# Acceleration decceleration

1. May 24, 2012

### waynexk8

Hi all,

Without sophisticated equipment, {I have an EMG} when lifting weight up and down, as in weightlifting/bodybuilding, would there be any way to tell when/if the weight starts to decelerate ???

I have a Smith Machine, this is a machine, that the weight goes up and down on two rods, on bearings, see below.

With this machine, I am able to lift if I want open handed, {to which I down, and your lifts suffer} if wanted, if I lowed then lifted 50% of my 1RM, {Repetition Maximum} as explosively as I could, open handed, the weight at the end of the lift will move slightly out of my hands, {say the positive was 20in and the negative was 20in} this mean in my opinion that there was no deceleration when in contact with my hands, is this correct please ???

If I then lift 80% of my 1RM, in the exact same manner as above, but the weight because it was heaver did not leave my hands, {If you do the equations for a machine lifting 80% at 2ms, the weight would leave your hands for about 4in, but because of the biomechanical disadvantages and advantages, this does not happen with the muscles of the body} but still, does that not mean there is not deceleration on my behalf ??? As I am trying to push the weight as explosively as I can, then after I push up explosively as I can, I then lower again. Next and more to the real, would there be a deceleration {I know there must be a deceleration, if I push up and then lower the weight} on my behalf ??? Meaning if I push up as explosively as I can, 80% for 20in that will take say .3 of a second, will I use less force than the weight ??? As if we go back and look at me pressing up 50% I would press up as hard as I can, thus the weight would move out of my hands, then, and only then would it start to decelerate, thus no deceleration on my behalf, then when it dropped into my hand I would then lower it, and repeat the lift.

The only problem here, I that I want to find if there is deceleration on my behalf lifting 80% closed handed ??? As is not my weightlifting/bodybuilding lifting up and down, just like if I lifted 80% 20in, and then there was a steel plate to which I hit, then, I would think not deceleration on my behalf {Mind you I do not lift to a steel plate, thus, it’s basically the same} then lower.

Wayne

Last edited by a moderator: Sep 25, 2014
2. May 25, 2012

### Staff: Mentor

Not really. An EMG measures muscle activation. You would need an accelerometer or a good high speed video camera.

Note, the rods will introduce some friction into the system, which will change some of the analysis that you and I have discussed in other threads.

No, this is not correct. The weight will decelerate whenever the force on your hand is less than the force of gravity plus the friction force. The weight leaves your hand when the force is zero. So there is a considerable range of force where the weight is both in contact with your hand and decelerating.

This has nothing to do with wether you are lifting with an open or closed hand.

3. May 25, 2012

### waynexk8

Maybe I could borrow an iPod, as that has an accelerometer, then dl this; http://liza.exelio.eu/

Would that be able to tell me when I was decelerating, as I am sure I don’t decelerate, well not in the normal decelerate, as the only way I think I decelerate, is because my arms cannot go any further, if you get what I am saying.

Yes totally agree, that the rods will introduce some friction, but I don’t see it damaging/altering the debate ???

Yes I understand that, that if I use less force than the weight that it will decelerate, but what if I went from using 100% to 99% then 98% and down to 85%, would not the weight still be accelerating ???

But if I lifted a weight, or a machine lifted a weight, 50% and then 80% and the weight was in lift 1 and 2, lifted with 100% force to full stretch, would not it be the length of my arm reach or the machine reach that at that moment makes for the immediate deceleration/stop ??? If so, then my force, {forget about the first acceleration for now} for 99% of the, sorry 100% of the lift would be 100% or as close as I could get to using my maximum force for that lift.

What I am saying, that if I am lifting 80% the way I do, as explosively/fast as I can, I don’t think there is in the first several rep, much deceleration at all. Could you maybe do a lift like this somehow and see what I mean, not sure if you could do/try this, or shall I video me doing a few lifts on the Smith machine maybe ??? As if I am lifting 80% with 100% I don’t see on the first several reps, if I am trying to move the weight up as fast as I can, how there can be much/any deceleration. Take a shot putter, now there is no deceleration on his behalf, this is what I am saying, the deceleration on the putt come after it leaves his hand, so what if he putted as usual, but at the instant that the putt left his hand, a steel plate appeared ???

Right, as then there is no force pushing it.

But has it got to be like this, as I said above about the shot putter and so forth. I don’t see why I can’t, or more why a machine could not push up for full range with 100% force ??? I don’t see a reason when a person, or more to the point a machine has to use less force and decelerate ???

It must have something to do with if I have open hands or not, as if I repped 20% with open hands, it would leave my hand well before I had finished the rep, or ROM {Range Of Motion} but if I had closed hands, I would still be exerting force on the weight, to which I would not open handed. What I am doing when closed hands, is taking more of the acceleration force, propulsive forces on my muscles.

So if I was to accelerate a weight of 80% for a whole .5 of a second, to a person moving a weight at a constant very slow speed of .5 of a second, would not I be using more force ???

Wayne

4. May 25, 2012

### Staff: Mentor

You need to learn to be more specific. What is 100% in this case? What is the weight?

As I said above, the weight will decelerate any time the force on the hand is less than the force of gravity plus the friction force. There is not enough information in "using 100%" to determine what happens. Provide the required information and try the calculation yourself.

80% of what? How much force are you using to accelerate? Do you mean "compared to a person" or "towards a person". 0.5 s is a time, not a speed. More force than what?

Last edited: May 26, 2012
5. May 26, 2012

### waynexk8

Just thought would say, will have to get back to your posts and PM’s tomorrow.

Wayne

6. May 28, 2012

### waynexk8

Sorry.

The weight is 80 pounds, too which that is 20% off my 1RM, {Repetition Maximum} so let’s just say I could use 100% or 100 pounds, if I then went from using 100 pounds to 99 pounds then 98 pounds and down to 856 pounds, would not the weight still be accelerating ???

Do you mean that if the force is a little under 80 pounds, then the weight is slowing/deceleration down ??? However, I could push up with from 100 to 81 pounds for the full rep/push, then there would be no deceleration on my part, as the only reason the weight would decelerate and stop, would be because my arm had reached its stretching length, and as I said, as of the biomechanical disadvantages and disadvantages, thought the ROM {range Of Motion} the weight would repped/pushed up with a constant force, thus it would not leave the hand. So basically as I said, If I could push up with from 100 to 81 pounds for the full rep/push, then there would be no deceleration on my part, as the only reason the weight would decelerate and stop, would be because my arm had reached its stretching length, there would just be an extra impulse/jolt on the bones and muscles as full ROM was reached.

Ok.

I am lifting up 80 pounds with a constant force of 100 pounds, for 20 inch, that will take .5 of a second. Person 2 is lifting the up 80 pounds with a constant force of 80 pounds, for 3.3 inch, that will take .5 of a second, the only deceleration on both parts, will be instantaneously, as both arms had reached out full ROM. Which please will use the more impulse.

Wayne

7. May 28, 2012

### Staff: Mentor

Neglecting the friction force, yes.

Yes.

Neglecting friction, yes. However, even ideally you cannot exert a constant force for very long. As soon as you get to the end of your reach, the force must change. It can change to 0, if you are throwing the weight. It can change to some force less than the weight if you are gently decelerating it. Or it can even change to a negative force if you are almost throwing it but not letting go.

This is impossible.

You can lift 70 lb with a constant force of 100 lb for 20 in, that will take .5 s.
You can lift 80 lb with a constant force of 113 lb for 20 in, that will take .5 s.
You can lift 80 lb with a constant force of 100 lb for 12 in, that will take .5 s.
You can lift 80 lb with a constant force of 100 lb for 20 in, that will take .6 s.

Assuming that you adjust your numbers to something possible, then the question is, which is going faster at the end? If the instantaneous decelerations at the end both bring the weights to a stop then the impulse is the same. If they are throwing the weights then the one that throws with a higher velocity has the higher impulse. The "instantaneous" force contributes to the impulse.

Remember the equation that relates impulse and velocity:
$m \Delta v = I_{lift}-I_{grav}$

Last edited: May 29, 2012
8. May 29, 2012

### waynexk8

Yes we are negtectiing friction, air resistance. Yes this is what I always said.

Right.

Glad you said that, as some seemed to think that had to be a deceleration. Hmm, just looked at your last answer, let’s get this right, so if I move 80% of my 1RM, with full acceleration to full ROM, and you move 80% of your 1RM, with a very slow speed for the same time frame, are you saying that both have the same impulse, or that I the person moving the weight at full acceleration for the whole time has the higher impulse ???

Yes.

Ok, as I always said, these number were not spot on, just a rough reference, let’s go for the 80 lb with a constant force of 100 lb for 20 in, that will take .6 s.

You seem to be saying that full acceleration to just holding a weight or moving it very slowly, uses the same impulse ??? I do not agree with that, F=ma is the force experienced by an accelerating object, and according to Newton’s laws of motion, force is equal to mass into acceleration. Now acceleration is the rate of change of velocity. This is called impulse which is the product of force and the duration for which it is applied. I am not talking off any deceleration here on my part, as I am not debating the time when the weights are at zero velocity/speed/acceleration/movement, I do not want to add in or take off anything after the weight has stopped moving, I am only concerned with the movement.

F = ma. A weight of 200kg with an Acceleration of 20m/s is 200 x 20 = 4000N, 40m/s is 200 x 20 = 8000N, if they both came to an instantaneously stop at the same time frame, 1 second, the N’s would still be 4000N and 8000N. The acceleration of the weight is directly proportional to the net force. Impulse is the hitting the barbell with my hand, the impulse in this situation is the average force exerted by the hand {arm and all muscles involved} multiplied by the time the hand and barbell were in contact. So the impulse given to the barbell by my hand causes a change in momentum/movement of the weight. As you told me, Impulse is the force over time, it is measured in Newton ,seconds, so a force of one Newton applied for one second will change the momentum/movement, a force of two Newton's applied for one second will change the momentum even more, and that’s what I have done, I have used more Newton’s in the same time frame, I am not interested in the time “AFTER” the rep, the time when momentum/movement is zero, “ONLY” when there is motion

A little problem, say your muscles have a tearing strain of a 100 pounds of force, so if I lift 80 pounds with a 100 pounds of force, in let’s say .3 of a second my muscles will tear. But if you lift the 80 pounds with 80 pounds of force, the muscles would not tear until they were totally fatigued, and that could take up to 30 seconds or more. So here is why I am saying that the forces and average forces, and the impulses and average impulses that I have produced in a very short time, cannot be made up or balanced out by the lower forces and average forces, and the lower impulses and average impulses in the same time frame. My first impulses are 100, 100, 100, 100, yours are 80, 80, 80, 80, my average is higher than yours, that’s why I fail faster, as I fail faster, you cannot average the other forces and impulse forces out, the maybe the only scenario that we can work out, is me failing 50% faster than you, and in that time I have used more force, and more impulse force yes ??? That’s why I use more energy and the EMG reads higher. Is not F=ma, where F is the force applied, m is the mass of the body and ‘a’ is its acceleration, acceleration is rate of change of velocity, so the weight and its velocity is its momentum, so the change in momentum of a body is a product of the force applied on it for some duration. So force effects the acceleration of an object {f=ma} while Impulse simply sets the velocity of an object. This means that same change in momentum/movement can be induced on a weight with a large force operating over a small time period and a small force operating over a long period of time. So the effect of a force is not just the amount of the force but also dependent upon the duration for which it is being applied.

Yes impulse and velocity are related.

Wayne

9. May 29, 2012

### Staff: Mentor

Wayne, there are very few important details:

1) how much is the weight
2) how long does it take
3) what is the change in velocity (in our previous discussions I had always assumed this to be 0, but with your discussion about throwing weights it is no longer certain)

Nothing else matters. With that information you can determine the impulse, without that information you cannot. If those three factors are kept constant then none of the other details about the force matter, the impulse is the same.

No way to tell from your description. Weight and change in velocity are missing.

The weight is 200 kg in both cases, the duration is 1 s in both cases, and the change in velocity (0 m/s) is the same in both cases, therefore the impulse is the same in both cases (~2000 Ns). What you are neglecting is the negative impulse from the sudden deceleration which is -2000 Ns for one and -6000 Ns for the other.

Yes. And the average force is ALWAYS equal to the weight if the barbell starts at rest and stops at rest.

In a rep the weight starts at rest at a low point, is lifted upwards to a high point where it is again at rest (at least momentarily), and is then brought back down to the same low point once again at rest. Correct? I have never discussed the time before nor after a rep, but this is my understanding of what a rep is.

I would prefer to return to discussing weightlifting and reps like normal, but we can continue talking about throwing weights if you prefer.

Yes, not only can you, physics demands it. Irrevocably and invariably. If those three important factors are held constant, then everything that happens in the middle must unavoidably average out. No amount of EMG readings or anecdotes about fatigue can change this simple physical fact.

Yes, exactly. That quantity, the force applied for some duration, is the impulse.

Yes, which is why impulse includes both.

Last edited: May 30, 2012
10. May 30, 2012

### douglis

Here we go again!

NO...either you lift the weight up and down for a minute or you just hold it you use the same impulse and the same force per second.

11. May 31, 2012

### Staff: Mentor

Hi waynexk8, you have made this comment enough times that I think that we need to address it head-on. Here is the proof that your claim is incorrect. Please walk through the proof and ask questions. This is something that you need to understand, so let's take the time to do it correctly.

Here is the mathematical proof, it relies only on 3 assumptions:
1) Newton's second law: $F_{net}=ma$
2) Free weights have only two forces acting on them, gravity and lift from the person: $F_{net}=F_{grav}+F_{lift}$
3) The lift starts and stops at rest: $v_i=v_f=0$

$$I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} ma \; dt = m \int_{t_i}^{t_f} \frac{dv}{dt} \; dt = m (v_f-v_i) = 0$$
$$I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} F_{lift} \; dt + \int_{t_i}^{t_f} F_{grav} \; dt = I_{lift} + \int_{t_i}^{t_f} mg \; dt = I_{lift} + mg(t_f-t_i) = 0$$
$$I_{lift} = -mg(t_f-t_i)$$

So the impulse of the lift depends only on the weight (mg) and on the time (tf-ti), not on the speed of the lift.

12. Jun 1, 2012

### waynexk8

If, and I do mean if, if you’re going to debate on a physics forum, please "try" and state/show your case for being right, and why you think that these are irrelevant here.

1,
The faster you move the weight, the faster you fail.
And more energy used.
You need to explain that I don’t use more impulse over the same time as you, but why I do use more energy, I say you use more energy, and you use more energy the more impulse you use, and this is a proven fact. As the faster you move the weight the more energy you use, and it's directly proportional. Example, if I use 200 calories {energy} to move a weight in 5 seconds, and then use 300 calories to move a weight in 5 seconds, the time I used 200 calories to move the weight, the weight might move 200m, and then the time I used 300 calories to move the weight, the weight would then move 300m, and as you should know, you HAVE to use a higher force and a higher/more impulse to move a weight further in the same time frame. The acceleration a of the weight is directly proportional to the net force and proportional to the mass. F = ma, so more full acceleration for .5 of a second, you HAVE to use more force, I am not talking about the time before acceleration, or after acceleration. The force at work on a car as it starts a race, f the car has a Mass of 500 pounds and an acceleration of 10m/s work out the Force pushing the car by multiplying the Mass by the Acceleration like this 500 x 10 = 5000N, if it’s moving SLOWER, 5m/s its 500 x 5 = 2500N. Please not that 5000 = more than 2500. If I accelerate a weight for 1 second using 10N and you move a weight for 1 second using 5N, I have used 10N for 1 second, you have used 5N for 1 second, how is 5 as high/more than 10 ???

2,
You need to explain why a very sophisticated bit of equipment, that has a computer that adds up things very fast and accurate, is as you think wrong ???

3,
Power,
At least you can understand that the faster moving uses more power.

Wayne

13. Jun 1, 2012

### waynexk8

I would say there are two proofs, paper equations and then a real World practical test/study, problem is, the real World practical test/study, as in energy, hitting failure and the EMG does not agree with equations, the equations are not wrong, just the way of doing these equations are wrong, something will be needed to be changed. It’s like when we did equation to see how far a weight would move if we lifted at a certain speed and stopped, it was 3 inch, but when we did the test the weight did not move, the equations were not wrong, it was because of the muscles biomechanical advantages and disadvantages thought-out the ROM, that they could not push will the full equations force for the whole rep. I this debate, there could be something completely different that has to be change, and nothing either to do with the human body. if the car has a Mass of 500 pounds and an acceleration of 10m/s work out the Force pushing the car by multiplying the Mass by the Acceleration like this 500 x 10 = 5000N, if it’s moving SLOWER, 5m/s its 500 x 5 = 2500N. Please not that 5000 = more than 2500. If I accelerate a weight for 1 second using 10N and you move a weight for 1 second using 5N, I have used 10N for 1 second, you have used 5N for 1 second, how is 5 as high/more than 10 ???
In this debate, we are not debating anytime before or after the has started to move.

$$I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} ma \; dt = m \int_{t_i}^{t_f} \frac{dv}{dt} \; dt = m (v_f-v_i) = 0$$
$$I_{net}=\int_{t_i}^{t_f} F_{net} \; dt = \int_{t_i}^{t_f} F_{lift} \; dt + \int_{t_i}^{t_f} F_{grav} \; dt = I_{lift} + \int_{t_i}^{t_f} mg \; dt = I_{lift} + mg(t_f-t_i) = 0$$
$$I_{lift} = -mg(t_f-t_i)$$

So the impulse of the lift depends only on the weight (mg) and on the time (tf-ti), not on the speed of the lift.[/QUOTE]

I cannot pretend to know I understand the equations, but if they were right, why do I fail faster using a faster velocity, why do I use more energy {calories} why does the EMG state I use more muscle activity, and more muscle activity = more force.

Here is why I fail faster, I fail faster because the faster I move the move tension I put on the muscles, thus the more strain on them, if the slower moving reps were to equal out as the equation seems to same, it says that’s the forces average out, if the forces averaged out, then the slow repping person would average out their forces thus tension on the muscles thus strain, and fail the same time as the faster moving person, but this does not happen, as its like this, 100, 100, 100, 100, 000 fast, 80, 80, 80, 80, 80, slow, 100 is higher than 80, 100 is higher than 80 force 80% of the whole time of the rep, so how can 80 done for just 20% make this up ??? In my opinion it does and cannot, if it did make this force up, then in the real would practical experiment as in lifting weight and seeing who hits momentary muscular failure faster, both would hit momentary muscular failure at the same time, because the slow would equal out balance out all the forces. But both reps do not hit failure the same time, the fast hits failure 50% faster, thus how can the forces be the same ??? How can they be the same if both don’t hit failure the same time ??? Here is another reason why, 100, 100, 100, 100, 000 fast, 80, 80, 80, 80, 80, note the fast does have to more force, and more impulse force the first 80% of the time frame, forget the last 20% for now, see it ???

The acceleration a of the weight is directly proportional to the net force and proportional to the mass . F = ma, so more full acceleration for .5 of a second, you HAVE to use more force, I am not talking about the time before acceleration, or after acceleration. If I use 10N of force for .5 of a second, and you use 3N of force for .5 of a second, I have used more force, and more impulse !!!

Newton's 2nd Law and the dynamics of Aristotle. According to Newton, a force causes only a change in velocity, {an acceleration} it does not maintain the velocity as Aristotle held.

So how can/is the object moving ??? We will say it’s a weight, I acceleration it using force, I move it at a constant velocity using force, I deceleration at with force, if I stopped using force, the weight would start accelerant until it hit the Earth, if it was dropped from a great height, it would accelerate until the air pressure was the same, then it would move at a constant speed. However if I was pushing a weight up, UI would HAVE to use force for both acceleration and constant speed, looks like both were right ??? If you have a 80 pound weight, and it’s travelling at a constant speed of 50m/s, Newton's second law, says there should be no force because there is no acceleration, but there is forces, the force of gravity, the force of friction, the force of momentum/movement, motion, a force pushed moved the weight, thus now the force moving the weight is the weight/mass of the weight itself, gravity and friction. As if it hits another object it will move the object, so there is force there but according to F=ma there is no force.

Sorry and thank you for your help agai8n dalespam, but it’s a little last for your other post sorry.

Wayne

14. Jun 1, 2012

### Staff: Mentor

I agree. That is why scientists continuously put the paper equations to the most careful and rigorous experiments that they can imagine, and great honors are bestowed on scientists which prove equations wrong and come up with better ones. The particular equations that we are discussing here have passed the most exacting tests imaginable for over 3 centuries of the most brilliant experimentalists. The real world tests completely agree with the equations I posted.

No, the way of doing the equations is correct. If you disagree then please point out EXACTLY where I made my mistake.

10 N*s is indeed more impulse than 5 N*s, nobody is disputing that.

What you don't seem to realize is that after you have applied 10 N*s and I have applied 5 N*s your weight is moving faster than mine. Over the course of the next second, if we both bring our weights to rest, then you must use LESS force than me such that our average force is the same.

Good to hear. I certainly have never mentioned anything about any time before the weight has started to move or after it had stopped.

The impulse is the same for fast or slow, the energy is higher for fast than for slow, therefore, the energy does not depend only on the impulse.

The impulse is the same for fast or slow, the EMG is higher for fast than for slow, therefore, the EMG does not depend only on the impulse.

The impulse is the same for fast or slow, you fail sooner for fast than for slow, therefore the failure time does not depend only on the impulse.

...

That would be correct IF the failure depended only on the average strain, therefore the failure does not depend only on the average strain.

Let's say that the weight is 75 lb and you lift for 1 s at 100 lb and I lift for 1 s at 80 lb. At the end of that second your weight is moving 11 ft/s and my weight is moving just 2 ft/s. Now, let's say we want to decelerate the weight in 0.5 s, in order to do that, you must use a force of 25 lb, and I must use a force of 65 lb. Any other force and the weight will not stop, i.e. the rep will not be at an end.
For me: 80 lb * 1 s + 65 lb * 0.5 s = 112.5 lb*s
For you: 100 lb * 1 s + 25 lb * 0.5 s = 112.5 lb*s

Your opinion is wrong, as proven mathematically, and the real world careful controlled experiments confirm the math. The average force is the same for fast and slow. Momentary muscular failure occurs sooner for fast than for slow. Therefore momentary muscular failure is not due only to the average force.

Yes. And your weight is moving faster than mine. Since the rep doesn't end until the weight is stopped again, this is not a complete description of a rep.

waynexk8, I really think that you should look at the mathematical proof I posted and try to understand it. You are just shrugging your shoulders and ignoring it instead of taking the opportunity to learn something. I cannot help if you don't tell me what part of it you do not understand. Do you understand the 3 assumptions?

Last edited: Jun 1, 2012
15. Jun 2, 2012

### waynexk8

You hit a Snooker ball with a force of 10N, the cue comes in contact with the ball for .1 of a second, it moves 1m hits another Snooker ball, the original Snooker ball stops, and the second Snooker ball goes 1m. You hit a Snooker ball with a force of 5N, the cue comes in contact with the ball for .1 of a second, it moves 1m hits another Snooker ball, the original Snooker ball stops, and the second Snooker ball goes 500mm. This will be the same as your muscles moving a weight, as the reaction, reaction, will then be the muscles moving the weight, and the reaction, reaction from the force and impulse will be the tension on the muscles.

I do not get why/how you are saying that both hits have the same force and impulse, average force and average impulse, when I stated the Ns were different but the time frame the same. As everything, the N, the time, and the distance the second Snooker ball moves points/says that more force and more impulse, more average force and more average impulse was used, there is nothing saying both were the exact same. Back later.

Wayne

16. Jun 2, 2012

### douglis

I have stated so many times why I think your examples are irrelevant that I'm sure it would be a waste of time to do it once again.But what the hell.....

I showed you studies that found that the greater fluctuations of the same average muscle force are more energy demanding.
For example,if the applied force varies from 0 to 10N in a second requires more energy than an applied force that varies from 4 to 6N in a second.Despite the fact that impulse and the average force is identical.
That's the ONLY reason that fast reps lead to failure faster.They're more energy demanding(due to their greater fluctuations of force) even though their average force per second is identical with slow reps.

Your example proves that you didn't even read or you didn't understand(most probably) the assumptions that DaleSpam is talking about.

We're talking about cases where the load starts and stops at rest hence there's zero change in momentum.In your example,in the first case the one ball hits the other with double speed than the second case(I assume that the 10N and 5N are average values) and in .1sec the balls stop in both cases.
So the first case has double change in momentum than the second.Nothing like our weight lifting example where the change in momentum is always zero.

Last edited: Jun 2, 2012
17. Jun 2, 2012

### Staff: Mentor

I have never said that, nor has anyone else. Please focus on the weightlifting scenario and stop bringing in irrelevant scenarios. They only distract you and the respondents.

If you really wish to discuss Snooker, then open a separate thread and keep the two topics separate.

18. Jun 2, 2012

### waynexk8

Back in full tomorrow, and the pm’s, SORRY for not getting back to the older, will do first thing before anything else.

Two weightlifters, one lifts the weight up and over his head and back down in 3 seconds, second only lifts it to his knees in 3 seconds. Are you telling me that both used the same force, same impulse/strength ??? If so, then how did the weightlifter have more impulse/stremgth to be able to lift the weight overhead against gravity and air resistance ??? Someone said that here you work out the average in two ways, maybe this is where we are losing each other.

Wayne

Last edited: Jun 2, 2012
19. Jun 2, 2012

### Staff: Mentor

I think that the phrase "the weight" means that both weightlifters are lifting the same weight.

I think this means that they both start at rest and stop at rest.

So the time is the same.

Let's use what we have learned so far and see if we can work this out.

First, we need to check our three assumptions:
1) Newton's second law: $F_{net}=ma$
Since we are on Physics Forums, let's go ahead and allow this assumption since it is pretty central to physics, even though it wasn't explicitly mentioned in the problem.

2) Free weights have only two forces acting on them, gravity and lift from the person: $F_{net}=F_{grav}+F_{lift}$
In the problem description there was no mention of any machine that would cause friction or any other force on the weight, so this one is a valid assumption also.

3) The lift starts and stops at rest: $v_i=v_f=0$
You explicitly mentioned this in the problem description.

So all three assumptions are valid. We know that whenever the assumptions are valid we can calculate the weightlifter's impulse by:
$I_{lift} = -mg(t_f-t_i)$

So for lifter 1 we have: $I_{lift1} = -mg(t_f-t_i) = -mg (3 s)$

And since the mass is the same for both and obviously gravity is the same for both, then for lifter 2 we have: $I_{lift2} = -mg(t_f-t_i) = -mg (3 s)$

Therefore $I_{lift1} = -mg (3 s) = I_{lift2}$

He used a higher peak force, and a lower minimum force, so that the average force was the same.

20. Jun 2, 2012

### douglis

Yes....we're telling you exactly that.Both used the same average force and impulse.

Last edited: Jun 2, 2012