Most force/strength used in the same time frame

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sophiecentaur

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So how would we describe the ability of a structure to resist a force if we've already used the word strength?
It's surely not too demanding to use the right (common) words if you want serious help about a scientific topic. Unless it's just a trivial query- which it clearly is not.
 
Doesn't really matter, unless you expect everyone on this forum to be able to ask questions using appropriate physics terms when they may be coming because they are not very familiar with physics. I believe anyone can infer that he means strength as a measure of your overall ability to apply force, so the more of that ability you use, the more strength you use. It's what is meant that counts, in my opinion.
 
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in other words, if you apply more than required to hold it up, and then never less than required to hold it up, it will keep moving up, since that is not the case there is more going on than listed in that expiriment
Of course there's more going on than listed.The subjects obviously applied less force than required to hold the weight up at the deceleration phase that occured in order to prevent the weight fly in the air.With the method they use the negative values of the propulsive force can not be measured or they chose not to record them.

The true mean propulsive force is 0 because the net delta V is zero. The average force is always equal with the load and...to use Wayne's terms...the "total applied force" is equal with gravity's impulse over the duration of the rep(force*time).

So to summarize...either you lift 100N once with 3/3 or 6 times with .5/.5 the total applied force is 600N*s in both cases.
 
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sophiecentaur

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Doesn't really matter, unless you expect everyone on this forum to be able to ask questions using appropriate physics terms when they may be coming because they are not very familiar with physics. I believe anyone can infer that he means strength as a measure of your overall ability to apply force, so the more of that ability you use, the more strength you use. It's what is meant that counts, in my opinion.
I think you are missing something important here. When an 'uninformed' person reads a thread, they may not distinguish what's good and what's bad information. Would you really want a School student, at an elementary level, to go away thinking that strength is the same as force by taking one of Wayne's posts as gospel? One of the difficulties that the non-Scientist has is that the commonly te accepted meaning of many words conflicts with how they are used, formally, in Science. This leads to any number of misunderstandings.

@Wayne - I'm not just 'picking on you' and I'm sure that you have a genuine wish to get something out of this thread. Your terminology is just one example of what I'm talking about. There are far far worse things to be seen elsewhere. :wink:

Only today, the moderators have actually pointed out the necessity of 'reporting' contributions that are potentially harmful to understanding. We owe it to people to keep a high standard where possible if we are to maintain our present level credibility. It's ok to be laid back about some things when you know your audience but it's really not fair to people if we don't keep our technical house in order. What is this forum for, when you get down to it?
 
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I think you are missing something important here. When an 'uninformed' person reads a thread, they may not distinguish what's good and what's bad information. Would you really want a School student, at an elementary level, to go away thinking that strength is the same as force by taking one of Wayne's posts as gospel? One of the difficulties that the non-Scientist has is that the commonly te accepted meaning of many words conflicts with how they are used, formally, in Science. This leads to any number of misunderstandings.

@Wayne - I'm not just 'picking on you' and I'm sure that you have a genuine wish to get something out of this thread. Your terminology is just one example of what I'm talking about. There are far far worse things to be seen elsewhere. :wink:
I have connection problems, I am online with a dongo, it’s very slow, so can’t say much.

Sorry about the wording sophiecentaur, will try better, buts it’s hard when I have not been shown. And I am genuine and want to learn.

One thing is no one will answer me is the distance, force problem I am having, could you ??? If impossible to lift 80% in .5 of a second to 1m, the time could be less, but I am sure I could lift that in the time, will try it on video>

1,
We both have a 100 pounds of maximum strength that we can bench press. We you 80% 80 pounds, I move the weight 1000mm in .5 of a second, I will be accelerating this weight for 60% for the ROM, {range of motion} of for the concentric rep, you will be moving the same weight in .5 of a second for only 166mm.

Question,
If I have moved the weight more distance in the same time frame, does that not that mean that I have used more force ??? {force and strength are the same I would have thought, but let’s just go for force now} As even if you just count my acceleration its 600mm, that far moiré distance than the slow.

2,
I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.

Question, basically the same as the above.

As a force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1 Newton, if you move it futher in the same time from you have to use more force, N’s.

Only today, the moderators have actually pointed out the necessity of 'reporting' contributions that are potentially harmful to understanding. We owe it to people to keep a high standard where possible if we are to maintain our present level credibility. It's ok to be laid back about some things when you know your audience but it's really not fair to people if we don't keep our technical house in order. What is this forum for, when you get down to it?
Wayne
 
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Equation 3 from that source points out that a linear impulse is equal to change in linear acceleration

If we take an force of 6.2N over 10.9 seconds, that is an impulse of 67.58N*s
This means we must see a change in momentum of 67.58kg*m/s The weight presumably started at a velocity of 0, so this means it is left with a velocity of .4m/s, thus, it has not yet stopped moving, if you stop applying force, it will take gravity another .04s to stop it, and during that time there is a negative propulsive force, an applied force less than the force required to hold the object up.

So in my opinion the data is not good, because they did not wait for it to stop to take their readings.

If you take 1672.2N(the applied force required to get a propulsive force of 6.2N) * 10.9+ 0N*.04s = 1672.2 and then divide that by the total time 10.94 = 1666.0859N avg which is really close to the 1666N that were required to hold it up.

For the faster rep, we have an avg propulsive force of 45.3N * 2.8s = 126.8N*s, so we know it is left with a velocity of .75m/s which would take gravity .076s to decelerate, so again, add the 45.3 to 1666 = 1711.3*2.8s = 4791.64N*s from the force applied
+0n*.076s =4791.64/(2.8+.076) = 1666.078 once again really close to the force required to hold it still in the first place.

So overall, sure, it is more force to get the same weight going faster, anyone will agree there, the problem is that unless you apply a force to pull the weight down toward you before it stop moving up, then comparing the force for the whole cycle should be the same

in other words, if you apply more than required to hold it up, and then never less than required to hold it up, it will keep moving up, since that is not the case there is more going on than listed in that expiriment
As I said, big thx for your help and time. I do not have much time, so would like to ask you AND THE OTHERS before I answer the above, the question that I asked before, as I am having difficulty understanding the above without this added in.

If I lift 80% up to 1m from a still start, and then lift 80% up from when it is being lowered under control at .5 of a second down, the weight will gather acceleration/ movement thus appear to be heavier than it is. Meaning if you put the 80% on a scales on the standing start it wound register 80 pounds. However if you could immediately put the scales under the weight at 1m when it had dropped 1m in .5 of a second, it would register far far far more.

So it would take more force to lift the same weight up and down on the second moving rep, than to just move it up and down from a still start.

Also, is not power in mechanics, the combination of forces and movement ??? I thought power was the product of a force on an object and the object's velocity ??? As we all will agree that the fast uses more power, {even I can work that out with some simple equations} so if power is the of a force on an object and the object's velocity, does that not mean more force more power ???

Wayne
 
it would register far far far more.

So it would take more force to lift the same weight up and down on the second moving rep, than to just move it up and down from a still start.
The process that causes it to register more is the "force-applied" impulse being shorter time, more force(higher reading), but this will still be equal to the impulse of gravity if you average in the time that no force was applied. Just before it is moving up again, it IS at a still start again. Any extra force applied to "get it going back up" from downward motion can be split between stopping it and starting again. The "stopping it part" is part of the downward rep.
 
@wayne
I am not sure how exactly your EMG machine works, but would you be able to test the muscle activity that results from holding a variety of weights at a variety of heights, holding it in place for a specified time? If so that would give you an idea of what a change in force-applied impulse gives in the EMG results.
 
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@wayne
I am not sure how exactly your EMG machine works, but would you be able to test the muscle activity that results from holding a variety of weights at a variety of heights, holding it in place for a specified time? If so that would give you an idea of what a change in force-applied impulse gives in the EMG results.
Again thx for your help. Yes I can do that. The dual EMG {Electromyography} is a machine that you put four pads on the muscles, and they read the eletrical activity in the muscles, more average evtivity would mean more force and strength used by the muscles. However I imagine you have done a little Google to see the basics of how it works.


Still on this dongo, and it very slow and does not hold a connection. This time I will copy and paste all and read in full and get back when my broadband is up and running fully. Wrote this before going online.

Douglas, you are telling me that a few people have worked this debate down on paper, and some say the forces are the same, first, I have put four electrical pads on my muscles that measure the muscle activity = muscles forces and strengths with a dual EMG, and in a real life experiment done several times in front of me, I have seen the reading that state there in more muscle activity in the faster reps every time. So do you not think that you have left something out of your equations maybe ??? Look, 40 + 40 x 0 + 1 = ??? I would say 81, as 40 + 40 = 80 X 0 = 80 + 1 = 81, but some say its 41 and some 1, just for fun could all please answer this, AS IT MIGHT HELP ME SEE IF I AM GOING WRONG HERE.

Thx all.

Wayne
 
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Look, 40 + 40 x 0 + 1 = ??? I would say 81, as 40 + 40 = 80 X 0 = 80 + 1 = 81, but some say its 41 and some 1,
Wayne
I see the problem is a little deeper than I thought...
 

sophiecentaur

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=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.
 
Yeah...I would say 41..and read it 40+(40*0)+1 = 40+0+1 = 41

I want you to do that expiriment because I think it will show that activity does not equal force, I think if you hold 20lbs for 1 second, it will be some amount that is less than double activity to hold 40lbs for 1 seconds.

In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for, rather than try to figure out how that answer is wrong, perhaps you should look for how that answer relates to what you're looking for.

What are you trying to find out about PS resistance training overall?
 
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bodmas? I learned please excuse my dear aunt sally (Parenthesis, exponents, multiplication, division, addition, subtraction)
 
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In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for
The answer for the applied force gives exactly what he was looking for.He just don't like the answer!
 

sophiecentaur

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bodmas? I learned please excuse my dear aunt sally (Parenthesis, exponents, multiplication, division, addition, subtraction)
Brackets 'of' division, multiplication, addition & subtraction was what they gave me. Then, later, it became BIDMAS, with the I for Indices.
 
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Hi,

Just did the static test, and some others, they were all with about 80% and for 15 seconds on the leg extension.

Static hold half way up. Average muscle activation = 92.1.

Slow. Average muscle activation = 159.1.

Fast. Average muscle activation = 191.7.

Wayne
 
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The process that causes it to register more is the "force-applied" impulse being shorter time, more force(higher reading), but this will still be equal to the impulse of gravity if you average in the time that no force was applied. Just before it is moving up again,
Hmm, see you point, but this time when there will be no force, will be very small, about ??? Say we semented the whole .5/.5 rep of 1 second in a 100 parts, the rest time maybe just 1 part, so how and why do we need to average in something so small, as this will also be in the slower rep, but for a little longer.

it IS at a still start again. Any extra force applied to "get it going back up" from downward motion can be split between stopping it and starting again. The "stopping it part" is part of the downward rep.
Wayne
 
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I see the problem is a little deeper than I thought...
Yet again you fail to answer.

=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.
After thinking about this, I make it 1, as what I did was x 80 x 0 the other way around, like this, 0 x 80, so if you x 80 by 0 it says at 80, but if you do it like this, 80 x 0, your timesing 0 by 80, but if you x anything by 0, its always going to be 0.

Not sure how you get 41 ??? But no matter, as Douglas did not even reply, it seems like he did not then understand the problem I was trying to get round.

Wayne
 
Static hold half way up. Average muscle activation = 92.1.

Slow. Average muscle activation = 159.1.

Fast. Average muscle activation = 191.7.
If the weight and time were the same, and they started and ended at v=0, then the force-applied impulses are the same in each. This shows that Average muscle activation does not directly correlate to fore-applied impulse. Do the same thing with different amounts of weight, each held for the same time, in the same position.

Also, is not power in mechanics, the combination of forces and movement ??? I thought power was the product of a force on an object and the object's velocity ??? As we all will agree that the fast uses more power, {even I can work that out with some simple equations} so if power is the of a force on an object and the object's velocity, does that not mean more force more power ???
Lets say we have a force of 1N, we apply that force at two velocities, one is 12m in 6s = 2m/s the other is 2m in 6 seconds = .33m/s, the power is 2(1) = 2watts for the fast one, and .33(1) .33watts for the slow one. The same force, different velocities.
 
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Yeah...I would say 41..and read it 40+(40*0)+1 = 40+0+1 = 41
Don’t get that, but no prob.

I want you to do that expiriment because I think it will show that activity does not equal force, I think if you hold 20lbs for 1 second, it will be some amount that is less than double activity to hold 40lbs for 1 seconds.
Ok, get you a bit more, will do the tests tomorrow with a few different weights, to late here now,

In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for, rather than try to figure out how that answer is wrong, perhaps you should look for how that answer relates to what you're looking for.
Great, like you attack, yes I think it’s great to change our angle of attack, and try and look at it from all different angles, and great you’re here, as someone very unbiased, and new to look at the debate.

What are you trying to find out about PS resistance training overall?
PS ???

wayne
 
Not sure how you get 41 ???
Because it is a well accepted convention that you do the multiplication before you do addition
so 40 + 40 x 0 + 1 you multiply 40 by 0 and have 40+0+1=41
 
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If the weight and time were the same, and they started and ended at v=0, then the force-applied impulses are the same in each.
Could not this be not true ??? As in my theory, when the fast reps are on their high peak forces, their accelerations, then on the low forces the decelerations, when they are on the low force decelerations, the slow force is still on its medium forces, and the medium forces of the slow reps, cannot make up or balance out the high peak forces, with their medium forces. The only way they could make up these forces, if the slow reps went on longer, but they do not.

Best I say this, as the faster reps will have what 12 v=0 starts and stops, too the slows just 2 v=0.
As the full debate is this; which rep/s use the most overall or total force or strength, using roughly 80% of the persons 1RM, and moving the weight 1m up and 1m down, 6 reps at .5/.5 = 6 seconds, moving the weight 12m in all, or 1 rep at 3/3 = 6 seconds, moving the weight 2m in all

As you know, an is the integral of a force with respect to time. When a force is applied to a rigid body it changes the movement of that body. A small force {slow rep} applied for a LONG TIME can produce the same movement change as a large force {fast reps} applied briefly, because it is the product of the force and the time for which it is applied that is important. The impulse is equal to the change of momentum. However both reps are done in the same time frame.


This shows that Average muscle activation does not directly correlate to fore-applied impulse. Do the same thing with different amounts of weight, each held for the same time, in the same position.
Will do.


Let’s say we have a force of 1N, we apply that force at two velocities, one is 12m in 6s = 2m/s the other is 2m in 6 seconds = .33m/s, the power is 2(1) = 2watts for the fast one, and .33(1) .33watts for the slow one. The same force, different velocities.
It’s getting late here so not thinking straight, as its 1.30. But to accelerate a weight up, and thus to reach different heights in the same time, you would have to use more Newton’s of force to reach the higher high, or move it at a faster m/s ???

Wrote this earlier.
I can work out the power, learnt this from a site, {hope its right} as seen below and to be honest, I thought it would be basically as simple as this, was I wrong.
Calculate how much power/strength I would be used on both rep speeds. Distance weight of 91 kg moved 1.85 M.

To determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/s = 892 kg.m/s or 892 N). Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm.

We can calculate that lifting a 91kg barbell overhead a distance of 1.85 m required 1650 J of work. You will notice that the time it took to lift the barbell was not taken into account.

Let us only add up the positive part of the lift.

Concept of power however, takes time into consideration. If for example, it took .5 seconds to complete the lift, then the power generated is 1650 J divided .5 s = 3300 J/s.

If it took 2 seconds to complete the lift, then the power generated is 1650 J divided 2 s = 825 J/s.

Wayne
 
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Because it is a well accepted convention that you do the multiplication before you do addition
so 40 + 40 x 0 + 1 you multiply 40 by 0 and have 40+0+1=41
Ok thx, will read that a few more times tomoorow, to late here now, half aslepp.

Wayne
 
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Why cant anyone asnswerr this please ???

1,
We both have a 100 pounds of maximum strength {or a machine is lifting} that we can bench press. We use 80% 80 pounds, I move the weight 1000mm in .5 of a second, I will be accelerating this weight for 60% for the ROM, {range of motion} of for the concentric rep, you will be moving the same weight in .5 of a second for only 166mm.

Question,
If I have moved the weight more distance in the same time frame, does that not that mean that I have used more force ??? {force and strength are the same I would have thought, but let’s just go for force now} As even if you just count my acceleration its 600mm, that far moré distance than the slow.

2,
I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.

Question, basically the same as the above.

As a force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1Newton, if you move it futher in the same time from you have to use more force, N’s.

Wayne
 
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Why cant anyone asnswerr this please ???

I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.

Wayne
Wayne...you got your answer.What's clear is that you don't understand the answer or you don't like it.

In both cases the applied force is equal with gravity's impulse over 6 seconds....so it's the same.

40 + 40 x 0 + 1 = ???
Yet again you fail to answer.
Yes...I failed to answer so I asked my 6 years old son and told me it's 41.
 

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