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Force free body diagram problem on gym equipment

  1. Feb 10, 2012 #1
    I'm looking at the force required to move the weights, i have a few sketches attached on a word document with an idea of how the apparatus works.

    Will the force required to lift the mass increase as the player moves forward.

    Any help is appreciated as i am lost with this problem at the moment.

    A video of the machine inn use can be seen at : http://www.youtube.com/watch?v=sO5FII1EmMU&feature=related

    Thanks
     

    Attached Files:

    Last edited: Feb 10, 2012
  2. jcsd
  3. Feb 11, 2012 #2

    Q_Goest

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    Hi Sully, welcome to the board. Are you an engineering student or just looking for the answer? I'll assume the latter... Forces are "vector quantities" meaning they have a magnitude (some measurable amount of force) that occurs in a given direction. So as you've shown, there is a weight that has a force in a direction vertically downward and a force by the person pushing on the apparatus that is horizontal. There is a point around which the weight rotates which will be in pure tension as long as the apparatus isn't moving. For this analysis I'll assume the weight isn't moving (static case). If the weight is actually moving, the forces are not equal to this static case, but they will be fairly close as long as the person isn't moving very fast. The slower the person moves, the closer the force will be to the static case.

    Attached is your file modified to show angle A. The horizontal component of force can be found from:

    tan (A) = Fh / Fw
    where Fh is the horizontal component of force that the person must exert
    Fw is the weight acting downwards

    So Fh = tan (A) * Fw

    Note that the weight Fw is actually the sum total of the weights plus the apparatus that's swinging around. The center of gravity of the combined apparatus plus weights may or may not coincide with the actual location of the weights due to the weight of the apparatus, so the angle A isn't necessarily the angle the weights make with the rotating pin, the angle is really the angle between the center of gravity of the rotating apparatus and the pin with respect to the vertical.
     

    Attached Files:

  4. Feb 11, 2012 #3
    Thanks very much for the help. i am studying engineering but have done very little in that area.

    For a project i'm investigating how the force a player has to exert as he moves through the full push because the resistance is increasing. I must create an excel program that will allow the user to enter a mass so at every stage along the push we can work out the opposite force. will the method using the vector quantities work for me to calculate the force the plaayer is exerting?
     
  5. Feb 11, 2012 #4

    Q_Goest

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    Yes. :)
     
  6. Feb 11, 2012 #5
    Cool, unfortunately im still struggling. I have drawn a free body diagram and attempted to solve the problem but im getting a value for Fh of over 3000N for a weight of 1000N on the machine. I have feeling im goin wrong somewhere. When Does this seem large or strange?

    I have attached the free body diagram of the weight before its lifted and after its lifted and at its max point.
     

    Attached Files:

    Last edited: Feb 11, 2012
  7. Feb 11, 2012 #6

    Q_Goest

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    So you have a horizontal force of 3000 N predicted given your second attachment? What angle A are you using between ab and vertical?

    Also, why do you have a 15 degree angle between cd and horizontal?
     
  8. Feb 11, 2012 #7
    Am i right in saying the following looking at the basic free body diagram?

    Fac = 1000/sin 15 =3863N ????

    Fh = Cos15 * Fac = 3732 ???
     
  9. Feb 11, 2012 #8

    Q_Goest

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    I'm afraid I don't know how you derived those...

    Maybe this would help: Let's ignore the linkage on the exercise machine with the exception of the link between the hinge point and the weight, ab. For this linkage, ab, let's imagine a rope instead of a bar of some kind. The rope can only be under pure tension. We can't bend it or it won't be straight any more like link ab So all we have left is a weight at the end of a rope that's held up by a pin. If it dangles straight down, the tension in the rope is obviously just the value of the weight.

    Now let's see if we can derive the horizontal force required to push this weight off to the side so instead of hanging vertically, the rope is rotated out to the side by some angle A (as I've marked in the attachment above). So the rope makes an angle A from a vertical line. Can you figure out what horizontal force is required to produce some given angle A and the resulting tension in the rope?
     
  10. Feb 13, 2012 #9
    Yes that makes sense. So at the start it will require very little force to start the motion from static and this will increase significantly as the angle increases?
     
  11. Feb 13, 2012 #10

    Q_Goest

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    Correct.
     
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