Here's an interesting question I thought of. Firstly, picture one of those card pyramids. Hopefully you know what I'm taking about when I say card pyramid, but basically you balance two cards up against eachother, then you balance two cards against eachother next to that; then, put a card as a platform ontop of the two crests, and use that platform to balance another two cards together ontop of that. You can make this bigger and bigger, but my question is:

What is the relationship between $µ_{static}$ and h?

(The coefficient of static friction and how tall the tower is)

Let's assume that you somehow build this pyramid perfectly, so the traingles formes are equilateral causing the angle of the cards to be 60˚ above horizontal. We are also assuming that all of the cards in the bottom middle are touching eachother, so it is only the outermost card that would "slip" if that makes sense.

The cards will have a given mass $m$ And the static friction force is of course this:

$$F_f=F_nµ_s$$ This must balance with the horizontal force of gravity: $$F_g=mg$$ As well as the $x$ component of the normal force from the other cards in the tower: $$F_x$$ The normal force is the problem for me here. How do you determine what portion of the normal force is acting on the horizontal component of that outer card and how much of it is being counteracted by the floor and between the middle cards? (To solve this you would also have to somhow take into account that closer to the top of the tower there are less cards).

This seems way beyond my knowledge, so I would be very interested in what people make of this.

Please tell me if any part of this is unclear. Thanks!

  • 2
    $\begingroup$ This question changed my life. $\endgroup$ – Tdonut Apr 29 '15 at 3:07

There are two parts to this question.

Part 1: will the card slide?

if I have a card at an angle, is there a limiting vertical force that will make it slide sideways? The force diagram looks something like this:

enter image description here

This is a bit like the "climbing a sliding ladder" problem, in which case there is going to be a limiting force F - once you exceed that force, the house of cards will slip and fall. We assume that the force $F$ is distributed among the $n$ cards in this layer, so the force supported by the end card is $F/n$.

We would want to balance the torques, so we get

$$F_f = \mu F_n\tag1$$ $$F_n = -(W_c + \frac{F}{n})\tag2$$ $$F_r = -F_f\tag3$$ $$\left(\frac{F}{n} + \frac{W_c}{2}\right)\ell \cos\theta = F_r \ell \sin\theta\tag4$$

This would indeed show that there is a value of $F$ for which the card would slip, namely when

$$\left(\frac{F}{n} + \frac{W_c}{2}\right)\ell \cos\theta > \mu \left(W_c + \frac{F}{n}\right)\ell \sin\theta\\ \left(\frac{F}{n} + \frac{W_c}{2}\right) > \mu\left(W_c + \frac{F}{n}\right) \tan\theta\\ \frac{F}{n} > \frac{W_c \left(\mu\tan\theta - \frac12\right)}{1 - \mu\tan\theta}$$

This does suggest there is a limit to the force - but we forgot one thing.

In the above equations, we assume that the lateral force $F_r$ which balances the torque of the vertical force (plus the weight of the card) must be entirely balanced by the friction at the bottom of the card. But that ignores friction at the top.

This is NOT, in fact, exactly like the "sliding ladder" problem - there is a twist: the card is supported from above by the card that is resting on it. This card provides lateral support (friction) which is in proportion to the force of the card above - with the maximum lateral force it will support equal to

$$\mu \frac{F}{n}$$

This means that the equation (3) above is not the force we need to balance - we only have to balance the weight of the card. And this means there is no limit to the height of the house of cards, as far as friction is concerned.

Part two - the load on the cards

So why do houses of cards tumble anyway? Mostly, this is a result of uneven distribution of forces; but even with a perfectly steady hand, you will reach another limit: the buckling strength of the cards.

When you put a longitudinal force on a rod, it will buckle when the force reaches the critical value

$$F_{buckling} = \frac{\pi^2 E I}{K L^2}$$

Where $I$ is the second moment of area, $E$ is the Young's modulus, and $K$ is a factor that depends on the way the ends of the rod are clamped. In the case of the house of cards, they are both free to rotate and K=1. It is hard to estimate $I$ and $E$ for a card, but we can find the size of the stack where this limit will be reached.

If we have a house of cards that is $N+1$ stories high, then the number of cards supported by the bottom tier is $2N + 2(N-1) + ... + 2$ for the upright cards plus (N + (N-1) + ... + 1) for the horizontal cards. The sum is

$$ \frac32 N (N-1)$$

This force will be distributed among 2(N+1) cards, so each card will support approximately $\frac34 N$ cards. This shows that as the house of cards gets taller, it will eventually reach a height where the cards in the bottom layer will buckle, and the whole thing collapses.

Some rough numbers:

cardboard Young's modulus: 200 N / mm^2 (source: http://www.slideshare.net/orkuthacker/paper-carton-analysis)

Other dimensions: (from http://en.wikipedia.org/wiki/Standard_52-card_deck)

thickness: 0.3 mm
width: 57 mm
height: 90 mm
weight: 1.8 g

Based on personal observation, I put the buckling strength of a playing card at about 2 N (higher than the above numbers would suggest... I think that the material that cards are made of must have a much higher modulus than the cardboard in the reference).

Solving for $N$ we find

$$F_b = \frac34 N m g\\ N = \frac{4 \cdot 2}{3\cdot 0.02\cdot 10} \approx 130$$ That means the limit on the height of a house of cards is approximately 130 levels. That is actually quite close to the Guinness world record of 131 levels. Given the amount of estimating I did, I'd say that is close enough.

Note that the technique used in the world record attempt was the "four cards on their side" style, rather than the "upright teepee" used in the above. That makes the calculations not so relevant (the load per card is greater than I calculated, but the card is stronger when it is placed on its side).

  • $\begingroup$ In the first part, how do you determine what "F" is based on the cards above it? $\endgroup$ – Me2 May 1 '15 at 13:53
  • $\begingroup$ @Me2 - I only determined that in the second part... assuming that the weight of all the cards above is divided equally among the cards. The point in part 1 was that the lateral force experienced is independent of F, since there is lateral friction at both the top and the bottom. $\endgroup$ – Floris May 1 '15 at 14:28

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