The question no one seems to know average force

[waynexk8]

Hi all,

Question 1,
If my 1RM {RM is the most I can lift, repetition maximum} is 200 pounds, and if I accelerate 160 pounds {80%} with a force of 200 pounds 80% of a mile or metre, or any distance, {it’s a weightlifting rep if you would like to know} and use the other 20% to decelerate and immediately reverse the direction, what is the average force please ?

As I am accelerating for the whole distance of 80% I would have thought it had to be 200.

As I find averages out like this, 200 divided by 1 = 200. Alternatively, 200 and zero {as there is zero acceleration but only deceleration} = 200 thus again divided by 1 = 200.

Question 2,
Here is where it gets a little tricky.

Above I have moved 200 pounds up and then down in .5/.5 and I can do this 6 times in 6 seconds. In addition, 5 more times the distance as in the one time I lifted it up and down.

So have I not in some way used my average force {strength} 5 more times ? And used more power {work energy} ?

As if you shot a putt at .5 of a second to a very slow 6 seconds, it is going to go further, thus using more force {strength} power {work energy} ?

Wayne

 

[waynexk8]

1 rep at 2/4 = 6 seconds, distance 1m each way, weight moved once, for 2m.

5 reps at .5/.5 = 6 seconds, distance 1m each way, weight moved 6 times, for 12m.

Thus, it must take more force {strength} and more average force {strength} on the .5/.5 rep speed rep.

As if you move something faster for the same distance you must you more force {strength} and energy {calories} and if you move something the same distance 5 times faster and for 5 times the distance, you must use more force {strength} and power {work energy]

Wayne

 

[waynexk8]

Hi there,

I am quite surprised that no one can tell me question 1, the average force ?

Wayne

 

[russ_watters]

Question 1 was a little vague and the answer probably depends on some specifics, but let me try some numbers for you:

Your 40lb additional force gives you 1/4g of acceleration upward, so you can cover 2 feet in half a second, at 4 fps final speed. 2 feet if 80% of 2.5 feet, so then you'd have to decelerate in .5 feet. That would be 2 fps average speed for 1/4 second, or 16fps/s of deceleration...which is .5g or 80lb (and 160-80=80).

So the average force over time is 160lb, over the .75 seconds total time.

 

[Gokul43201]

Hi all,

Question 1,
If my 1RM {RM is the most I can lift, repetition maximum} is 200 pounds, and if I accelerate 160 pounds {80%} with a force of 200 pounds 80% of a mile or metre, or any distance, {it’s a weightlifting rep if you would like to know} and use the other 20% to decelerate and immediately reverse the direction, what is the average force please ?

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

Reason: You start from rest, accelerate, then decelerate to rest. The net change in momentum is therefore zero, which is equal to the net impulse delivered. Therefore, the average net force on the object is also zero. But since the net force is equal to the vector sum of your upward force and the downward weight of the object, the average upward force should equal the average downward force, i.e., the weight.

\Delta \mathbf{p} = \mathbf{I} = \mathbf{F}_{av}(net)\Delta t = 0

\implies \mathbf{F}_{av}(net)=0=\mathbf{F}_{av}(up)-\mathbf{F}_{av}(down) = \mathbf{F}_{av}(up) - \mathbf{W}

\implies \mathbf{F}_{av}(up) = \mathbf{W} = 160lb

 

[waynexk8]

Ok thx.

Right here is where it gets a little more interesting and difficult.

Above I have moved 160 pounds up and then down in .5/.5 {.5s concentric .5s eccentric} thus you have told me the average force is 160 pounds. And I have moved the 160 pounds up and down in 2/4,{2s concentric 4s eccentric} and you have told me the average force is 160 pounds.

But what is the average force if I move the same 160 pounds up and down 6 times in 6 seconds at the .5/.5

As it’s being moved 5 times the distance, and it’s moving at 5 times the speed, as it was the one time it was lifted up and down.

So have I not in some way used my average force {strength} 5 more times ? And used more power {work energy} ?

As just shove a penny on the table with small force {strength} and then shove a coin with a BIG force {strength} and you will see the BIG shove force {strength} the penny goes far further. Thus, I must be using more force and more average force and power {work energy}.

Wayne

 

[russ_watters]

If the deceleration exceeds the acceleration due to gravity, then you'd have to pull the weight down.

 

[waynexk8]

If the deceleration exceeds the acceleration due to gravity, then you'd have to pull the weight down.

Not sure what you are saying there.

However, what I am saying is we found both the .5/.5 and the 2/4 to be of the same average force. But what happens to this average force when we add 5 more .5/.5

As you will be able to rep a weight up, and down, 6 times in the same time frame as you can do 1 rep up and down at 2/4

Thus if the average force is the same on both one of each reps, the 1 rep at .5/.5 and the one rep at 2/4, adding 5 more reps at .5/.5 must up the force and average force.

As it will take more force {strength} to do 6 reps {repetitions} rather than 1 rep. {at any speed]

Wayne

 

[waynexk8]

Rep means repetitions, as in weightlifting up and down, or the concentric and eccentric.

.5/.5 means reps, .5 concentric and .5 eccentric.

Still nobody on this one ?

Basically, all I want to do, is find out how much more force {strength} and more average force {strength} I used doing 6 reps at .5/.5 to one rep .5/.5 and one rep at 2/4.

We have worked out that one .5/.5 rep accelerated for 80% of the concentric, using 200 pounds of force {strength} and one 2/4 rep at a constant speed using 160 pounds for 100% of the concentric, using 160 pounds of force {strength}

As there must be more force {strength} and more average force {strength} used on doing 6 reps at .5/.5 to 1 rep at .5/.5 as we have used 5 times more time, 5 times more reps, and covered 5 times more distance.

BUT HOW DO WE WORK OUT HOW MUCH MORE FORCE IS USED ?

Wayne

 

[waynexk8]

Cant not we use f/p or n/m ? as when multiplying feet and pounds, the resulting unit of measurement is foot-pounds. In the metric system, the unit of measurement used to express work is normally netwon-meters.

As one foot-pound force is the amount of energy expended when one pound-force acts through a distance of one foot along the direction of the force.

In my example it could be better to usde the metric system, as each rep is 1m by 1m.

Wayne

 

[Gokul43201]

Ok thx.

Right here is where it gets a little more interesting and difficult.

Above I have moved 160 pounds up and then down in .5/.5 {.5s concentric .5s eccentric} thus you have told me the average force is 160 pounds. And I have moved the 160 pounds up and down in 2/4,{2s concentric 4s eccentric} and you have told me the average force is 160 pounds.

But what is the average force if I move the same 160 pounds up and down 6 times in 6 seconds at the .5/.5

It's still the same as the weight (160lb). The argument used above continues to apply, since the initial and final states are still rest.

So have I not in some way used my average force {strength} 5 more times ? And used more power {work energy} ?

As just shove a penny on the table with small force {strength} and then shove a coin with a BIG force {strength} and you will see the BIG shove force {strength} the penny goes far further. Thus, I must be using more force and more average force and power {work energy}.

Wayne

Push a penny gradually along a table for 1sec. Then, using the same force, push the penny along the table for 6 seconds. Clearly, the penny travels a greater distance the second time, but the extra work done does not come from a greater force, it simply comes from applying the same force for longer.

 

[waynexk8]

I think I said that a little wrong.

You shove the penny with a force of 200 pounds for 1m that takes .5 of a second, then immediately when you stop shoving the penny at 80% of 1m {0.8m} how far would the penny go. Let’s say the penny weights 160 pounds. {Yup I want to too ROL}

You then shove the penny with a force of 160 pounds for 1m that takes 2 seconds, then immediately when you stop shoving the penny at 1m how far would the penny go.

Wayne

 

[waynexk8]

Can not we work the force out like this ?

As in the 6 reps at .5/.5 you move the weight 6 times, thus 6 times the disstance as to the 2/4.

Power = Force x velocity,

Slow rep shorter distance.

Work = 100Ib x 1ft
Work = force x distance = 100ft-lb.

Fast reps, same time frame, but far longer distance.

Work = 100 x 6ft
Work = force x distance = 600ft-lb.

Take of 20% for the deceleration and the transition from positive to negative = 480ft-lb

A STAGGERING 380% MORE FORCE USED.

Wayne

 

[Gokul43201]

Can not we work the force out like this ?

As in the 6 reps at .5/.5 you move the weight 6 times, thus 6 times the disstance as to the 2/4.

Power = Force x velocity,

Slow rep shorter distance.

Work = 100Ib x 1ft
Work = force x distance = 100ft-lb.

Fast reps, same time frame, but far longer distance.

Work = 100 x 6ft
Work = force x distance = 600ft-lb.

Take of 20% for the deceleration and the transition from positive to negative = 480ft-lb

A STAGGERING 380% MORE FORCE USED.

Wayne

No, that's 380% more work done (not force used).

 

[waynexk8]

No, that's 380% more work done (not force used).

But is not power {work} part of force ?

Ok how can I work out how much force and average force was used, as it has to be more ? As this is driving me mad.

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 ms?
Ok what about this,

Slower speed,
a force of 25kg is required to accelerate a body with a mass of 50 kilograms at a rate of .5 ms
F=ma
Force = Mass x Acceleration
50 x .5 = 25kg

Faster speed,
And a force of 150kg is required to accelerate a body with a mass of 50 kilograms at a rate of 3ms
50 x 3 = 150kg

But that does not seem to work out, as how can 25kg move 50kg ?

Wayne

 

[Gokul43201]

But is not power {work} part of force ?

Work, power and force are three completely different things. They are all related to each other, but that doesn't mean anything here.

Ok how can I work out how much force and average force was used, as it has to be more ? As this is driving me mad.

No, the average force is not any greater for a million reps as it is for one rep. You've seen the proof, and if you're asking this question again it means you did not understand the proof.

 

[waynexk8]

Work, power and force are three completely different things. They are all related to each other, but that doesn't mean anything here.

No, the average force is not any greater for a million reps as it is for one rep. You've seen the proof, and if you're asking this question again it means you did not understand the proof.

Hmm, ok sorry,

However surely I must be using more force to do 1 rep at .5/.5 to 2 reps or in my case 6 ? Thus how do we work that out please ? As I am not using the same force each time I lift, I am using a new/different force each time.

What about the F=ma, Force = Mass x Acceleration ? Can not we use that.

Wayne

 

[waynexk8]

I still can not believe no on e can help me with this ? Surely there is a way to work out how much force is used ?

Wayne

 

[waynexk8]

One Slow rep at 2/4 = 6 seconds = 160 pounds of average force {strength} and 160 pounds of gross overall force {strength}


Six fast reps at .5/.5 = 6 seconds = 160 + 160 +160 +160 +160 +160 = 160 pounds of average force {strength} BUT 960 pounds of gross overall force {strength}

This = 500% MORE OVERALL GROSS FORCE {STRENGTH}


Wayne

 

[Gokul43201]

However surely I must be using more force to do 1 rep at .5/.5 to 2 reps or in my case 6 ?

As I've mentioned before, this is wrong. You do not use more force to do more reps. You do more work, but you do not use more force.

I still can not believe no on e can help me with this ? Surely there is a way to work out how much force is used ?

Wayne

Yes, there is a way, and I showed you how to calculate it. Unfortunately, you do not want to accept that answer. And it appears you are instead looking for someone to provide a solution that fits in with your preconceived - but incorrect - ideas about force and work.

The only thing I can suggest is to start over from the beginning, and work your way up from the definitions. You have some deep seated misconceptions about what a force is, and you need to fix them.

If on the other hand, you believe you are correct, and that my answer is wrong, then you must try to show me the mistake in my proof. If there is such a mistake, then fixing it will likely lead us to a correct answer.

 

[waynexk8]

As I've mentioned before, this is wrong. You do not use more force to do more reps. You do more work, but you do not use more force.

Hi Gokul, and thanks for helping me.

Maybe be I am either confusing force, or explaining it wrong, shall I try again.

When I lift 100 pounds for 1 set, and for 1 rep at .5/.5
I have to use a force {strength} but when I lift a 100 pounds for 1 set, but for 6 reps at .5/.5 I have to use more force {strength} As I have lifted the 100 pounds 5 more times, and 5 more times the distance, and it took 5 times more time.

As if you go pick up a 100 pounds onto a wall, you will know its far harder to pick it up 6 times rather than 1 time, thus I have used more force {strength}


I also know the power {work energy calories} is more, and know how to work this out, but I want the extra force {strength} used.


I tried using Force = Mass x Acceleration, but that did not seem to work ?

Yes, there is a way, and I showed you how to calculate it. Unfortunately, you do not want to accept that answer. And it appears you are instead looking for someone to provide a solution that fits in with your preconceived - but incorrect - ideas about force and work.

The only thing I can suggest is to start over from the beginning, and work your way up from the definitions. You have some deep seated misconceptions about what a force is, and you need to fix them.

If on the other hand, you believe you are correct, and that my answer is wrong, then you must try to show me the mistake in my proof. If there is such a mistake, then fixing it will likely lead us to a correct answer.

I think you only showed me how to calculate the average force {strength} however I wanted the overall force {strength}

It’s like I have £100 and you have £50 to find the average we add 100 + 50 = 150 and divided by 2 = 75. But that not our full/overall amount, as its 100 + 50 = 150. This if I lifted a 100 pounds 6 times would I x 100 by 6 = 600 pounds of full/overall force {strength} used.

Wayne

 

[Gokul43201]

Wayne, the terms force, work, power, etc. have very specific definitions in Physics. Just because you have to put in more "effort" on 6 reps than on 1 rep does not mean you are using more force. The work done is 6 times greater, and the impulse delivered by you is 6 times greater, but the force applied is the same.

And terms like "overall force" and "strength" do not exist in Physics. If you wish to calculate something, it needs to have a definition, or you need to define it in terms of already well-defined quantities.

 

[waynexk8]

Wayne, the terms force, work, power, etc. have very specific definitions in Physics. Just because you have to put in more "effort" on 6 reps than on 1 rep does not mean you are using more force. The work done is 6 times greater, and the impulse delivered by you is 6 times greater, but the force applied is the same.

And terms like "overall force" and "strength" do not exist in Physics. If you wish to calculate something, it needs to have a definition, or you need to define it in terms of already well-defined quantities.

Ok thx. Yes think you hit the nail on the head there, as I am using more strength, or more of the same strength or force, or as you say effort, but do not exist in Physics.

Ok what about,

F=MA. One Newton is defined as the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s. 1 kg exerts a force of approximately 9.8 N.

MANY Ns WOULD YOU NEED TO ACCELERATE 50kg MASS 1 M/S/S,
AND HOW MANY Ns WOULD YOU NEED TO ACCELERATE 50KG MASS 6 M/S/S,

Slow rep is moving 50kg at a constant 2 seconds for 1 metre. It takes 2 seconds to travel 1m. Fast rep is accelerating 50kg 1 metre every .5 of a second. And does this 4 times in the same time frame as the slow rep. Thus, it has moved 50kg 3 times the distance in the same time frame as the 2-second slow rep.

As 2 seconds is 4 times slower than .5 I divided .5 by 2.

Slow rep, moving at 2s per 1m
F=50kg x .125 = 6.25N

Fast rep, moving at .5s per 1m
F=50 kg x .5m = 25N.

Thus I need and have used {lets take off the 20% for the deceleration phase on the faster reps} N18.75 = 1.875kg MORE force {strength} on the fast reps to accelerate if faster in the same time frame !

So for doing 12 reps = 22.5kg

SO ON MY FAST REPS ACCELERATING AT .5/.5 {THAT’S .5 OF A SECOND FOR THE CONCENTRIC AND .5 SECOND FOR THE ECCENTRIC} AND SLOW REPS AT 2/4 USING A 100 POUNDS FOR 12 REPS, I HAVE USED 49.5 POUNDS MORE FORCE {STRENGTH} NOT COUNTING THE ECCENTRIC.

?

Wayne

 

[waynexk8]

In simple terms.

I cannot see in my none physics way of looking at how two people of the same strength, or say clones, and one moves a 100 pounds 6m in 6 seconds, and then the other clone move 100 pounds only 1m in 6 seconds.

I cannot see how the clone that moved the 100 pounds further, 5m further in the same time frame did not use more force {strength} ? Some might then say, but the clone who moved it further used more work, however you need more force {strength} to do more work in the same time frame.

If you and me were the same strength, and we both tried to move the 100 pounds, and I moved it 6m and you 1m in the same time frame, I would say to you, you just did NOT use as much force {strength} effort as I did.

Wayne

 

[Dale]

Hi waynexk8,

I verified Gokul43201's statements numerically. He is completely correct (as his math showed that he would be). As long as the initial and final velocity is 0 then the time averaged force is equal to the weight. It does not matter how much time is taken nor does it matter how far up or down the weight is moved, the average force is constant.

For example, let's say for simplicity that you are using maneuvers where you lift up at one constant force for the first part of the maneuver and then a second force constant force for the remainder. If you want to lift a 160 lb weight a distance of 3 ft in 1 s with an initial force of 200 lb then you have to use 200 lb for .75 s and 40 lb for .25 s. That averages out to (200 lb .75 s + 40 lb .25 s)/1s = 160 lb. Now, if you instead want to use an initial force of 220 lb then you have to use 220 lb for .5 s and 100 lb for .5 s. That averages out to (220 lb .5 s + 100 lb .5 s)/1s = 160 lb.

I think the problem is that average force is simply not a quantity that you are really interested in, because Gokul43201 is correct that over any maneuver where the weight begins and ends at rest the average force always equals the weight.

 

[DavidSullivan]

I have a question regarding this discussion, which might get at the crux of what Wayne is asking.

Correct me if I’m wrong Wayne, but I don’t think he’s asking about the average force of the system as a whole (person, weights, and gravity), but specifically about the average force of the person lifting the weights (only). There are two forces in play here – gravity and the force the person exerts on the weight to lift it. Given that context, the force is not equal (as far as I know – correct me if I’m wrong) if you change the amount of time you used to lift the weight:

Assume I lift the weights twice from rest, once slowly and once quickly, with constant force. I measure the acceleration for the first repetition (r1) and find the net acceleration is 1 m/s^2 (10.8 m/s^2 – 9.8 m/s^2). I measure the net acceleration for the second repetition (r2) and find it is 2.0 m/s^2.

Assume the weight is 100kg, then the force used for r1 (F1) is 100kg * 10.8 m/s^2 = 1080N and the force used for r2 (F2) is 1180N.

Assume I lift the weight 1 meter each time and I ignore the forces in bringing the weight back down because the force exerted by the person is not a conservative force. (The person expends additional energy to bring the weight back down.) The work for r1 = F1 * 1m = 1080J and the work for r2 = 1180J.

So, if a person “pushes” an object at rest “harder” to cover the same distance in a shorter period of time, then the force the person exerts (only) will be larger, correct?

 

[waynexk8]

First thanks all for the help.

I have a question regarding this discussion, which might get at the crux of what Wayne is asking.

Correct me if I’m wrong Wayne, but I don’t think he’s asking about the average force of the system as a whole (person, weights, and gravity), but specifically about the average force of the person lifting the weights (only). There are two forces in play here – gravity and the force the person exerts on the weight to lift it. Given that context, the force is not equal (as far as I know – correct me if I’m wrong) if you change the amount of time you used to lift the weight:

Assume I lift the weights twice from rest, once slowly and once quickly, with constant force. I measure the acceleration for the first repetition (r1) and find the net acceleration is 1 m/s^2 (10.8 m/s^2 – 9.8 m/s^2). I measure the net acceleration for the second repetition (r2) and find it is 2.0 m/s^2.

Assume the weight is 100kg, then the force used for r1 (F1) is 100kg * 10.8 m/s^2 = 1080N and the force used for r2 (F2) is 1180N.

Assume I lift the weight 1 meter each time and I ignore the forces in bringing the weight back down because the force exerted by the person is not a conservative force. (The person expends additional energy to bring the weight back down.) The work for r1 = F1 * 1m = 1080J and the work for r2 = 1180J.

So, if a person “pushes” an object at rest “harder” to cover the same distance in a shorter period of time, then the force the person exerts (only) will be larger, correct?

DavidSullivan,

What was wrong with my F=MA ? What I thought I was working out was the accelerating force of the person moving the weight ?

WHAT you said here; So, if a person “pushes” an object at rest “harder” to cover the same distance in a shorter period of time, then the force the person exerts (only) will be larger, correct?

Sounds more than interesting, and exactly what I am after. As yes, I want the average force of the person lifting the weights (only). However, are you working out the force {strength} of the person or the power ? As its force I want

Will answer more later. Have to go somewhere now, but back in any hour from now, its 21.47 1+GMT here now.

Wayne

 

[Dale]

I don’t think he’s asking about the average force of the system as a whole (person, weights, and gravity), but specifically about the average force of the person lifting the weights (only). There are two forces in play here – gravity and the force the person exerts on the weight to lift it. Given that context, the force is not equal (as far as I know – correct me if I’m wrong) if you change the amount of time you used to lift the weight:

Even if you change the amount of time the average net force remains the same, namely 0. Remember, we are starting and stopping at v=0. So vi=0, vf=0, and Δv=vf-vi=0. Average acceleration is Δv/Δt, which is 0/Δt=0 regardless of Δt.

Obviously we are not interested in the 0 average net force, but in the force exerted by the person. Since the average force of gravity is equal to the weight and the average net force is equal to 0 then the average force exerted by the person is equal to the weight (but pointed up instead of down).

The average force is equal to the weight regardless of the duration or the distance, provided you start and stop at v=0. The math is unambiguous.

 

[waynexk8]

Hi waynexk8,

I verified Gokul43201's statements numerically. He is completely correct (as his math showed that he would be). As long as the initial and final velocity is 0 then the time averaged force is equal to the weight. It does not matter how much time is taken nor does it matter how far up or down the weight is moved, the average force is constant.

Hi there DaleSpam,

But when I am repping, I do not think I come to a stop at all ? See below video. As am I not doing continuous movement, just like the pistons on an old steam engine, they never come to a stop, they go around and around.

http://www.youtube.com/user/waynerock999?feature=mhw4

QUESTION 1,
If I am only accelerating for 80% of the ROM {range Of Motion} or the rep, why do we count the other 20% ? As I am not acceleration here, just decelerating, so why should we count no force ? I mean just say we only counted the 80%, and I was using a 100 pounds of force, would not the average force be 100 pounds ? What would be the average force if I was a machine that could accelerate a load up to 99% of the rep, and then immediately reverse direction ?

QUESTION 2,
I thought my F=MA would do this. But what gets me is that I am in my fast rep accelerating constantly, but the person on the slow rep is not accelerating at all, but is only moving at a constant speed, using only enough force to just move the weight, as going 1m in 2 seconds is quite slow.

I mean are we doing the equations right, as one person is acceleration, and the other not, are not there a big difference ?

For example, let's say for simplicity that you are using maneuvers where you lift up at one constant force for the first part of the maneuver and then a second force constant force for the remainder. If you want to lift a 160 lb weight a distance of 3 ft in 1 s with an initial force of 200 lb then you have to use 200 lb for .75 s and 40 lb for .25 s. That averages out to (200 lb .75 s + 40 lb .25 s)/1s = 160 lb. Now, if you instead want to use an initial force of 220 lb then you have to use 220 lb for .5 s and 100 lb for .5 s. That averages out to (220 lb .5 s + 100 lb .5 s)/1s = 160 lb.

Yes get that.

I think the problem is that average force is simply not a quantity that you are really interested in, because Gokul43201 is correct that over any maneuver where the weight begins and ends at rest the average force always equals the weight.

Hmm yes your right there, as I want to find the extra force {strength} as too me, if we both use a 100 pounds of force for 1 rep, but as I can do 5 more reps in the same time frame, I have used 100 x 100 x 100 x 100 x 100 x 100, I have used that same force as the other person but 5 more times. Some might call this power {work} but you must need more force {strength} to do more power {work} ? As the second thirst forth etc, reps do not go up with the same force {strength} used on the first rep.

Wayne

 

[waynexk8]

I am still not sure what was wrong with my F=MA ? As a physasist wrote the below, so I thought I would have a go.

jeffpinter wrote:
Assume that your 1RM is 200 pounds. In order to complete the rep you must generate a force slightly above 200 pounds, but let's call it an even 200, which is the maximum force you can generate.

And let's assume that the load for your 20RM set is 150 pounds, probably a reasonable estimate, which is 75% of your 1RM.

This % is not uncommon for those who train Rogue HIT, but is unlikely for those who haven't, so it may still offer questions.

Now, if you take that 150 pounds and lift it very slowly, so it takes maybe 5 seconds to complete the lift, there will be negligible acceleration. This means that the force you generate is 150 pounds all the way to the top, which is what is required to overcome the weight of the bar due to gravity alone.

However, if you were to lift it to the top in exactly one second, the acceleration would not be negligible, and would equal 2 ft/sec2 (assuming you move it one foot). This acceleration requires an additional 12 pounds of force, for a total force of 162 pounds.

This is arrived at using basic physics which I will not detail, unless you want to see the calculations. It is simply Newton's 2nd Law at work - F=ma. The force needed to accelerate an object is proportional to the acceleration.

BIO-FORCE wrote:
Likewise, lifting it in 3/4 second requires a total force of 222 pounds.

Is this a mis-type? Would the 3/4 second be greater than the 1/2 sec reps?

Wayne