# When will the pringles can fall?

When I'm pushing a tall, cylindrical object stood on its end (like a Pringles can) I can push it near its bottom, and it will slide forwards, or push it near the top, and it will topple over. So, there must obviously be a point somewhere on the side of the cylinder where it stops moving forwards and begins falling forwards.

My question is, how can I calculate where this point occurs?

It might be something fairly obvious, like: "above the center of mass" or something like that, but I'm guessing that it's something more complicated, taking into account friction, etc.

This knowledge will be very useful in my life, as the next time someone asks me to "slide along the beer" I can whip out my pocket calculator, my ruler and measuring instruments, and finally stop being such an embarrassment. I swear, it falls over every single time.

• Do you understand torque? – Jerry Schirmer Apr 25 '15 at 18:01
• It's going to depend (in a rather complex way) on the coefficient of friction between the can and the surface. – Hot Licks Apr 25 '15 at 22:32
• Related to physics.stackexchange.com/q/95234/37364 – mmesser314 Apr 25 '15 at 23:38

Let's figure it out. First let's figure out the force you would need to slide the object. In order to slide it you'd need a force that overcomes friction. $$F_{\text{slide}} > \mu m g$$ where $\mu$ is the coefficient of friction.

In order to tip it, you'd need to cause a torque that will cause the can to rotate about its far bottom corner. If you press on the can with a force $F$ at some height $h$ on the object, that will cause a torque about its pivot point of $Fh$, the only other torque you're competing against is gravity, which if we assume your object is more or less homogenous will be centered at the center of mass of the body, which should be a distance $w/2$ from the pivot point if $w$ is the width of the base. So our criterion for tipping is

$$F_{\text{tip}} > \frac{mgw}{2h}$$

The object will tip rather than slide if the threshold for tipping is lower.

$$\frac{mgw}{2h} < \mu mg$$ or simply $$h > \frac{w}{2\mu}$$

And we can test it out quick. I've got an empty paper cup on my desk. I've also got some graph paper. I can measure the width of the base of my cup in boxes and get $$w = 9 \frac 12 \text{ boxes}$$

It remains to determine the coefficient of friction. For that I opened the flap of my book until the cup just started moving. Doing this I can extract the coefficient of friction. Since, balancing forces, we have

$$mg \sin \theta = \mu mg \cos \theta \implies \mu = \tan \theta$$ and I can measure the tangent of the angle I have to move the book, by measuring the width of the flap in boxes (24 boxes), and the height in boxes that I lift the edge of the flap (7 boxes), giving me a coefficient of friction of $$\mu = \frac{7}{24} \sim 0.3$$

so now I can estimate the critical height to push the cup to get it to tip

$$h = \frac{w}{2\mu} = \frac{ 9.5 }{ 2 \times 0.3 } \sim 16 \text{ boxes}$$

and I have to say, that is within a box of the right answer.