1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The question no one seems to know average force

  1. Feb 25, 2010 #1
    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
     
  2. jcsd
  3. Feb 25, 2010 #2
    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
     
  4. Feb 26, 2010 #3
    Hi there,

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

    Wayne
     
  5. Feb 26, 2010 #4

    russ_watters

    User Avatar

    Staff: Mentor

    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.
     
  6. Feb 26, 2010 #5

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

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

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

    [tex]\implies \mathbf{F}_{av}(up) = \mathbf{W} = 160lb[/tex]
     
    Last edited: Feb 26, 2010
  7. Feb 26, 2010 #6
    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
     
  8. Feb 26, 2010 #7

    russ_watters

    User Avatar

    Staff: Mentor

    If the deceleration exceeds the acceleration due to gravity, then you'd have to pull the weight down.
     
  9. Feb 26, 2010 #8


    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
     
    Last edited: Feb 26, 2010
  10. Feb 27, 2010 #9
    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
     
    Last edited: Feb 27, 2010
  11. Feb 27, 2010 #10
    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
     
  12. Feb 27, 2010 #11

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

    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.
     
  13. Feb 28, 2010 #12
    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
     
  14. Feb 28, 2010 #13
    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
     
  15. Feb 28, 2010 #14

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, that's 380% more work done (not force used).
     
  16. Feb 28, 2010 #15
    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
     
  17. Feb 28, 2010 #16

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  18. Mar 1, 2010 #17
    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
     
  19. Mar 1, 2010 #18
    I still can not belive no on e can help me with this ??? Surely there is a way to work out how much force is used ???

    Wayne
     
  20. Mar 1, 2010 #19
    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
     
  21. Mar 2, 2010 #20

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The question no one seems to know average force
  1. Average Force (Replies: 5)

Loading...