most force/strength used in the same time frame


by waynexk8
Tags: force or strength, frame, time
Zula110100100
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#19
Nov24-11, 10:57 PM
P: 253
No, in my example the height was 1m for both, slow and fast, I did not work it out for slow, short distance.


"I believe the point being made by Douglis is that in order to have 0 momentum on the top and bottom the sum of the impulses must be 0,"

Not sure the point here to be honest.
However at both transitions the zero movement is so small of a time, do we need to bother with this ??? As at that time, which would be a Milly of a second, would have no force/strengths ???
The point here is that if it goes to 0 at all, then all of the energy put into it was taken back out at the other end, that is the only way to get it to 0 at the point you want

otherwise it would still have momentum in one direction or another,

But we do have movements in both directions, its only the grip of the hands and the muscles forces/strengths that stops this on both transitions ???
I don't lift a lot of weights, but I was figuring you do not push them up so fast that you stop them with the other side of your hands wouldn't that be horrible jerk on your tendons? like, if you did an open-hand benchpress, would it fly out over your hands, or just raise up and stop without leaving your hand?

3/3 = 1m, has to use more force/strength than the .5/.5 = 1m, then yes, however the faster reps at .5/.5 are done 6 times
Alright, lets go with 80lbs, thats about 36kg, force of gravity is 352.8N, gravities impulse over 3 seconds is 352.8N * 3s = 1058.4N*s if it stops at the 1m mark at 3seconds, then you MUST have applied 1058.4N*s over that time period, it could have been 1058.3N for 1 second or 529.2N for 2 seconds, either way you applied 1058.4N*s of impulse.

The same is true on the way up, whatever impulse was applied must have been equaled by gravity, or else it would be in motion at the top, and we would not be looking at a 3/3 1m model, we would have different numbers.

So for the 1 rep, over 6 seconds, the force applied must have been equal to gravities impulse over that 6 seconds, or else we would still be in motion and not on the model.
352.8*6s = 2116.8N*s of impulse

for a .5/.5 1m, 6 reps, we are looking at 352.8*.5 = 176.4N*s impulse 1 way, 352.8N8s impulse round trip, so 6 over those is equal to the previous answer of 2116.8N*s, since we still have 6 seconds of gravity acting on 36kg of mass. and as per the models, we still have a beginning and ending momentum of 0.

change in momentum = force * time

This is because momentum and force are really just velocity and acceleration, respectively, with mass taken into consideration

momentum = mv
force = ma

anyone would agree that v = at, so momentum = force * t is really the same thing.

Also you fail faster in the faster reps, in my opinion that would say you are using more force/strength faster ???
In my opinion it is because you apply less force against gravity, and allow gravity to pull it more quickly

Fast = average work = 295 iV
Slow = average work = 177 iV

The fast was 118 IV or 61% higher than the slow, as of MORE average/total force/strength is used.
I would REALLY like to stress that your body is a bunch of levers and your appendages bend and can feel different at different angles, and it might take more energy at different angles, also, the slower you go, the more you will be exercising balancing muscles, we cannot really take that into account without know a LOT about the physics of your body, that is why I say the actual way lifting weight works is pretty much unanswerable, not unanswerable, but I sure don't know enough of what goes on in your body.

The answer I am helping you with are for the FORCES, it would be correct for the forces measured on a force-plate, I think, but it has no DIRECT link to how tired you get, or how much muscle activity is going on, there are alot of other things at play here, This is a simplification to certain aspects, in this case, the "Force applied" to the weights.

It is very well possible that more signal is required to make your muscle apply more force over a shorter time, I don't know how muscles work that well, I would take that specific question to a biology forum, The fact of the matter though is that whatever is required to get your body apply that force, the result is an equal amount of acceleration(force)*time
douglis
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#20
Nov25-11, 02:27 AM
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Quote Quote by Zula110100100 View Post
So for the 1 rep, over 6 seconds, the force applied must have been equal to gravities impulse over that 6 seconds, or else we would still be in motion and not on the model.
352.8*6s = 2116.8N*s of impulse

for a .5/.5 1m, 6 reps, we are looking at 352.8*.5 = 176.4N*s impulse 1 way, 352.8N8s impulse round trip, so 6 over those is equal to the previous answer of 2116.8N*s, since we still have 6 seconds of gravity acting on 36kg of mass. and as per the models, we still have a beginning and ending momentum of 0.

change in momentum = force * time
Zula...I admire your patience!Nice explanation anyway.I don't have anything to add.
God knows how many times the above has been explained to Wayne.

So Wayne...you got your answer once again.I'll wait for your next thread after a few weeks asking the same thing.
waynexk8
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#21
Nov25-11, 08:17 AM
P: 399
Hi everyone,

No time to read anything in work.

Here is a study/test showing the there was more force/strength in the faster lifts as what I am saying, but D. says they are leaving something out, but he canít or will not explain to me what, in scientific terms or laymanís. Could anyone here please say whets been left out if anything.

http://docs.google.com/viewer?a=v&q=...ObsxJUbHpDqdXQ

If you note, the graph, table 2, normal 170kg produced 17.9N force/strength to the slow 6.2N. Thatís about 180% MORE force/strength, and in less time, the slow force/strength was calculated in 10.9 seconds, and the fast in 2.8 seconds.

Imagine the slow person had 6.2N of force/strength or should I say that ammount of tension on his muscles for 10.9 seconds, the fast person had 17.9N of force/strength or should I say that ammount of tension on his muscles for 2.8 seconds.

But if the fast person had done the fast for the same amount of time, he would have had 17.9 of force/strength used by his muscles for 10.9 seconds. This as I said is a huge difference, 180% more force/strength used by the muscles, or should I say tension on the muscles from the force/strength used, in the same time frame.

Wayne
sophiecentaur
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#22
Nov25-11, 09:11 AM
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@Wayne
Why do you keep using the two terms force and strength together? They just don't mean the same thing. 'Force' (an active word) is something that changes shape of an object or produces acceleration. 'Strength' (a passive word) refers to how an object will deform when a force is applied. If you really want a scientific answer then you could at least use scientific words appropriately in your questions.
douglis
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#23
Nov25-11, 09:42 AM
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Quote Quote by waynexk8 View Post

If you note, the graph, table 2, normal 170kg produced 17.9N force/strength to the slow 6.2N. Thatís about 180% MORE force/strength, and in less time, the slow force/strength was calculated in 10.9 seconds, and the fast in 2.8 seconds.

Wayne
No...read what they write.The 17.9N and 6.2N are the mean propulsive forces....which as they explain represent the additional to the load force required for the positive acceleration.
For the negative acceleration the propulsive force takes negative values which are not measured in the study.
The average propulsive force is zero and the force equal with the weight regardless the speed.
Zula110100100
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#24
Nov25-11, 12:07 PM
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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
Zula110100100
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#25
Nov25-11, 12:38 PM
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@sophiecentaur
In layman's terms, strength and force are very similar, a strong person can supply more force than a weak person, a force with a large magnitude is a strong force, a force with a small magnitude is a weak force, so strength is like the magnitude of force in a way
-degree of intensity or concentration

I agree with the problem in wording it "used more strength/force" because it's not something you use up per se

I guess it is that you think if he comes here to ask a question he should have it in the correct format for physics, and he assumes he can use terms how HE uses them and we being the ones knowledgeable in physics will know the physics terms to work out the problem he is outlining. It depends on what you feel the point of this forum is.
sophiecentaur
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#26
Nov25-11, 02:36 PM
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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.
Zula110100100
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#27
Nov25-11, 03:26 PM
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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.
douglis
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#28
Nov25-11, 04:24 PM
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Quote Quote by Zula110100100 View Post
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.
sophiecentaur
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Nov25-11, 05:17 PM
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Quote Quote by Zula110100100 View Post
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.

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?
waynexk8
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Nov25-11, 06:21 PM
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Quote Quote by sophiecentaur View Post
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.
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 moirť 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.

Quote Quote by sophiecentaur View Post
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
waynexk8
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#31
Nov25-11, 06:42 PM
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Quote Quote by Zula110100100 View Post
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
Zula110100100
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#32
Nov25-11, 11:40 PM
P: 253
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.
Zula110100100
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#33
Nov26-11, 05:15 AM
P: 253
@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.
waynexk8
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#34
Nov26-11, 08:01 AM
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Quote Quote by Zula110100100 View Post
@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
douglis
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#35
Nov26-11, 09:03 AM
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Quote Quote by waynexk8 View Post
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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#36
Nov26-11, 12:17 PM
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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.


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