocity, and for how long in 10ths or whatever in time would that higher force have to be untill the normal acceleration forces of the lift at rest. Thank you for your time and help. Wayne
 P: 399 Sorry, etited the above slightely, as the machine only has a maximum force to use of a 125 pounds. However if the weight has been falling that fast, and is stopped fast, the force on the machine will be well over the 125 pounds, even thou when it does slow down stop and relift, the weight will keep going down for a few mm’s, the force then on it will be far higher than the 125 pounds. Wayne
 Sci Advisor HW Helper PF Gold P: 6,038 Say Wayne, 1.) Bodies in motion cannot immediately (instantaneously) come to a stop; they must stop (decelerate) over a time interval. The impulse force depends on that time interval. 2.) why is your force measured in USA units and distances/velocities in SI units? You'll have to convert one to the other. 3.) If the time interval to stop is rather small, net impulse force will be very high; if the time is large, impulse force will be much less. 4.) If you are lifting a 100 pound body from rest with your machine rated at 125 pounds, to a constant speed of 2m/s (say 6.5 ft/s), then the max net force is 25 pounds, the acceleration is about 8 ft/s (per Newton 2), and the time interval during the accelerating phase must be about 1 second to stay within the limits of the machine.
P: 399
Impulse/force in pounds for the time frame

 Quote by PhanthomJay Say Wayne, 1.) Bodies in motion cannot immediately (instantaneously) come to a stop; they must stop (decelerate) over a time interval. The impulse force depends on that time interval.
Hi PhanthomJay, thank you for your time and help, and to anyone else who joins in the debate.

Yes agreed, there would have to be a deceleration from the eccentric motion to the concentric, and in this/that time frame there would be the highest force used by the machine, but more important, in this very short deceleration phase there would be also the highest force “on” the machines parts/components. I am unable to work out the extra force from the weight coming down as of the acceleration component, or am I able to work out that force that will be on the machine, or the time frame that this deceleration too acceleration would take. I am no physicist, however this is one of a very large and long debate, all I can give are the weights, distances and other times frames.

 Quote by PhanthomJay 2.) why is your force measured in USA units and distances/velocities in SI units? You'll have to convert one to the other.
SORRY about that, I am from the UK, and a lot of us work in both, however it was wrong of me to put both down, imperial or metric is fine.

 Quote by PhanthomJay 3.) If the time interval to stop is rather small, net impulse force will be very high; if the time is large, impulse force will be much less.
Yes very true, as the weight is being moved quite fast at .5 of a second concentric, and point .5 of a second eccentric, the net impulse force will be relatively very high. As we all know, the deceleration usually takes less force, however in this example it will take more, as the machine is try first to stop the fast downward motion of the weight, with its appeared extra weight as of the acceleration, then to acceleration it again, thus throughout the whole deceleration the machine is trying to accelerate the weight.

Some people are saying that the force on the machine will be as much as 300 pounds for a Milly second, I would not say that much. However, if it able to be worked out, it would be “very” interesting to find out it’s as ??? 180 pounds for .2 of a second, or what the real numbers are.

 Quote by PhanthomJay 4.) If you are lifting a 100 pound body from rest with your machine rated at 125 pounds, to a constant speed of 2m/s (say 6.5 ft/s), then the max net force is 25 pounds, the acceleration is about 8 ft/s (per Newton 2), and the time interval during the accelerating phase must be about 1 second to stay within the limits of the machine.
Ok, see what you are saying, could we go for the weight being lifted for just 500mm up, and 500mm down instead of 1m up and 1m down ??? Not sure if I should be saying this, as it may complicate matters, but actually it’s a Man lifting the weight, and a Human muscle on average can lower 40% more than it can lift its one time maximum. However if you do not need to know this, and can would it out for just the machine that would be better. I said a machine, as I did not want to get involved that a Man could not lift at constant forces, meaning because of the biomechanical advantages and disadvantages thought the ROM {Range of Motion} its impossible to life like a machine.

Wayne
 Sci Advisor HW Helper PF Gold P: 6,038 It really depends on the nature of the movement. If you were to raise the weight halfway up (1/4 m) in 1/4 second at constant acceleration, the acceleration would be 8 m/s^2, and since Fnet=ma, where a is 8 and m is about 4.5 kilos, then Fnet = 36 N up, which means that since the weight acts down at about 450 N, you got to push up at 486 N, or about 110 pounds. For the remainder of the upward thrust, you've got to decelerate to 0, and the weight helps you to decelerate, so the net force is 36 N down, which means you only need to apply a force of 414 N up, or about 90 pounds. Same on the downstroke, 90 pounds to the halfway point and 110 pounds back to the beginning. But it is more likely in weightlifting that the acceleration phase takes place in a much shorter time period and distance, say 0.1 second for a distance of say 1/8 m; then its constant controlled velocity from this point until you decelerate near the top. In which case your initial acceleration is 25 m/s^2, Fnet = 110 N, or your force is 450 + 110 = 560 N, or about 125 pounds. These are just rough numbers for talking purposes only, but the bottom line is that the initial liftoff and final dropoff requires the greatest force, and perhaps over 150 pounds if you accelerated quickly over an even shorter time.
P: 399
 Quote by PhanthomJay It really depends on the nature of the movement. If you were to raise the weight halfway up (1/4 m) in 1/4 second at constant acceleration, the acceleration would be 8 m/s^2, and since Fnet=ma, where a is 8 and m is about 4.5 kilos, then Fnet = 36 N up, which means that since the weight acts down at about 450 N, you got to push up at 486 N, or about 110 pounds. For the remainder of the upward thrust, you've got to decelerate to 0, and the weight helps you to decelerate, so the net force is 36 N down, which means you only need to apply a force of 414 N up, or about 90 pounds. Same on the downstroke, 90 pounds to the halfway point and 110 pounds back to the beginning.
Very interesting.

So am I right in saying, that on the above example, if we split the whole upward motion into 10 parts, and I accelerated up for 60% that it “may” look like this if we could plot my force somehow;

110
110
110
110
110
110
90
70
50
30

Then zero for the transition from positive to negative.
Total, just for a reference number 900.

 Quote by PhanthomJay But it is more likely in weightlifting that the acceleration phase takes place in a much shorter time period and distance, say 0.1 second for a distance of say 1/8 m; then its constant controlled velocity from this point until you decelerate near the top. In which case your initial acceleration is 25 m/s^2, Fnet = 110 N, or your force is 450 + 110 = 560 N, or about 125 pounds. These are just rough numbers for talking purposes only, but the bottom line is that the initial liftoff and final dropoff requires the greatest force, and perhaps over 150 pounds if you accelerated quickly over an even shorter time.
Even more interesting.

So am I right in saying again, that on the above example, if we split the whole upward motion into 10 parts again, and I accelerated up for 60% But came off the eccentric to the concentric where the peak forces are, that it “may” look like this if we could plot my force somehow;

150
125
125
125
110
110
90
70
50
30

Then zero for the transition from positive to negative.
Total, just for a reference number 985.

As you “said” those numbers are just rough numbers for talking purposes only, but would the above be a rough estimation only ??? As some people are saying that if you lift the same weight “very” slow for say 166mm to very fast for say 1000mm in the same time frame, that the same total or overall force will be used by or and on the muscles, we all know that you have to use more energy doing anything faster in the same time frame, and as you fail far far far faster when doing faster repetitions, my opinion is that’s it’s you have to use more total or overall force ??? My EMG {a machine that reads the electrical activity in the muscles, the force or strength that the muscles are using} machine also gave me a higher readout force the faster repetitions.

First is the most important, as it shows the EMG average muscle activity, or muscle force/strength used, as we know, the higher the average, the higher the total/overall force/strength used.

1, Fast 409, Slow 349,
2, Fast 437, Slow 346,
3, Fast 0.1, Slow 0.3,
4, Fast 0.6, Slow 0.7,
5, Fast 1104, Slow 1114,
6, Fast 146.0, Slow 193.4,
7, Fast 175.0, Slow 173.0,

1/ WRK This is the work average for the session measured in [µV]
AVG microvolts. The average readings will vary from one patient to
another.

2/ RST This is the rest average for the session measured in µV - Below
AVG 4 µV a muscle is beginning to rest.

3/ AVG This is the average onset of muscle contraction measured in
ONST seconds, readings below 1 second can be considered normal.

4/ AVG This is the average muscle release measured in seconds,
RLSE readings below 1 second can be considered normal.

5/ W/R This is the average peak value measured in µV - The value will
PEAK vary from one patient to another.

6/ WRK This is the average muscle deviation when contracting the
AVDV muscle. Readings of below 20% of WRKAVG can be
considered adequate, below 12% can be considered good.

7/ RST This is the average muscle deviation when the muscle is at rest.
AVDV Below 4 µV a muscle is beginning to rest.

MORE thank a big thank you for your time and help.

Wayne
 P: 253 There is little value in guessing some possible values, because these guesses are made without knowledge of whether they are even likely or possible. It seems you are understanding the concepts just not the values of the numbers. The only cure for this, I believe, is with actual proven data, not with guesses as to what it could be if it were like this or that. Perhaps it would surprise you, or perhaps not, but guessing is no answer.
P: 148
 Quote by waynexk8 Very interesting. So am I right in saying, that on the above example, if we split the whole upward motion into 10 parts, and I accelerated up for 60% that it “may” look like this if we could plot my force somehow; 110 110 110 110 110 110 90 70 50 30 Then zero for the transition from positive to negative. Total, just for a reference number 900. Wayne
Your numbers are just figments of your imagination.
What part of PhanthomJay's answer didn't you understand?
He told you that for the first half you have to apply force equal with 110 pounds and for the second equal with 90 pounds.

Just to prevent you from posting for pages and pages until the thread will be closed...I'll ask three questions:

1)Do you understand that when you lift a weight the average force is always equal with the weight?
2)Do you understand that when you lift a weight the average acceleration is always zero?
3)Do you understand that when you lift a weight the applied force over the duration of the lifting(or else the "total force" as you call it) is always equal with gravity's impulse?
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P: 6,038
 Quote by waynexk8 Very interesting. So am I right in saying, that on the above example, if we split the whole upward motion into 10 parts, and I accelerated up for 60% that it “may” look like this if we could plot my force somehow;
Assuming you accelerate up half the distance to the top and reach a speed of 2m/s, then accelerate down from a speed of 2m/s to a momentary stop at the top, then continue accelerating down as you lower the weight to midway to 2m/s, and finally accelerating up as you lower to its start point, then the numbers read more like this:

UPSTROKE
110
110
110
110
110
11090
90
7090
5090
3090

DOWNSTROKE
90
90
90
90
90
110
110
110
110
110

 Then zero for the transition from positive to negative.
When you are talking force, the force you apply is always positive (upward), whether on the up or down stroke. The force you apply is not zero until you fully release (let go of) the weight (or drop the weight if you're a lunk. )
 Total, just for a reference number 900.
the total of these up forces divided by 10, or down forces divided by 10, or up and down forces divided by 20, gives you the average force during the cycle, or 100 pounds average force. This would be the force required if you lifted up and down very slowly at constant speed without accelerating. In the example acceleration case, the force you apply varies from 110 to 90 pounds.
 Even more interesting. So am I right in saying again, that on the above example, if we split the whole upward motion into 10 parts again, and I accelerated up for 60% But came off the eccentric to the concentric where the peak forces are, that it “may” look like this if we could plot my force somehow;
more like, for the upstroke in the example I gave,

150125
125100
125100
125100
110100
110100
90100
70100
50100
3075
 Then zero for the transition from positive to negative. Total, just for a reference number 985.
again average force = 100 pounds.
 As you “said” those numbers are just rough numbers for talking purposes only, but would the above be a rough estimation only ???
Yes, rough, actual values depend on how you choose to do the lifting. Move it slowly at constant speed for the duration, and your applying more or less 100 pounds of force throughout
 As some people are saying that if you lift the same weight “very” slow for say 166mm to very fast for say 1000mm in the same time frame, that the same total or overall force will be used by or and on the muscles, we all know that you have to use more energy doing anything faster in the same time frame, and as you fail far far far faster when doing faster repetitions, my opinion is that’s it’s you have to use more total or overall force ???
You have to use more force during the accelerating phase up, so if you start off slowly you are using 100 pounds of force, but then as you pick up speed the force you apply rises to 125 or 150 or more, until you decelerate at the top using much less force at the final transition from upward to downward thrust. But you are correct, for most of the duration you are applying a force much larger than 100 pounds, so the average force is also higher than 100 pounds. Better to move slowly in a controlled manner.
 My EMG {a machine that reads the electrical activity in the muscles, the force or strength that the muscles are using} machine also gave me a higher readout force the faster repetitions.
yes, good machine!
 First is the most important, as it shows the EMG average muscle activity, or muscle force/strength used, as we know, the higher the average, the higher the total/overall force/strength used. 1, Fast 409, Slow 349, 2, Fast 437, Slow 346, 3, Fast 0.1, Slow 0.3, 4, Fast 0.6, Slow 0.7, 5, Fast 1104, Slow 1114, 6, Fast 146.0, Slow 193.4, 7, Fast 175.0, Slow 173.0, 1/ WRK This is the work average for the session measured in [µV] AVG microvolts. The average readings will vary from one patient to another. 2/ RST This is the rest average for the session measured in µV - Below AVG 4 µV a muscle is beginning to rest. 3/ AVG This is the average onset of muscle contraction measured in ONST seconds, readings below 1 second can be considered normal. 4/ AVG This is the average muscle release measured in seconds, RLSE readings below 1 second can be considered normal. 5/ W/R This is the average peak value measured in µV - The value will PEAK vary from one patient to another. 6/ WRK This is the average muscle deviation when contracting the AVDV muscle. Readings of below 20% of WRKAVG can be considered adequate, below 12% can be considered good. 7/ RST This is the average muscle deviation when the muscle is at rest. AVDV Below 4 µV a muscle is beginning to rest. MORE thank a big thank you for your time and help. Wayne
I'm not sure of all these numbers, but it shows that the actual motion and muscular activity is complex. In particular, for item 1, it shows that more force is needed when moving 'fast'. I think I'll try this at the gym this weekend....but using 20 pound weights!
P: 399
 Quote by douglis Your numbers are just figments of your imagination. What part of PhanthomJay's answer didn't you understand? He told you that for the first half you have to apply force equal with 110 pounds and for the second equal with 90 pounds. Just to prevent you from posting for pages and pages until the thread will be closed...I'll ask three questions: 1)Do you understand that when you lift a weight the average force is always equal with the weight? 2)Do you understand that when you lift a weight the average acceleration is always zero? 3)Do you understand that when you lift a weight the applied force over the duration of the lifting(or else the "total force" as you call it) is always equal with gravity's impulse?
By even asking those questions you don’t yet see that there are not just the ONE force at work, there are TWO, at least, thus you have to add then up, you seen to be only adding the one force up ??? why. If you go over to BB.com, and please look at what I have named a few of my threads, then you might get it.

I have tried to tell you over and over, that {see what PhanthomJay states as well} when you lift a weight with more velocity or and acceleration, you have to use more force, and the more force you apply, the more the {not sure if I am saying this right, however I am try} weight will give an opposite reaction force, as to every action there is always an equal and opposite reaction, this opposite reaction, then puts more force on the pushing force, the muscles.

Or we could say; The g-force, associated with an object in its acceleration. So we have accelerations produced by gravity, and non-gravitational forces, proper accelerations. I explained this in my Clay example, in other simple words, the faster you push up on and object, the more it pushes down on you. From a site; G-forces, when multiplied by a mass upon which they act, are associated with a certain type of mechanical force in the correct sense of the term force, and this force produces compressive stress and tensile stress. Such forces result in the operational sensation of weight.

As there are two or more forces in action here, the force “from” the muscles, and the force
“on” the muscles from the weight being moved.

Wayne
P: 399
 Quote by PhanthomJay Assuming you accelerate up half the distance to the top and reach a speed of 2m/s, then accelerate down from a speed of 2m/s to a momentary stop at the top, then continue accelerating down as you lower the weight to midway to 2m/s, and finally accelerating up as you lower to its start point, then the numbers read more like this: UPSTROKE 110 110 110 110 110 11090 90 7090 5090 3090
However PhanthomJay, what about the opposite reaction forces, the force “ON” the muscles from the weight itself being moved ??? As if I lifted a 100 pouinds very slowely, there would be no extra forces on the muscles as I call it, the opposiot reaction forces, but there are all the time reactuion forces on the muscles from the weight itself when moving faster.

Now it’s gets even more interesting. As many times I have thought about the deceleration, and would have thought your numbers were more correct, as when I do lift weight in this very explosive style, I would say it’s far more like that. As I “try” to accelerate up all the way, but as its harder to accelerate the more you accelerate, it’s getting harder, thus I have to use less force, and my body is sub concisely telling me I have to decelerate and stop and do the transition to the opposite direction soon in a Milly second.

 Quote by PhanthomJay DOWNSTROKE 90 90 90 90 90 110 110 110 110 110 When you are talking force, the force you apply is always positive (upward), whether on the up or down stroke. The force you apply is not zero until you fully release (let go of) the weight (or drop the weight if you're a lunk. ) the total of these up forces divided by 10, or down forces divided by 10, or up and down forces divided by 20, gives you the average force during the cycle, or 100 pounds average force. This would be the force required if you lifted up and down very slowly at constant speed without accelerating. In the example acceleration case, the force you apply varies from 110 to 90 pounds.more like, for the upstroke in the example I gave, 150125 125100 125100 125100 110100 110100 90100 70100 50100 3075again average force = 100 pounds. Yes, rough, actual values depend on how you choose to do the lifting. Move it slowly at constant speed for the duration, and your applying more or less 100 pounds of force throughout You have to use more force during the accelerating phase up, so if you start off slowly you are using 100 pounds of force, but then as you pick up speed the force you apply rises to 125 or 150 or more, until you decelerate at the top using much less force at the final transition from upward to downward thrust. But you are correct, for most of the duration you are applying a force much larger than 100 pounds, so the average force is also higher than 100 pounds.
Yes this is what I have always said, as these MUST be a higher average force, as you are developing more power, and using more energy, thus WHAT else could you be using more of if you’re travelling far far far more distance in the same time frame and using more energy, it MUST be more “FORCE” if not what ???

A question please, if you use more power, does that mean you use more overall or total force ??? Total or overall forces in the meaning as, if you only had a 100 pounds of force, or strength in my case, before your force runs out, and it can and does in the human body quite fast if you are lifting 80%

 Quote by PhanthomJay Better to move slowly in a controlled manner.yes, good machine!I'm not sure of all these numbers, but it shows that the actual motion and muscular activity is complex. In particular, for item 1, it shows that more force is needed when moving 'fast'.
Yes in my and many other people opinions you have to use more force to lift faster, it’s like trying to lift 100% with just 80% it will just not go up, as you’re not using enough force.

 Quote by PhanthomJay I think I'll try this at the gym this weekend....but using 20 pound weights!
Sounds cool,

Could I ask another question, it’s sort of on impulse. As my force lifts the weight faster and for more distance in the same time frame, some people think that when you lift slow, that when my force is on the deceleration, its where the slow moving people make up or balance out the forces, however I think no, as if this was true, then they should then also make up the difference in distance moved, but they don’t, meaning the lower force don’t make up or balance out the high force when they are on lower forces as of the decelerations.

That I also think I may have proved that the slow lifting forces cannot make up or balance out the fast lifting forces with my clay theory. Put a piece of clay between your hand and the weight, lift a weight up and down one time for 3 seconds each way, then lift the weight up and six times at .5 of a second each way. The clay will be far more squashed on the faster reps, thus more force, or more important MORE tension is going onto the muscles ??? As the clay will feel all the forces of the slow and fast reps.

Wayne
P: 148
 Quote by waynexk8 By even asking those questions you don’t yet see that there are not just the ONE force at work, there are TWO, at least, thus you have to add then up, you seen to be only adding the one force up ??? why. If you go over to BB.com, and please look at what I have named a few of my threads, then you might get it. Wayne
I don't know what forces you fantasize...but there're only two in this example.The force applied by the muscles and the weight.
When you accelerate the muscle force is greater than the weight and when you decelerate is less than the weight.The acceleration is always balanced by the deceleration(average acceleration=zero) since you start and end at rest and on average you apply equal force with the weight.
End of story.

The magnitude that you call "total force" or "effect of force over time" can be described by the impulse and is always equal with gravity's impulse over the duration of the lifting regardless the lifting speed.
P: 399
 Quote by Zula110100100 There is little value in guessing some possible values, because these guesses are made without knowledge of whether they are even likely or possible.
Yes I suppose your right in a way about the guessing, not sure it’s actually guessing mind you, and you and the other Physicist are very clever, and if you know all the variables Physics should be able to locate an answer. Also by chatting about the subject, we call all learn more.

 Quote by Zula110100100 It seems you are understanding the concepts just not the values of the numbers.
Not sure what you mean there, as the values are the different forces thought-out the ROM {Range Of Motion} of the concentric and eccentric. When I try a fast explosive rep I try to use as much force/strength as I can, however it’s basically impossible to keep this up for 100% of the time, as its too demanding.

 Quote by Zula110100100 The only cure for this, I believe, is with actual proven data, not with guesses as to what it could be if it were like this or that. Perhaps it would surprise you, or perhaps not, but guessing is no answer.
Yes you are quite right there. As I thought my EMG reading would be enough to show some people they were wrong in there numbers, as they were, or could not be adding in all the variables into the equations, D. does not seem to understand the opposite reaction on the pushing force from the weight being pushed. Then there was my Clay scenario, that also showed in practice that the fast would use more force, thus put more tension on the muscles.

However, I will try and get data from the Myotest, or the Apt from the IPod, or buy a force plate; also we have the device/machine you said you were hoping to make. Yes they might surprise me or perhaps not.

Thank you for your input and help.

Wayne
 P: 399 Would like to add to this last question. A question please, if you use more power, in the same time frame with the same weight, does that mean you use more overall or total force ??? Total or overall forces in the meaning as, if you only had a 100 pounds of force, or strength in my case, before your force runs out, and it can and does in the human body quite fast if you are lifting 80% Does also in a machine when the energy runs out. Wayne
P: 399
 Quote by douglis I don't know what forces you fantasize...but there're only two in this example.The force applied by the muscles and the weight.
I have not much time; however there are many more than two forces at work ??? The muscles doing the pushing, the weight resisting back more and more the faster you go, gravity on both the force and the weight, air resistance {however we agreed to ignore this one, but was that the right thing to do ???}

Also you do not seem to believe EMG readings that are use all over the World.

 Quote by douglis When you accelerate the muscle force is greater than the weight and when you decelerate is less than the weight.The acceleration is always balanced by the deceleration(average acceleration=zero) since you start and end at rest and on average you apply equal force with the weight. End of story.
What I am saying is, that when the fast high force go from the high forces of the accelerations, to the lower forces of the decelerations, that the constant medium forces of the slow, can NOT make up or balance out here those higher forces, and you have no way of knowing this. As of more work done, more energy used, more distanced travelled, and the high forces themselves. This has verified with my clay experiment and EMG.

 Quote by douglis The magnitude that you call "total force" or "effect of force over time" can be described by the impulse and is always equal with gravity's impulse over the duration of the lifting regardless the lifting speed.
The EMG and many things disagree with this.

Wayne
P: 148
 Quote by waynexk8 What I am saying is, that when the fast high force go from the high forces of the accelerations, to the lower forces of the decelerations, that the constant medium forces of the slow, can NOT make up or balance out here those higher forces, and you have no way of knowing this. As of more work done, more energy used, more distanced travelled, and the high forces themselves. This has verified with my clay experiment and EMG. Wayne
Same average and equal with the weight force means by definition that the upper fluctuations of force are exactly balanced by the lower fluctuations.
If you don't have the intelligence to understand this...what's the point of the discussion?
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P: 6,038
 Quote by waynexk8 However PhanthomJay, what about the opposite reaction forces, the force “ON” the muscles from the weight itself being moved ??? As if I lifted a 100 pouinds very slowely, there would be no extra forces on the muscles as I call it, the opposiot reaction forces, but there are all the time reactuion forces on the muscles from the weight itself when moving faster.
If your muscles are pushing on the weight with a force of 100 pounds, the weight is pushing in the opposite direction on your muscles with a force of 100 pounds. If your muscles are accelerating the weight and pushing on it with a force of 150 pounds, the (normal force of) the weight is pushing in the opposite direction on your muscles with a force of 150 pounds. There is no addition of action-reaction forces here. This is newton's 3rd law where equal and opposite force pairs act on different objects. Note that there are only 2 forces acting on the barbell; it's weight, acting down, which is always equal to 100 pounds, and the muscle force acting up, which is 100 pounds when it is not moving of moving at constant speed, or more than that when you are accelerating it (changing its speed to a higher value), or less than that when you are decelerating it (changing its speed to a lower value).
 Now it’s gets even more interesting. As many times I have thought about the deceleration, and would have thought your numbers were more correct, as when I do lift weight in this very explosive style, I would say it’s far more like that. As I “try” to accelerate up all the way, but as its harder to accelerate the more you accelerate, it’s getting harder, thus I have to use less force, and my body is sub concisely telling me I have to decelerate and stop and do the transition to the opposite direction soon in a Milly second.
Your body is not telling you to decelerate it; the laws of physics demand it if you are to slow it. Your body tells you that your muscles are getting strained with too much force, and it will tell you that even if you were to hold the weight still and motionless for a long period of time, without speed , acceleration, or deceleration, only a force of 100 pounds on your muscles.
 Yes this is what I have always said, as these MUST be a higher average force, as you are developing more power, and using more energy, thus WHAT else could you be using more of if you’re travelling far far far more distance in the same time frame and using more energy, it MUST be more “FORCE” if not what ???
Sorry, i misspoke, the average force (for what it's worth) is always just 100 pounds. Even though for most of the time the applied force is 150 pounds while accelerating, there is very little force that you exert when decelerating (the weight does the workfor you), so the average force averages out, if you know what I mean (in eight equal intervals on the upstroke, force you apply might be 125, 150,150,150,150, 25, 25, 25; average = 800/8 = 100 pounds). You can be moving at constant speed slowly (say 1m/s) or at constant speed quickly (say 3m/s),and the force you exert is still the same 100 pounds for both cases. Your muscles are tiring more easily in the high powered fast case, because internal energy is being lost from the complex muscular activity..if your muscles were robotic, it wouldn't matter..or a machine getting powered electrically...but humans are not machines, they need chemical/biological/food energy to operate. Don't confuse the two.
 Could I ask another question, it’s sort of on impulse. As my force lifts the weight faster and for more distance in the same time frame, some people think that when you lift slow, that when my force is on the deceleration, its where the slow moving people make up or balance out the forces, however I think no, as if this was true, then they should then also make up the difference in distance moved, but they don’t, meaning the lower force don’t make up or balance out the high force when they are on lower forces as of the decelerations.
You are not correct. However, the power required is greater when you are moving faster with the same applied force.

 That I also think I may have proved that the slow lifting forces cannot make up or balance out the fast lifting forces with my clay theory. Put a piece of clay between your hand and the weight, lift a weight up and down one time for 3 seconds each way, then lift the weight up and six times at .5 of a second each way. The clay will be far more squashed on the faster reps, thus more force, or more important MORE tension is going onto the muscles ??? As the clay will feel all the forces of the slow and fast reps. Wayne
More force goes into the muscles only during the accelerating phase. For the slow case, you might have 110 pounds at the beginning during the accelerating phase...for the fast case, it might be 150 pounds during the accelerating phase, so more squooshh...and clay doesn't rebound. The averrage force is still 100 pounds for both cases. But yes, at some point during the fast case, your muscles are receiving (and giving) more force during the accelerating phase, because it requires more force to accelearte to the higher speed in the same distance. Now I bet your really confused.
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 Quote by PhanthomJay If your muscles are pushing on the weight with a force of 100 pounds, the weight is pushing in the opposite direction on your muscles with a force of 100 pounds. If your muscles are accelerating the weight and pushing on it with a force of 150 pounds, the (normal force of) the weight is pushing in the opposite direction on your muscles with a force of 150 pounds. There is no addition of action-reaction forces here.
Sorry I meant the normal action-reaction forces.

 Quote by PhanthomJay This is newton's 3rd law where equal and opposite force pairs act on different objects.
Yes.

 Quote by PhanthomJay Note that there are only 2 forces acting on the barbell; it's weight, acting down, which is always equal to 100 pounds, and the muscle force acting up, which is 100 pounds when it is not moving of moving at constant speed, or more than that when you are accelerating it (changing its speed to a higher value), or less than that when you are decelerating it (changing its speed to a lower value).
So the weight of the weight acting down is what I call the force of gravity ??? And we are not counting the force of air resistance as it’s to small ??? Ok.

 Quote by PhanthomJay Your body is not telling you to decelerate it; the laws of physics demand it if you are to slow it.
Yes understand and totally agree.

 Quote by PhanthomJay Your body tells you that your muscles are getting strained with too much force, and it will tell you that even if you were to hold the weight still and motionless for a long period of time, without speed , acceleration, or deceleration, only a force of 100 pounds on your muscles.
Right.

I still have a like problem with this deceleration. There was a “good” study that a person used 80% and the deceleration was for 48% however the full concentric took 1.5 seconds to go about 22 inch. My repetition speed is 3 times more than that.

Here is my problem and questions, or more the way I lift or see both lifts that we are debating.

1,
If I was to lift a weight that was 80% {that’s 80% of my 1 repetition maximum} for a ROM {range of motion} of 20 inch, but just push up as hard as I could and not try to slow down, like if I was shooting the putt, or throwing a stone, thus then would have to be using more force than if I needed to decelerate ??? We are doing a little experiment on if I decelerate a lot, a bit or not at all, maybe Zula will fill you in on this.

2,
That’s the way I envisage I do lift these weights on my fast repetitions, {will do some new videos if you like ???} I explode up and keep on pushing and pushing; I don’t or did not seem at all I was or had to decelerate, and until me and D. Found that study, we was trying to work this out with me decelerating for only 10 to 20% of the ROM. So could not I fully accelerate up, and then immediately reverse ??? As that’s what it seems I do.

As there is no offloading, as we worked out that if you accelerate 80% up, can’t remember the distance, but if you immediately stopped the weight would move 3 inch, however it does not, it does if a machine pushes it, but not the Human body, as of the biomechanical advantages and disadvantages thought the ROM, meaning the body just can’t produce full constant force thought-out the range of motion.

Will get back to the rest, been very busy and sleepy.

Big thank you for your and the or members help and time.

Wayne

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