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How to calculate the force needed to get out of bed?

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  1. May 30, 2015 #1
    I know from physics 101 that f = ma. And if I want to lift an object straight up i have to multiply its mass times gravity (acceleration) and I get the force I need to require to overcome it.

    But what about lifting your body out of bed? Just the part where you physically get up from the lying down position to the sitting up position. In other words, just for lifting your torso up from horizontal to a vertical?

    I know it has something to with the the angle but Im not sure how to incorporate it into f=ma.

    Thanks
     
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  3. May 30, 2015 #2
    My initial guess is on center of mass. It is a point where all the external forces acting on a body can be considered acting at that point.

    Well, initially, center of mass has one position and after the process, it has another which is higher than usual.

    So, you need the same force mg to lift you off the bed.

    I am not so sure about this treatment.
     
  4. May 30, 2015 #3

    mfb

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    Your body is not a single point, so giving an answer as a single force value is problematic.
    If you model the torso as rigid object that pivots around the hip, you can calculate a force. Initially the whole torso will move directly upwards, which allows to calculate the force as you described. That is not the same as forces inside your body to lift you up, as you have to include lever arms and other issues there.
     
  5. May 30, 2015 #4

    phion

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    It might make more sense to calculate the energy required to get out of bed, considering the biological and chemical aspect of waking the **** up.
     
  6. May 30, 2015 #5
    No Im not interested in Energy at all.

    Lets imagine that you dont even use the arms. Let's just consider lifting the mass out of bed.

    Even if its a rough calculation, how would I go about doing it?
     
  7. May 30, 2015 #6

    mfb

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    Make a model of what moves where and how the energy changes as function of the position, force (in the direction of motion) is then the derivative of this.
     
  8. May 30, 2015 #7
    Ok here are my drawings. 4 scenarios.

    A: Simplistic F = ma of lifting an object up from the ground.

    B: Force at an Angle = Not sure how but if the same object was being lifted up, but at an angle, the force would be different? Like in those examples of a horse pulling a cart vs the same horse pulling that same cart but from a sideroad, at an angle. I kinda remember we needed to consider the angle somehow.

    What I really want to understand is the getting out of bed example.

    C: Getting up out of bed requires you to lift a weight from 0 angle to 90 degrees (sitting up). This requires a certain force.

    D: I want to udnerstand the basic C scenario in order to understand how that would change when inclining the bed, so the person needs to sit up, once again to 90 degrees from the bed, but this time the bed is inclined 20 degrees.

    IMG_5892.JPG
     
  9. May 30, 2015 #8

    CWatters

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    Actually it involves rotating a mass from the horizontal to the vertical position so it requires a torque rather than a force. The torque is at a maximum when the trunk is horizontal (just starting to rotate off the bed.)
     
  10. May 30, 2015 #9
    Thanks CWatters.

    So I ran into this online:

    http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

    Im having trouble understanding what everything is in this diagram. The red line moves thru theta (dunno how to make that on the keyboard of my mac), to where the dotted black line is. Im guessing thats my body.

    F is the Force moving that lever. Except that in my case, to that force I will have to add the force of gravity as well, right? because in that diagram I think its considered as being on a flat surface, like twisting a monkey wrench.

    How would I enter the values?
     
  11. May 30, 2015 #10

    CWatters

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    According to...
    http://www.exrx.net/Kinesiology/Segments.html

    Your legs and feet account for about 17% of your mass. Perhaps it would be reasonable to assume that the part being rotated (trunk, head, arms etc) accounts for the other 83%.

    Lets say you weigh 100Kg then 83% is about 83kg. If the centre of mass of the moving part is say 0.4m from the pivot point then the required torque is 83 *9.8 * 0.4 = 325Nm.

    If you don't like my assumptions you can easily change them.

    PS: The above implies something needs to hold your feet down.
     
  12. May 30, 2015 #11

    CWatters

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    No the red line doesn't move through theta. The diagram is showing how to calculate the length of the "lever arm" when it's not at 90 degrees to the applied force.

    I'll make a diagram that shows what I meant in post #10. Back soon.
     
  13. May 30, 2015 #12
    Ok but that website also has a calculator at the end for force, i think.

    I did the following, although I like your assumptions better:

    I weight 72 kg, so thats = 72 * 9.8 = 705N.

    I considered 50% weight in the torso (yours seems more logical), so I went with a rough 350N.

    Force Applied 350N
    Distance: 1.7m (my height)
    Theta: 90 degrees

    I got stuck calculating the lever arm. For some reason the online calculator sets my lever arm automaticaly to 1.7m and gives me a value of 595 Nm.
     
  14. May 30, 2015 #13

    CWatters

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    Ok so this shows how to calculate the max torque.

    Torque = force * length of lever arm.

    As you sit up the length of the lever arm reduces until the centre of mass is over the pivot point. The length of the lever arm and the torque is at a maximum when you are horizontal as shown.

    Sitting up.jpg
     
  15. May 30, 2015 #14

    CWatters

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    That would imply a pivot point at your feet.
     
  16. May 30, 2015 #15

    russ_watters

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    Edit: deleted, think I missed your point. You were referring to the torque being great enough to lift your feet instead of your torso. Agreed.
     
    Last edited: May 30, 2015
  17. May 30, 2015 #16
    Yes CWatters, so changing 1,7 to 0.85 I get 297.5Nm.

    I know Im leaving out the "someone holding your feet down" counter force and the fact that the lever arm changes.

    So but is that correct? I guess it makes sense that you require less force to lift 1/2 your body up to a sitting position that to lift your entire body.

    Now I have 2 lingering questions:

    A/ If the bed was at an incline, as in scenario D from my image. What would change is the angle, from 90 to 90+20, correct?

    B/ If I wanted to consider the changing length of the lever arm as I sit up, I would need to use calculus?
     
  18. May 30, 2015 #17

    CWatters

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    Sure. What I meant is... If 83% of your mass is above the pivot point (your hips) then your feet will lift up rather than your head.
     
  19. May 30, 2015 #18

    CWatters

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    0.85 is on the high side. It should be the distance from the pivot to the centre of mass of the moving part. If the pivot point is at your hips (as in doing sit ups) then I would expect the centre of mass of the top half of you body to be around chest height (hence my estimate of 0.4).
     
  20. May 30, 2015 #19

    CWatters

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    Yes. However since gravity acts vertically the torque will still be at a maximum when the trunk is horizontal. eg raising the legs 20 degrees won't affect the max torque required.

    Basic trig might be all you need. For example if you lift up to the 45 degree position the lever arm is 0.4 Cos 45 = 0.28 and the torque falls to 83 * 9.8 * 0.28 = 227 Nm.

    Sitting up 45.jpg
     
  21. May 30, 2015 #20
    Ok I didnt understand either of your answers. Im slow. :-)

    A/ I didnt mean raise just the legs. I meant that the entire bed, instead of being flat (0 degrees), the bed is now inclined 20 degrees. Its like making situps in a gym, where you have those boards that you can incline to make the situp workout harder.

    How does that factor in? I thought it just meant the torque angle was greater by 20 degrees?
     
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