Most force/strength used in the same time frame

[waynexk8]

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 thanks 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]

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]

@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]

@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 thanks 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 electrical 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]

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]

=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.

 

[Zula110100100]

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?

 

[Zula110100100]

bodmas? I learned please excuse my dear aunt sally (Parenthesis, exponents, multiplication, division, addition, subtraction)

 

[douglis]

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]

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.

 

[waynexk8]

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

 

[waynexk8]

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

 

[waynexk8]

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

 

[Zula110100100]

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.

 

[waynexk8]

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

 

[Zula110100100]

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

 

[waynexk8]

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, learned 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

 

[waynexk8]

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

 

[waynexk8]

Why can't 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

 

[douglis]

Why can't 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.

 

[sophiecentaur]

Why can't 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

You seem to want answers about this without having to make any effort to know the basics of Science or even primary School Arithmetic. We just can't help you on your terms. You have had many answers (the same answer put in different ways, actually) in a attempt to get the message over to you. Yet you refuse to give an inch (or even 2.54cm) so I am unsubscribing from this thread. I suggest you return the the pub / club / locker room and have your conversation with people who have no use for real Science but who just want an hour's worth of idle, anecdotal chit chat.

 

[waynexk8]

You seem to want answers about this without having to make any effort to know the basics of Science or even primary School Arithmetic. We just can't help you on your terms. You have had many answers (the same answer put in different ways, actually) in a attempt to get the message over to you. Yet you refuse to give an inch (or even 3.54cm) so I am unsubscribing from this thread. I suggest you return the the pub / club / locker room and have your conversation with people who have no use for real Science but who just want an hour's worth of idle, anecdotal chit chat.

Hi sophiecentaur,

I don’t understand what you’re saying, to me the questions I asked on distance are just basic questions, and I put them down as plain as I can. I will give an inch or more. I am not from the pub or locker room.

I have made videos too which you have seen that you fail faster in the faster reps, thus why else would you fail if it was not because of you had used more force.

I bought an EMG that shows more muscle activity in the faster reps, = more force used.

I asked quite plain and polite questions on distance, and also proved with this little scenario that more force was used.

You place a piece of clay between your hands and the weight, THE CLAY WILL REPRESENT THE FORCE AND THUS TENSION FROM AND TO THE MUSCLES. The clay would be squashed MORE when doing the faster reps in the same time frame, YES ? What you miss is that the high peak forces of the faster reps, can and do not make up or be balanced out by the medium forces of the slower reps when the faster reps are on the deceleration.

HERE IS WHAT A PHYSICIST ON ANOTHER FORUM SAID,

1. The first man uses the most force to do that, since he must push stronger for the weight to move a larger distance within the same time interval (acceleration is greater). Power is larger, energy is larger.

2. The fast man again. Since the time is smaller, there is more acceleration required and thus, higher magnitude of force. Power would be larger, energy would be larger.

3. Same as 2 since this is merely 4 times the work they are doing.

I honestly have not tried in any way to offend you, if I have I am all to sorry, that was not my intentions, I did say I would try to learn. What gets me is why you or some others can’t answer my distance questions ? I don't get why people are getting uptight, all I ever wanted is to ask polite questions, and trieed my best to lay them out right.

If you can in my subjects, I would try to help you.

Wayne

 

[waynexk8]

Think this may help the physicists here understand more, and no, this is not in any way to sound sarcastic, as I need to explain in more my layman’s terms.

Wayne,

As asked above, could you please clarify what your argument is? Are you saying that the maximum possible forced produced by a muscle will occur with faster vs. slower contractions?

How can I put this, say a muscle has a 100 pounds of force/strength to use up in a rep or in a set, the faster reps, I am saying that they use say a 100 of that total overall force/strength to the slow using say 70%, thus more total overall tension will go on the muscles, when doing the faster reps.

Does not my more distance moved in the same time frame, more EMG activity, muscles fail faster in the faster reps, and some more things show that.

Also, in regards to your physicist, what is his background in physiology? From a mathematical standpoint some of the things you have said are true.

Thank you, which ones please ? Now we are getting somewhere.

However, when you are presented with what happens from a physiological standpoint your muscles would not be able to produce the increase in force required for the math to work. In other words there are physiological reasons why you cannot move a heavy weight very quickly and why when you move a light weight quickly muscular force is reduced. You need to apply the math within the constraints of muscle physiology. The person you are talking with needs to know both physics and physiology to understand the problem.

Well said. As because of the biomechanical advantages and disadvantages thought the ROM, the muscles cannot push up with full force like a machine could, but I don’t mind keeping it as a machine, as it still holds true. I mean how can a machine push with a force a weight 1000mm to 166mm in the same time frame, and not use more total or overall force. Even if you just could the accelerations of the machine pushing fast, that’s say 600mm the machine has accelerated the weight to 166mm of the slow, and SOME here are trying to tell me that they use the same overall force accelerating the weight 600mm to 166mm ?

Your above examples violate what the muscle is capable of doing in the physiological world so they are not worth commenting on.

Yes I know that, but are just basic examples, I think all here know that ? Big thanks for your time and help.

Wayne

 

[waynexk8]

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.

Add in the acceleration conponants please, How can the applied force be equal with gravity's impulse over 6 seconds all the time ? Add in time and distance.

Or try and tell me how you move a weight 166mm and I move a weight 1000mm in the same time frame, and I accelerate for about 600mm of that distance. Your trying to tell me that you use the same force moving a weight far less disstynce in the same time frame



Yes...I failed to answer so I asked my 6 years old son and told me it's 41.[/QUOTE]

Let’s see you try and prove this, with basic numbers

40 + 40 x 0 + 1 = ?

Is not 40 + 40 = 80 ? Now we x 80 by 0, and whatever we x 0 by it will always be 0 right ? So we are now at 0 right ? So add 1 to 0 = 1 right ?

Wayne

 

[douglis]

Add in the acceleration conponants please, How can the applied force be equal with gravity's impulse over 6 seconds all the time ? Add in time and distance.

Or try and tell me how you move a weight 166mm and I move a weight 1000mm in the same time frame, and I accelerate for about 600mm of that distance. Your trying to tell me that you use the same force moving a weight far less disstynce in the same time frame

Irrespective of any of the details along the way (acceleration, deceleration, fraction of time spent accelerating, distance covered, etc.), the average force is always equal to the weight of the object, so long as you start and finish at rest.

Let’s see you try and prove this, with basic numbers

40 + 40 x 0 + 1 = ?

Is not 40 + 40 = 80 ? Now we x 80 by 0, and whatever we x 0 by it will always be 0 right ? So we are now at 0 right ? So add 1 to 0 = 1 right ?

Wayne

Instead of trying to prove nonsense learn to do the mathematic acts at the right order or find a kid to show you.

 

[Zula110100100]

@wayne, That is basically what we have been saying, the answer we gave you in not taking into account physiological factors. That is the whole reason you simplified it to a machine, in the machine, the OVERALL impulse is equal.

In the case of the clay, yes, the fast one would make more of a dent in the clay due to higher peak forces, but as you said, this is a greater acceleration and a greater veloicty in a small time, as a result, in order for gravity to cause it to stop again it must act longer, so when you average them together, they are equal.

Also, as mentioned before, you get different mechanical advantage at different anges with your arms.

 

[douglis]

I really can't figure out what Wayne can't understand.
If we compare 6 reps with .5/.5 and 1 rep with 3/3 the only difference is that the 6 reps have higher fluctuations of force.But in both cases the same average force(the weight) is applied for 6 seconds.

What's so hard in the above to understand?Why does he keep posting the same irrelevant examples?Do we have to use superior maths for something obvious even to a 10 years old child?

 

[waynexk8]

I really can't figure out what Wayne can't understand.
If we compare 6 reps with .5/.5 and 1 rep with 3/3 the only difference is that the 6 reps have higher fluctuations of force.But in both cases the same average force(the weight) is applied for 6 seconds.

What's so hard in the above to understand?Why does he keep posting the same irrelevant examples?Do we have to use superior maths for something obvious even to a 10 years old child?

How do you work out my EMG states you are wrong ?

And how do you work out my clay example proves you wrong ?

Irrespective of any of the details along the way (acceleration, deceleration, fraction of time spent accelerating, distance covered, etc.), the average force is always equal to the weight of the object, so long as you start and finish at rest.

Could you do me a favour please, could you tell me again how you work out this average force. Let’s say I am moving a weight at 1m/s and move it for 1 second, I then move a weight 100m/s and move it for 1 second, how do you work out the averages ? AND WHAT YOUR SAYING IS THAT YOU USE THE SAME TOTAL OR OVERALL FORCE TO MOVE THE SAME WEIGHT 1M IN 1 SECOND, AND TO MOVE IT 100M IN ONE SECOND ?

You always do the same thing; you say something without an explanation.

Don’t you understand you are moving against gravity, which is a force, and the faster you move against gravity the more force you need, yes you need less force for deceleration, but if you’re covering more distance in the same time frame, you HAVE to have more total or overall force.

And for the Hundredth time, I am not on about average force; I am on about total or overall force.

My more distance in the same time frame.

Are you saying the force, the Newton is wrong ? It is equal to the amount of net force required to accelerate a mass of 1kg at a rate of 1m/s.

Are you saying that to accelerate a mass of 1kg at the rate of 1m/s, or 5m/s or 10m/s, all you need is the exact same force as 1 Newton ? As that’s what it sounds like. So the question why are you saying this ?
F = ma let's us work out the forces at work on objects by multiplying the mass of the object by the acceleration of the object.
Example:
The force at work on a Formula 1 car as it starts a race! If the F1 car has a Mass of 600kg and an Acceleration of 20m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 20 = 12000N

Wayne says, If the F1 car has a Mass of 600kg and an Acceleration of 40m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 40 = 24000N. But your saying that’s wrong ?


If the car has a mass of 700kg And a driver pushes the car with an acceleration of 0.05m/s/s Then:
F = MA
Force = 700kg x 0.05m/s/s
Force = 35kN (kiloNewtons)

Wayne says, If the car has a mass of 700kg And a driver pushes the car with an acceleration of 0.05m/s/s Then:
F = MA
Force = 700kg x 1m/s/s
Force = 700kN (kiloNewtons)
But you’re saying that’s wrong ?

http://www.racemath.info/forcesandpressure/what_is_f=ma.htm

Instead of trying to prove nonsense learn to do the mathematic acts at the right order or find a kid to show you.

D. basically it was a sort of trick question for you, as the way you think now is VERY wrong.

LOOK, you CANT change the order of the question to suit you, WHY, and when I say WHY, I mean please give me an answer as why you THINK you can change the other of the question ? IF you change the order of the QUESTION, it NO longer because the same question, thus you CAN NOT changed the dammed order of the question, I can’t believe anyone can THINK they can answer a question like this, with changing the order of the question.

He is what happens when you change questions, ONE, YOU HAVE NOT ANSWERED THE ORIGINAL QUESTION, HAVE YOU ? YES OR NO ?

TWO, I ask you this question, 4 + 4 = 8 x that number by 2 = 32, YOU might think I am wrong, but ho not, well I am not, well I am but I am now doing what you did, {and what you’re doing in our big physics debate} I changed the question to suit myself, I changed it to this, 4 x 4 = 16 x that number by 2 = 32.

I cannot believe anyone could try and change the question, have you a good explanation for this ?

Back later for you post Zula110100100.

Wayne

 

[douglis]

How do you work out my EMG states you are wrong ?

Your EMG is not suitable for lifting that involves SSC...I explained why.

And how do you work out my clay example proves you wrong ?

Your irrelevant clay example just proves that higher peak force is produced with fast lifting.Nothing more.

Could you do me a favour please, could you tell me again how you work out this average force. Let’s say I am moving a weight at 1m/s and move it for 1 second, I then move a weight 100m/s and move it for 1 second, how do you work out the averages ? AND WHAT YOUR SAYING IS THAT YOU USE THE SAME TOTAL OR OVERALL FORCE TO MOVE THE SAME WEIGHT 1M IN 1 SECOND, AND TO MOVE IT 100M IN ONE SECOND ?

Wayne

I must have said this more than a hundred of times.
I work out the average from the net impulse delivered.Either you move the weight 1 or 100m the change in momentum is zero so the net impulse delivered is zero too.Concequently the average net force is zero and the applied force equal with the weight.
End of story.
It's obvious that you're not able to understand the answer or you don't want to.Either way continue alone.I'm fed up with that nonsense.

 

[waynexk8]

D, are you saying the average, total, overall forces here are the same ? I am.

https://www.physicsforums.com/attachment.php?attachmentid=783&d=1079995820

But on the above I rep once and you rep once, so let’s show you where you ARE wrong on averages, or should I say thinking as averages are the same, the total overall force is the same. Too which it is NOT.

Slow reps,
1 + 1 + 1 + 1 + 1 = 5 / BY 5 = 1

Fast reps,
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10 / BY 10 = 1

NOW don’t you get it ? YES the averages ARE the same in both speeds, BUT please NOTE the faster reps had a higher total or average force, that’s what I am trying to tell you, DON’T TAKE TO MUCH NOTICE ON THE AVERAGE.

Wayne