# Built on Facts

## Pig sliding down a ramp

#### May 20th, 2008 · 4 Comments

My intro physics text in my undergraduate education was by Halliday, Resnick, and Walker. I personally thought it was a quite good text, but what particularly stood out was its obsession with penguins. Probably a dozen or more problems involved penguins in very improbable situations. It was pretty funny. But there’s another problem I remember that involved a pig, which is a nice thing for undergrad physics students to think about. I won’t quote it directly as I don’t have the book with me (and I’m not sure how copyright works for physics problems anyway), but here’s the gist:

A pig slides down a ramp which is inclined at an angle 45 degrees above the horizontal. It takes twice as long to slide to the bottom of the ramp as it would have if there were no friction. What is the coefficient of friction?

It’s an interesting problem because it doesn’t give the length of the ramp, the time of the slide, or anything like that. You have to think a bit about how to approach it. In fact, let’s also pretend we don’t know the angle, and just call it θ. So let’s call the unknown length of the ramp x and we’ll figure out how long it takes to slide down without friction. We know that the acceleration is g sin(θ). Therefore the distance the frictionless pig slides in a time t is

Which implies that the time required to slide a distance x is

Well and good. But what about the case with friction? Well, aside from the force down the ramp from gravity, there’s a frictional force pointing upward. From the definition of the coefficient of friction μ = (friction force)/(normal force), we see that the total acceleration acting on the pig down the plane of the ramp is

You don’t see? Think of it in terms of forces, and then divide by m to find the acceleration. Notice that m doesn’t appear in any of the acceleration expressions. Thus the sliding time has nothing to do with how heavy the pig actually is. Plug that into our equation for sliding time, and we get that the sliding time for the pig with friction is

And that’s (we’re told), double the time of the frictionless time. Thus

Ugly, but all we’re after is μ. So solve for it!

Notice the x cancels, therefore the length of the board is irrelevant as well. Interestingly, this result also means that a smaller angle requires only a small frictional force to cause the sliding time to be twice as long. But this is what we expect, because a small angle means it will be sliding longer in the first place, giving friction a longer time to act. A silly problem to be sure, but one that’s an interesting and subtle exploration of some basic introductory classical mechanics!

### 4 responses so far ↓

• 1 anon1234 // May 26, 2008 at 3:04 pm

But if the pig is rolling without slipping, then you have to calculate 0.5 * I * omega**2 for the pig.

What is I for a pig, rotating around its long axis?

Matt replies: In this case we assume it’s not rolling, just sliding. If it were rolling, we’d have to do exactly as you suggest and take into account the rotational inertia. As the problem stands there’s not enough information to find it, so it would either have to be given or we’d need more information. But once we had it we could do as you suggest and use energy, or stay fully in the force picture and put it in terms of torque: τ = Iα and a = rα, etc.

• 2 anon1234 // May 27, 2008 at 8:36 am

Thanks. Here’s another question, somewhat more serious: Back in 101, I noticed that some problems could be solved much more easily by resolving kinetic energy into its X and Y components. For instance, the canonical cannonball has K0(x) and K0(y), and if you set K0(y) equal to m*g*h(max) then the problem falls apart quite nicely.

My instructor was not pleased at this; he did not want me to regard K as a vector, and implied that, while it might work for trivial problems, it would certainly fail catastrophically later.

Was he right? Is this simply one of those purists-vs-engineers debates, or are there really places where this trick would break down?

Matt replies: Kinetic energy is not a vector, so it doesn’t have components. But velocity is a vector, and so you can think of there being a kinetic energy associated with each velocity component. As long as that’s kept in mind there shouldn’t be a problem. Problems will show up the second someone tries to do actual vector operations on kinetic energy. For example, if a particle of kinetic energy K is traveling at 45 degrees to the x-axis, the kinetic energy along the x-axis isn’t K cos(45). The energy associated with the y-motion isn’t K sin(45) either. Even worse the sum of the two isn’t K, because K (cos(45) + sin(45)) ≠ K.

However, those trig operations work just fine on v since velocity is a vector. Plug those into the kinetic energy equation and things will work out fine. In fact you might notice the kinetic energy “components” would contain and sin2θ and cos2θ terms, which add to 1 no matter what theta is. This preserves the Kx + Ky = K requirement.

• 3 CCPhysicist // Jun 22, 2008 at 3:10 pm

I noticed that about the latest H+R. I am sure it is a result of the popularity of two penguin movies back when the book was being revised.

As if I would pick that version over superior books just because it had penguins in it….

As for the other questions:

1) Assume a cylindrical pig, adding hemispherical caps if you feel like it.

2) Since KE is not a vector, that solution is worth zero points for failing to grasp the basic difference between scalars and vectors. For example, on a brachistochrone-shaped ramp, Ky is zero at the bottom.

• 4 Ira Marquez // Nov 12, 2008 at 7:42 pm

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