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Calculating force ?

  1. Nov 14, 2009 #1
    Calculating force ???

    I am trying to work out the force used on lifting a 91kg barbell for .5 of a second for 1.85M. And holding a 91kg barbell statically for 5 seconds.

    Below is my attempt ???
    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/s² = 892 kg.m/s² or 892 Newtons).

    Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm. 1 Newton meter “Nm” is equal to 1 joule, hence 1 joule is the work done when a force of 1 N moves through a distance of 1m in the same direction as the force.

    As no distance is used on the static hold, the force used is 892N ???

    The concept of power however, takes time into consideration. If for example, it took four seconds to complete the lift, then the power generated is 1650 J divided 4 s = 412.5 J/s. If on the other hand it only took 2 seconds to complete the lift, then the power generated is 1650 J divided 2 s = 825 J/s. Hence, the person who can lift the barbell in two seconds is more powerful than the one who lifted the barbell in four seconds.

    So while the work done remains constant, more power is developed when mechanical work is done more quickly. Indeed, power can be thought of as how quickly or slowly work is done.

    Thus for the person holding the barbell statically for 5 seconds we need to divided 892N by 5 = 178.4 Joules

    However BY lifting the same 91kg for 1.85m in .5 of a second, first you x 892 N x 1.85 m = 1650 Nm, then divided 1650 by .5 = 3300 joules per second.

    Wayne
     
  2. jcsd
  3. Nov 14, 2009 #2
    Re: Calculating force ???

    As no distance is used on the static hold, the force used is 892N ???

    Thus for the person holding the barbell statically for 5 seconds we need to divided 892N by 5 = 178.4 Joules.



    Looking back at the above two points, I must be wrong. As it must take more something to hold the weight static for 5 seconds to 1. Surely it take more force and energy ??? But it can not take more power {work} as its a static hold.

    Wayne
     
  4. Nov 14, 2009 #3
    Re: Calculating force ???

    As far as I see, everything is correct, except for the above.

    As you said:
    work is equal to Force x distance

    Therefore the work done by holding a barbell statically for 5 seconds is 892 x 0 = 0

    Hold it statically for 30 seconds....892 x 0 is still 0

    Holding a barbell in place requires just about the same force as leaving it on the floor. No work is being done. You can leave it for 5 seconds, or for a million years....work is going to be 0, and power is going to be 0.
     
  5. Nov 15, 2009 #4
    Re: Calculating force ???

    Hi Lsos, and thx for your answer.

    However what I would really like to know is how much the force the static hold will produce for the 5 seconds ??? Would I maybe x the 892 ??? Which would then be = 4460, as to hold a weight for longer must produce/take more force/strength ???

    I think the above x the 892 is wrong, but I am just putting porkers in the fire.

    Wayne
     
  6. Nov 15, 2009 #5

    HallsofIvy

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    Re: Calculating force ???

    Are you clear on what you are asking? The force necessary to hold an object at a given height is its weight. For an object with mass 91 kg, that weight is 91g= 91(9.8)= 98 Newtons, approximately. And that is constant- it doesn't change over time so It makes no sense to say "how much the force the static hold will produce for the 5 seconds". It does NOT take more force to hold something for a longer time. That's why you can set something on a table for as long as you please.

    (Physiological note: If you are holding something out in front of you, your arm muscles have to keep applying that same 982 N force. That's why you get tired, not because it requires more force.)
     
  7. Nov 15, 2009 #6
    Re: Calculating force ???

    Hi HallsofIvy,

    Hmm, see what you are saying, however it must take more of something to hold a static out for the different seconds??? Like strength/energy, as power the rate at which work is performed or energy is converted.

    Is there not some way in physics that could/should be able to work this out please ???

    Wayne
     
  8. Nov 21, 2009 #7
    Re: Calculating force ???



    Hi all,



    And thx for the above.



    I now find that it seems you cannot compute what I asked. I just thought there would be some way ???



    Wayne,

    I am not exactly sure what you are asking. When the lift involves a displacement, the performance can, as you described, be quantified by the work that was done by the muscles or the power that was produced. When the task does not involve a displacement, the performance can be expressed as the impulse (F x t), which is a reasonable estimate of the energy required to hold the position of the weight against gravity. However, there it is not possible to compare the magnitudes of work (or power) and impulse for the two tasks.

    I hope this helps.


    Roger M. Enoka, Ph.D.
    Professor and Chair
    Department of Integrative Physiology
    University of Colorado






    Wayne
     
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