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Calculate the force needed to lift

  1. Dec 29, 2009 #1
    I made a machine a few years ago, there seems to be something wrong with it. I need the force to push the object to be equal to the weight 1:1. but it seems that it needs more force to move it. You can see the details in the below diagram.

    I suspect the mistake is in the angel and that the length of the vertical and the horizatal are not equal.

    I need to know the current ratio and how to calculate it, I would also like to know the mistake I did to make it 1:1.

    You can find the diagram in the attachment.

    Please explain in lame english terms.

    I did not understand " Your horizontal distance is 6*sin(80) = 5.9 ft, while your vertical distance is 5 ft. For the torque to balance, you need W*5.9 = F*5, thus F = (5.9/5)*W."
     

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    Last edited: Dec 29, 2009
  2. jcsd
  3. Dec 29, 2009 #2

    Doc Al

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    Yep. Your horizontal distance is 6*sin(80) = 5.9 ft, while your vertical distance is 5 ft. For the torque to balance, you need W*5.9 = F*5, thus F = (5.9/5)*W.
     
  4. Dec 29, 2009 #3
    I need the info to be explained in lame english terms. I dont remmber my physics. so please explain in detail.

    Please explain the mistake and if I am to make it again what should I change. The equation that is needed to calculate the current setup.

    If I would rebuild the machine. Shouls I make the angel 90 degrees? and the the vertical and the horizantal 1:1 as in equal to each other?
     
  5. Dec 29, 2009 #4

    Doc Al

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    If you want the applied force to equal the weight, then you need to make the horizontal distance from the weight to the pivot equal to the vertical distance from the force application point and the pivot. Currently, the horizontal distance is 5.9 ft and the vertical distance is 5 ft. (Realistically, you should also account for the weight of this apparatus, which needs to be lifted as well as the weight.)
     
  6. Dec 29, 2009 #5
    That is a good explanation.

    Does it matter that the bottom beam is about 6 inches above the pivot point? There are 2 other support beams above the bottom one. The middle one is 90 degrees it is 6ft long. Do they make any changes?

    Should I rebuild the machine with 90 degrees at the bottom beam, and the vertical and horizontal being equal to get 1:1 ratio?

    The weight of the machine when empty is about 70lbs at the back, when the weight is added it becomes about 320lbs. can u tell be the force needed to lift that, and show me it in an equation where I can change the numbers to get the proper calculation?

    is it (5.9/5)*320?

    Shouldn't be more since the middle beam is 6ft long, I made a mistke, the bottom one is more. sorry.

    does it count if the middle beam is at 90 degrees, this is not the usual situation since it it is above the pivot point.
     
    Last edited: Dec 29, 2009
  7. Dec 30, 2009 #6

    Doc Al

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    No.
    What matters is the horizontal distance from where the weight is applied to the pivot. If the horizontal beam 6 ft long, then that distance is 6 ft. I thought your diagram showed the lower (angled) beam as 6 ft long, which gives a horizontal distance of 6*sin(80) = 5.9 ft.

    No.
    Since you need to take into account the weight of the apparatus, you'll need the vertical distance a bit greater than the horizontal.

    It depends on the location of the center of mass of the apparatus.

    So the middle beam is 6 ft long, not the bottom one.

    No, only the horizontal distance matters.
     
  8. Jan 3, 2010 #7
    I understand it 100%, I new question about the weights and the distance. If I correct the length of he vertical to be 60ft and 60ft. 1:1 but the weight is round and connected as shown in the below Pic.

    You said b4 its about the distance from the vertical. if its about a 1000 lbs and about 100" wide, will the distance make a diffrance or will de round weight at the front cancel the wieght at the back?
     

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  9. Jan 3, 2010 #8

    Doc Al

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    So your apparatus has changed from 6 ft to 60 ft?

    I'm not quite sure what you are asking here. But if your round weight is symmetric, you can consider its weight as acting at its center.
     
  10. Jan 3, 2010 #9
    I changed the size just to see all possibelities and future upgrades. So from what I understand. The distance on the end of the weight from the pivot does not matter as long as it is cemetrical. Round and 100% balanced. so even if the weight extends 50" at the back, it wont make a change as long as it round.

    Is there a theory or an equation that u can show me on this pls. U said B4 it about the vertical distance. so what makes it ok if the weight is round?

    Pls explain.
     
    Last edited: Jan 3, 2010
  11. Jan 5, 2010 #10

    Doc Al

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    What matters is where the added force from the weight is applied. Since the weight is symmetrical, its center of mass is at its center. So all that matters it the horizontal distance of the weight's center to the pivot.

    For the purpose of calculating torques, the gravitational force on any object can be treated as acting at the object's center of mass.
     
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