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Calculating max weight object supports without counterweigh

  1. Oct 19, 2015 #1
    In tower cranes counterweights are placed on the counter jib to prevent the crane from falling over. Whereas on mobile cranes the counterweights are placed directly behind the cabin.
    The crane in itself also acts as some counterweight.
    My question to you is:
    How much weight can an object support on one side, without using counterweights, before falling over?
    E.G. a 40 cm arm with an 8 Kg weight is added to a 20 Kg box. Will the box support the weight without tipping to the weighted side?
     
  2. jcsd
  3. Oct 19, 2015 #2

    BvU

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    Depends on the size of the box. You want to make a balance about the potential tipping axis.
    On the left you have 20 kg x half the length of the box
    On the right you have 8 kg x 0.4 m

    Half the length applies if the box has uniform mass density (horizontal distance between center of mass and tipping axis -- the lower right)

    Make a force diagram to see.
     
  4. Oct 19, 2015 #3
    I have made a sketch as example. 123.png
     
  5. Oct 19, 2015 #4

    BvU

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    So on the lefet you have 20 x 0.125 kg m and on the right 8 x 0.4 kg m. Sum is 0.7 kg m to the right will keel it over !
     
  6. Oct 20, 2015 #5
    I'm having some trouble seeing how you got 0.7 kg m at the end??
    Also just curious, does three dimensional properties (l x h x w) have any effect on this?
     
  7. Oct 20, 2015 #6

    BvU

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    20 x 0.125 kg m minus 8 x 0.4 kg m is minus 0.7 kg m.

    Of course. If you lay the box on its side, you get 20 x 0.25 kg m minus 8 x 0.4 kg m is plus 1.8 kg m and the contraption will stay put.
     
  8. Oct 21, 2015 #7
    Ohhh... I thought since you said sum in your second message I had to add the two together.
    This makes sense.
    If the 8kg weight was directly against the side of the box, it would also stay up, right? (20 x 0.125 kg m) - (8 x 0.125 kg m) = 1.5 kg m.
     
  9. Oct 21, 2015 #8

    BvU

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    Yes, sum They are vectors and try to rotate in opposite directions, so one has a minus sign. Analogous to pulling to the left and pulling to the right.
    Nice try, but that case the 8 kg would try to rotate to the right with 8 kg times the horizontal distance of the center of mass of the 8 kg weight to the potential tipping point. Which is not the .125 m if it's an iron weight
     
  10. Oct 21, 2015 #9
    So lets make the weight 10 cm long.
    (20 x 0.05 kg m) - (8 x 0.05 kg m) = 0.6 kg m. Is it right this time?
    Can you then adjust this formula to calculate counterweight needed?
     
  11. Oct 21, 2015 #10

    BvU

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    It would be (20 x 0.125 kg m) - (8 x 0.05 kg m)

    See it as a see-saw or a weighing scale with uneven arms. As long as m1x1 - m2x2 > 0 it won't topple over point P


    Topple1.jpg
     
  12. Oct 21, 2015 #11
    Thank you for your help. I grasp it now.
    (20 x 0.125 ) -(8 x 0.05) = 2.1 kg m. Won't topple over.
     
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