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Moving objects vertically with horizontal pressure

  1. Mar 12, 2012 #1
    If you have some relatively small boxes next to each other you can pinch them with your fingers and lift them vertically. I would like to do this with big boxes, but I don’t know how to calculate certain things. I would like to make a frame with 4 plates on it which pressure the boxes horizontally so they can be picked up vertically. (Most likely I will also put a plate under the boxes to support but let’s pretend I will not). Picture:


    Assume there are 25 boxes (5x5) and they weight 150kg total. How to calculate the pressure needed to be able to pick up all the boxes vertically with the horizontal pressure provided by the cylinders? Assume there is also no acceleration and the boxes only have to be pinched so they can be picked up and they won't fall.

    Friction coefficient box on box μs = 0,5
    0,5 Friction plate on box μs = 0,4

    I know it’s impossible to exactly calculate it because of the many, many other factors. But all help would be appreciated.
  2. jcsd
  3. Mar 12, 2012 #2


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    The quick response to this one is that you can't do it. UNLESS the boxes or the container can flex in some way.

    Why can't you use the vertically acting ram that's already there?

    Or is this a FRICTION question? Did you want to know what sideways force is needed to stop the from boxes dropping out? That would depend upon the specifics of the set up and you would need to specify the 'coefficient of friction'.
  4. Mar 12, 2012 #3
    I can't use the vertical ram, because I would like to use this method to pinch the boxes. Hold them in the air for a while and then unpinch them again. So I want the boxes to be in the air.

    Yes basically that is what I need to know. What sideways pressure do I need to stop the boxes from dropping out.
  5. Mar 12, 2012 #4


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    OK. You will need to know the coefficient of friction and that would tell you the force needed to support their weight, without them sliding. Having worked out that force, you would need to decide whether the boxes would be strong enough to withstand this force.

    The sideways force needed would be given by
    Force = weight times coefficient of friction
    You can see how, if these boxes were really slippery, you couldn't ever hold them up. So we need to know more.

    Is this a real-world problem? You could find the coefficient of friction by finding out the minimum slope needed for the boxes to slide down a ramp made of 'box material'.
  6. Mar 12, 2012 #5
    But there are different boxes and different types of cardboard used for the boxes, which causes different friction coefficients. So far I found out the friction coefficient between the used cardboard on cardboard differs from μs =0,464 to 0,673, that's why i wanted to use 0,5 for the calculation.

    Vertical force = 150 kg x 10 = 1500N

    Force = 1500 x friction coefficient
    Force = 1500 x 0,5
    Force = 750N
    so 750/4 = 188N per plate

    If only taken the friction between boxes and their weight into account, would it have to be something like this?
  7. Mar 12, 2012 #6
    Assumming that your picture is a plan view of your boxes, that the 25 boxes are lying on the ground with no stacking, that the boxes are totally rigid, and that the force Fn supplied by the cylinder is transmitted with no loss of force between boxes, then you could calculate it this way.

    Take only one row of 5 boxes in your x-plane and forget the cylinder acting in the y-plane.
    Each box weighs 6 kg.

    Take 5 boxes as an ensemble.
    Your total mass of 5 boxes is now 30 kg.
    The x cylinder has to supply a force Fn in the x-direction, which acts on both ends of the 5 box assembly.
    The static friction between plate and the end boxes has to support the weight of all of the 5 boxes, so you can calculate the force Fn.

    Now take 3 inner boxes of the 5. The friction between box to box, on both ends of the 3 box ensemble, has to support the weight of 3 boxes, so you can calculate the horizontal force Fn needed also.

    The greater Fn is the force the cylinder must provide.

    Caveat: This is not taking into account the flex of the boxes nor beam action that will occur which could be substantial depending upon the dimensions of a box. Also if you add a cylinder in the y-direction, it could also occur that some middle box may not receive enough 'pressure' and drop out if all the boxes are not aligned perfectly.
  8. Mar 12, 2012 #7


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    You have calculated the horizontal force ok and, assuming you are only pressing from one side and the boxes don't distort, than this is the force you need. (You get the one on the other side 'for free', cos it's equal and opposite). Most friction calculations assume that the friction force is independent of area, so your first answer would be the one to work with. (The "per plate" is not what you need)
  9. Mar 12, 2012 #8


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    Shouldn't that be

    Required force = weight divided by coefficient of friction
  10. Mar 12, 2012 #9
    Ok, that would give me the following:

    With 5 boxes Fn in the x-direction:

    Fn = F x friction coefficient plate >> box
    Fn = 300 x 0,4
    Fn = 120N

    3 inner boxes of the 5:

    Weight 3 boxes is 18 kg

    Fn = F x friction coefficient box >> box
    Fn = 180 x 0,5
    Fn = 90N

    If only those factors are taken into account I would need 120N. Something like this? Why would I only need the grater Fn and not the two forces together?
    So with this calculation I would need 120x5 = 600N for all 25 boxes?


    So according to that calculation I would need a total of 750N force. Can I say I would need to press with 750/2 = 375N with both cylinders to get the 750N?
    Last edited: Mar 12, 2012
  11. Mar 12, 2012 #10


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    By George - I think you're right. Low coefficient would require an enormous force. Durrrrr!

    Did you get that, bladd? Sorry
  12. Mar 12, 2012 #11


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    Did you read the above correction? You'll need to correct your initial figures - sorry.

    Why are you including 'both cylinders'. I thought the boxes were supposed to hang there without the one pushing upwards????

    You do not need to be considering all the interior surfaces - they all have the same force on them, caused by the horizontal ram - likewise the opposite side of the holder. What counts is the force against the outsides because the inner interfaces do not need to support the whole weight - just the inner boxes.

    You do not 'divvy up' the force to get an answer. Is that counter-intuitive?
  13. Mar 12, 2012 #12
    Plate pushes on box 1 with a force of 120 N ( and box 1 pushes on plate with a force of 120). Box 1 pushes on box 2 with a force of 120 N ( and box 2 pushes back with a force of 120)
    And so it goes all down the line with a force of 120 N on all boxes.
  14. Mar 12, 2012 #13
    Something is wrong in your calculations.
    See the following:

  15. Mar 12, 2012 #14
    I learned that Ff = μ x Fn. But now that we need Fn we have to Fn = Ff / μ, I get it.

    So that means:

    Force = 1500 / friction coefficient
    Force = 1500 / 0,5
    Force = 3000N

    I wanted to use 2 cylinders to pinch the boxes like drawn in the picture in the first post, meaning I could divide the total force between two cylinders. Buw now that I think of it. If I press with 1500N on the boxes with cylinder 1 and cylinder 2 would start pressing with 1500N, it would bend/break the boxes so I should be using only 1 cylinder, right?

    According to 256bits method:

    With 5 boxes Fn in the x-direction:

    Fn = F / friction coefficient plate >> box
    Fn = 300 / 0,4
    Fn = 750 N

    3 inner boxes of the 5:

    Weight 3 boxes is 18 kg

    Fn = F x friction coefficient box >> box
    Fn = 180 / 0,5
    Fn = 360N

    Meaning for all 5 rows: Fn = 750 x 5 = 3750N.

    Are those calculations correct? Which calculation is more reliable?

    Edit: I see it's basically the same, I only used a friction coefficient of 0,5 for the first method. If i take 0,4 I get:

    Force = 1500 / friction coefficient
    Force = 1500 / 0,4
    Force = 3750N

    Which is the same. So without external factors, my cilinder would need to push the boxes with a force of 3750N to another plate in orther to let them hang in the air?
  16. Mar 12, 2012 #15


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    OH - is that picture a PLAN VIEW?????

    In that case it makes sense now. haha.

    3000N makes sense and this is the force that either or both rams need.
    As I said, the friction force is independent of area - counter intuitive, possibly, but don't think in terms of 'sharing the forces'.
    What counts is the force on the outsides of the boxes. Total weight / μ will keep them all up there. Forget the inside boxes - you will have already included their weight in the Total.

    Pros and cons for using one or two rams could be to do with the strengths of the cartons, which may be better off when pressed equally in both directions. You certainly would need a fairly tight fit so the boxes didn't spread too much sideways.
  17. Mar 12, 2012 #16
    The reason one does it for 5 box in one direction only is to calculate the force on box to box and box to plate and to simplify the conceptual understanding of the problem.. If the coefficient of friction between boxes is different (much lower) between that of box/plate, that could end up as the determining factor for the applied force, so you have to do the calculation just to be sure.

    Just be sure your boxes are lined up correctly and corners do not overlap. You do not want any one box from a line of 5 interferring with another line of 5, or these calculations go out the window, and you will have some boxes getting more of their share of the force than needed and others not getting enough force to be able to be picked up this way.

    If you go the double cylinder route, then as Sophiecentaur mentions it would be best used to keep the boxes in a tight knit group so they do not squish out sideways.
  18. Mar 13, 2012 #17
    Thanks for the help guys!
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