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Homework Help: Cards hanging off the end of a table problem

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A set of 59 playing cards of length L = 15 cm are stacked on the edge of a table so that the length of the overhang of the top card with respect to the table is maximized. A stack of only 6 such cards is illustrated. This problem requires you to draw an accurate Free Body Diagram of each card (or, at least of about 4 of the cards). To do these correctly, you will need to understand normal forces and torque.

    a) What is the overhang distance of the tip of the top card?
    b) How many cards are needed to extend the total overhang of the top card to at least 28 cm?
    c) What is the offset distance between cards 30 and 31, assuming that the top card is number 1?

    2. Relevant equations

    We are supposed to be using series and the ones I was given are:
    (the sum should be read from left to right = bottom to top)

    i=0 Ʃ n-1 (ar^i) = a(1-r^n)/1-r

    i=0 Ʃ n (ar^i) = a(1-r^n+1)/1-r

    if abs(r) <1

    i=0 Ʃ ∞ (ar^i) = a/1-r

    i=1 Ʃ ∞ (ar^i) =ar/1-r

    3. The attempt at a solution

    I've been trying to find a series to work this out for hours. A few I've come up with but are unsuccessful I think:

    1) L/2 + L/4 + L/8 + L/16 + ... + L/2^n

    2) i=0 Ʃ n-1 (L/2)(1/2)^i

    3) (L/2)(1+ 1/2 + 1/4 + 1/8 +... +1/2^n-1
    I got this one from the first one, but thought it might make a difference with the L/2 out of the picture
    These are all for part a. I don't want to start the other parts until I get a grasp on what I'm doing.

    But really anything that could help with any of them is appreciated. Thanks!
  2. jcsd
  3. Jan 30, 2013 #2


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    Did you do what the problem said to do? That is, draw free body diagrams?
  4. Jan 30, 2013 #3
    right i forgot I did that

    for the first card:
    normal force up
    m1g down 1 for first card

    second card:
    normal force up
    m2g +m1g down

    third card:
    normal force up
    m3g + m2g + m1g down


    I'm probably doing that completely wrong but basically, each card is slightly less than half hanging over the one below it. The torque applied is due to gravity plus the weight of the cards above it.
  5. Jan 30, 2013 #4


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    Where are these forces acting on each card?
    How do you know that? Try it yourself, and see how many cards you can stack that way.
  6. Jan 30, 2013 #5
    oops typed that wrong. What I meant was each card is centered on the edge of the card underneath it, but with slightly more of the card on the one beneath it, so that more mass of the card is on the one below it, rather than hanging off the edge.

    When I said "half hanging over" I meant half of the distance the on top is hanging plus the original length. so if half of the top card is over the edge, then 1/2 + 1/4 of the next card is hanging over and 1/2 + 1/4 + 1/8 of the next card, etc
  7. Jan 31, 2013 #6


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    That's not the same as you have written below. The above would mean that with n cards the top card extends n/2 beyond the edge - clearly unstable.
    Yes, that's stable, but doesn't extend as far as it could. The position of the top card in relation to the second is ok. Now consider the top two cards as one. Where does that need to be in relation to the third card? Then the top 3 in relation to the fourth...?
  8. Jan 31, 2013 #7


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    That's just what you said before.
    This doesn't make sense to me; it sounds contradictory.

    Start by considering just 1 card, then go to 2, then 3 and 4. These cases should be easy to draw and easy to analyze. Don't assume an answer in advance (you seem to have assumed a harmonic series is involved).
  9. Jan 31, 2013 #8
    Okay, started over from the beginning.

    Thinking of a single card, it can have less than half of its mass hanging over the edge of a table. The forces acting on it are the normal force from the table pushing up, and the torque pushing down. The torque is equal to gravity x distance from edge of table. Since the torque is only applied to half the card, 1/2mg*r = 1/2mg - N. This is probably totally wrong, but hopefully not.
  10. Jan 31, 2013 #9


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    Forces and torques are not the same thing. The forces on the card are the normal force and gravity, both acting at the center of mass. This net force may or may not create a non-zero torque.
    The torque is applied at the center of mass; it affects the whole card.
  11. Jan 31, 2013 #10
    wow I really need to brush up on the basics. Thanks for being patient.

    So in order to have a card hanging off the table, but not falling off, the center of mass would always have to be on the table, so that the normal force from the table still equals the gravitational force, even though part of the card is hanging off the table.

    Am I going in the right direction here?
  12. Jan 31, 2013 #11


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    For one card, yes. How will you generalise that to the stack?
    For the purposes of this question, don't worry too much about 'not quite' half. We're dealing with a theoretical limit, so take it all the way to the limit point. E.g. for one card, exactly half off the table.
  13. Jan 31, 2013 #12
    Right, so for the second card the center of mass is further to the right of the card, and the one below that has a center of mass even further to the right

    PS forgot to attach the picture

    Attached Files:

  14. Jan 31, 2013 #13


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    I don't know what you are calling first and second etc., or how many cards in each context.
    Consider just two cards. How would you arrange them? Then consider three.
  15. Jan 31, 2013 #14
    I would put the center of mass of the first card on the front edge of the second card.

    So the second card effectively becomes the table edge.

    The I would put the center of mass of the 2 cards on the edge of the third card, etc
  16. Jan 31, 2013 #15


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    You got it. So by how much does the nth card from the top overhang the next?
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