The question no one seems to know average force

[waynexk8]

Sure, I will go over it.

Fine.

As you said above f=ma so
50 kg 1 m/s² = 50 N
50 kg 6 m/s² = 300 N

OK here is the first mistake. Your units are not correct, remember what you said above, force (N) is mass times acceleration (kg m/s²), not mass times distance (kg m) as you have here.

What you need to do is to calculate the acceleration. Acceleration is the change in velocity divided by the change in time, and the velocity is the change in distance divided by the change in time. So let's go back up to the top. Your mass is moving at a constant 2 s for 1 m, so v = Δx/Δt = 1m/2s = 0.5 m/s. Now, you say that the speed is constant so the change in v is Δv = 0, so the acceleration is also 0.

Now, in reality the speed is not constant, but it starts and stops at 0 so the average acceleration is 0.

Again, you have similar problems with units, you cannot equate Newtons and kilograms.

Hi, and thanks for your time and helping me.

So you are saying the below is wrong ?

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 4.

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.

However, I found the below and checked on there and they seem right ? Unless I am calculating the wrong thing ? I will read what you wrote again, but getting a little sleepy, and as you know its best not to go over things then, as its getting late here, its gone 01.00

http://www.ajdesigner.com/phpforce/force_equation.php

Wayne

 

[waynexk8]

This is correct except that you cannot convert joules into pounds. You could convert it into calories (1 kcal = 4187 J)

If you are doing 4 fast reps in the same time as 1 slow rep then it has moved 4 times the distance in the same time.


All correct.

All correct except for the conversion to pounds.

I wouldn't take off for the deceleration, you are still doing work. And the time is 0.5 s for a fast rep. So you for the power you should have 490.5J/0.5s = 981 J/s.

Sorry, that is all I have time for now, but I am sure you get the idea. You are going 4 times as fast so you are doing the same amount of work in 1/4 the time which implies 4 times the power, despite the fact that the average force is the same.

Thx get what you say there, and see how silly I was trying to change Js to pounds.

Wayne

 

[waynexk8]

I WAS JUST THINKING, Maybe that we all are forgetting that the slow reppers are going at a constant speed, and NOT accelerating, BUT I am NOT going at a constant speed, as I AM accelerating all the time, and there is a BIG difference between going at a Slow constant speed, and accelerating at a VERY FAST ACCELERATING SPEED.

THUS IF THE BELOW IS RIGHT, HOW CAN EVEN THE AVERAGE FORCE BE THE SAME ? IF I AM ACCELERATING AND THE OTHER AT A CONSTANT SPEED, ARE NOT THINGS VERY DIFFRENT ?

They push a 100 pounds at 2/4 {we are just counting the concentric} for 1m with a constant extra force {strength} of 6.25N. That works out they are using 1.4 pounds extra for the whole duration of 2 seconds.

I can push that same 100 pounds at .5/.5 for 2m, accelerating with an extra force {strength} of 25N. That works out I am using 5.6 pounds extra for the whole duration of 2 seconds. Take off 20% for the deceleration = 4.4 pounds. But I am just having mini breaks, the eccentrics which you are not having.

Therefore, in 2 seconds of lifting for the slow 1 rep, and me lifting for 2 seconds fast, I have used OVER 200% more force {strength}

Thus if we did 12 reps you would use an extra 16.8 pounds, and I would use an extra 52.8 pounds, or 36 pounds MORE force {strength} that they in one set, and that's a VERY lot that's as I said over 200% more in each rep and in each set.

Wayne

 

[Dale]

THUS IF THE BELOW IS RIGHT, HOW CAN EVEN THE AVERAGE FORCE BE THE SAME ? IF I AM ACCELERATING AND THE OTHER AT A CONSTANT SPEED, ARE NOT THINGS VERY DIFFRENT ?

Don't forget that acceleration and force are vector quantities. So if you use 2000 lb force to accelerate it really fast to the left and then 2000 lb force to accelerate it really fast to the right (which you have to do) then the average force is 0. Similarly with up and down forces and accelerations. No matter what you do, over any cycle the average force is equal to the weight (but pointed up).

 

[waynexk8]

Thx for the above, not much time now DaleSpam.

If you move a load faster over a given distance you are producing more force/power, it is impossible for this not to be the case. Moving 100 pounds a distance of 2 feet in 1 second requires more force/power than moving 100 pounds a distance of 2 feet in 2 seconds.

Agreed.

In my example, the rep was 1m concentric and 1m eccentric.

However, something is still bugging me on these average forces, but think I have worked out where we are all going wrong, no time now. But why are we counting the decelerating phase of the fast concentric, thus why are we talking 20% off ? I know saying the following may seem a little silly, but in a drag race, they only count the force or in their case speed/time that was done in the ? Mile they raced, they did not count the decelerating part of it. Got to go later.

Wayne