## December 14, 2003

### One Hundred Interesting Mathematical Calculations, Puzzles, and Diversions: Number 23: The Muddy Parent Problem

#### One Hundred Interesting Mathematical Calculations, Puzzles, and Diversions: Number 23: The Muddy Parent Problem

Suppose a 250-pound parent climbs a muddy 45-degree hill...

1: Calculate the force tending to make the parent slide down the muddy hill.

Answer to 1: The first force to be taken into account is the force of gravity. We represent the force of gravity by an arrow, G, pointing downward toward the center of the earth. The earth is pulling the parent downward. By how much? By 250 pounds--that's what it means to say that this is a 250-pound parent.

But the parent does not fall straight down through the hill. The molecules of dirt are pushing on the parent, counteracting the force of gravity. This force is represented by an arrow, H, pushing the parent away from the hill. We add up the forces G and the forces H by putting the tail of the H arrow at the head of the G arrow. The sum of these two arrows ("vector addition") is the arrow S--the total force that is tending to make the parent slide down the hill.

How big is this force arrow S? Well, we know that the force arrow G is 250 pounds. We know that the arrow H makes a 90 degree angle with the slope of the hill, and thus with the arrow S. And we know that the angle between force arrow G and force arrow S is 45 degrees--that's what it means to say that it is a 45-degree hill.

The arrows G, S, and H thus form a triangle, with one angle equal to 90 degrees and a seconed angle equal to 45 degrees. Since the angles of a triangle add up to 180 degrees, the third angle must also be 45 degrees. The triangle is an isoceles triangle--a triangle with two angles, and also two sides equal. This tells us that the force arrow S must be the same length as the force arrow H, since they are the two equal sides.

Because the third angle of the triangle is 90 degrees, it is a right triangle. Pythagoras discovered that, in a right triangle, the square of the length of the third side is equal to the sum of the squared lengths of the other two sides. Writing |G| for the length of the long "gravity" side, and similarly for the other sides:

|G|2 = |H|2 + |S|2

But since S and H are the same length:

|G|2 = 2(|S|2)

We know that |G| is 250 pounds:

2502 = 62500 = 2(|S|2)

Dividing by 2:

31250 = |S|2

Taking the square root:

177 = |S|

177 pounds. That's the strength of the force trying to make the parent slide down the hill.

2: Why doesn't the parent slide down the muddy hill?

Answer to 2: Because of friction! There is a force F of friction on the parent's shoes, counteracting the sliding force S. The force F of friction can counteract up to 200 pounds of sliding force, on that hill, on that day, at that hour.

3: What happens if, just before the parent reaches the top of the muddy hill, a child at the top of the muddy hill loses control of America's Silliest Dog, which then jumps on the parent who is one step from reaching the top of the hill?

Answer to 3: The sliding force S is 177 pounds. When 40 pounds of dog-force is added, the total force pushing the parent down the hill is 217 pounds--more than the 200 pounds that the friction force can counteract. The parent's feet slip out from underneath, and the parent slides down the hill feet-first belly-down.

4: How many pounds of mud does the parent get on his clothes during this misadventure?

Answer to 4: At least six pounds of mud.

Coda: "Are you all right? How many broken bones do you have? We need to get you to the emergency room! You are 43 years old!

Posted by DeLong at December 14, 2003 08:21 PM | TrackBack

you sure that hill was 45 degrees? it doesn't seem steep when you draw it, but for context, the West Face of KT-22 is something like 37 degrees, and that's about as steep as most people can ski.

Posted by: wcw on December 14, 2003 01:11 PM

wcw beat me to it. 45 degrees is very steep.

Posted by: theCoach on December 14, 2003 01:23 PM

This hill is very steep. (OK. It's 30 degrees rather than 45. But I'm going to ask a Thirteen-Year-Old to take sin(pi/6)?)

Posted by: Brad DeLong on December 14, 2003 01:45 PM

Applause! Brilliant, and with the strict analysis gradually losening towards the end, it is actually correct enough.

(However. With "friction", one usually means a force that is directed along the contact plane between two bodies with a *maximum* strength that is proportional to the force normal to the contact plane. Hence, with "friction" on the scene, the weight of the person (person+dog) on the hillside is redundant - the only parametera deciding if the person slides or not is the steepness of the hillside and the proportional constant of friction between the person and the ground.

If this was to appear in a mechanics textbook, it would have been about the maximum "shear stress" that could appear in the mud without (layers of mud) sliding (on each other). And formally correct up to and including point 3.)

Posted by: Mats on December 14, 2003 01:55 PM

45 degrees is OK for pedagogic purposes but 40 pound dog-force is too much of an approximation, unless the dog is about 56 pounds.

And the 250 pound parent is not setting a good example for a thirteen year old! Gosh that's over 100 kilograms, no? (Well, maybe the parent is 6 foot 6.)

Posted by: Bulent Sayin on December 14, 2003 01:57 PM

Taking sines builds character.

Posted by: Cosma on December 14, 2003 02:02 PM

Dog is 60 pounds. Parent, alas, is 5'10"

Posted by: Brad DeLong on December 14, 2003 02:04 PM

Taking sines builds character.

Posted by: Cosma on December 14, 2003 02:14 PM

Sin 30º is 0.5, that is not a difficult quantity, or at least much less than square root of 2 divided by 2 ( 0.707.... ).

Bulent, the static weight is not what matters, but the variation on linear moment. Just like a hammer, remember. Now I've no intention to try to solve that in imperial units, no wonder I love SI.

DSW

Posted by: Antoni Jaume on December 14, 2003 02:15 PM

Taking sines builds character.

Posted by: Cosma on December 14, 2003 02:19 PM

45 degrees is the limit if the hill is of "dry concrete" and the shoes are "rubber"-shoes. At least according to this table of proportionalities between normal forces and maximum friction force:

http://www.physics.ucf.edu/~saul/Common/06-Forces/FrictionCoeffs.html

Note how problematic the notion of friction is in the real world: once you lose grip friction becomes lower! [good (bad) for skiing on snow (mud) though]

Posted by: Mats on December 14, 2003 02:21 PM

To Cosma:

"Sorry. I rephrase myself:

45 degrees is OK for simplification purposes..."

...I was going to say... but Antoni Jaume took away from me that one too! I am devastated!

Wonder if I should go back to teaching but this time teach basic science so that I could refresh my cerebral tracks on basic math and physics?

Posted by: Bulent Sayin on December 14, 2003 02:28 PM

Correcting point 3. by using "shear strength" of the mud rather than "friction" between shoe and mud:

Motivation, observation: shoe is not sliding over mud (which would have been preferable), mud is sticking onto shoe, upper layer of mud is sliding over lower layers of mud.

Assume: "shear strenghth" of the mud is 177 psf (pound-force per square foot), contact area between shoes and mud is 1 square foot.

Solution: Shear stress in mud is 177pounds/1 square foot = 177psf and mud stays put. Add the dog => sliding force increase with contact area constant, so shear stress increase over limit and mud slides.

Re-examination: Measured Shear strenghts in some muds is 30-7410 according to http://www.flysfo.com/about/runway/docs/SF-Oakland-Appendixb.pdf

Posted by: Mats on December 14, 2003 03:14 PM

In part 3, if the 60 pound dog was delicately placed onto the parent, then all you'd have to worry about is the parent's feet supporting 310 pounds of parent + dog instead of 250 pounds of parent, and your calculation (217 pounds) would be about right. But this dog jumped onto the parent, which means that there was probably quite a bit of force knocking the parent backwards. Conservation of momentum is very relevant when bounding dogs are hitting you.

Posted by: Dan on December 14, 2003 03:29 PM

on a dry surface the co-efficient of friction is higher until the object is moving, which means once you start to slide it,s difficult to stop.

Posted by: big al on December 14, 2003 03:34 PM

Next is the slipping ladder problem, how high, before the feet slip.

Posted by: big al on December 14, 2003 05:06 PM

Never forget the energy of rotation which has to be added to the falling phat person.

Where does it come from, and where does it go?

The theory is to be found in Feynman's Brazilian essay. The numbers, of course, are to be found empirically.

Posted by: David Lloyd-Jones on December 14, 2003 05:54 PM

Where the hell is Anne to fill the paramedic's role and console our patient?

Posted by: calmo on December 14, 2003 07:58 PM

What, are you guys trying to make these 13-year-old kids head's explode?

What we going to add next, the fact that friction is actually not a simple, linear function of force? And that it is actually force per unit area, not just force? I mean, how much surface area do Dad's shoes have? If Mom does it, and wears stilleto heals (or better yet, crampons!), she has a better chance of not sliding. And that all of that is just nasty approximation gleaned fron experimental data, not any governing theory?

That might be a bit much for middle-school aged kids. This problem is good for 13 year olds. Actually, when I tutor physics for 18-45 year olds taking College Phyics I, people are still having great difficulty with problems of this type.

Posted by: Timothy Klein on December 14, 2003 10:56 PM

What, are you guys trying to make these 13-year-old kids head's explode?

What we going to add next, the fact that friction is actually not a simple, linear function of force? And that it is actually force per unit area, not just force? I mean, how much surface area do Dad's shoes have? If Mom does it, and wears stilleto heals (or better yet, crampons!), she has a better chance of not sliding. And that all of that is just nasty approximation gleaned fron experimental data, not any governing theory?

That might be a bit much for middle-school aged kids. This problem is good for 13 year olds. Actually, when I tutor physics for 18-45 year olds taking College Phyics I, people are still having great difficulty with problems of this type.

Posted by: Timothy Klein on December 14, 2003 11:01 PM

Huf! I am completely .... I just caved in! My recollection was friction force was independent from surface area!

Hey 'ow 'bout a bit of quantum aspects of the sliding parent problem then? They didn't tell us much about quantum physics when I was freshman in 1970 -- even though we had superb professors of physics (and superb professor of economics too).

Posted by: Bulent Sayin on December 14, 2003 11:12 PM

>>My recollection was friction force was independent from surface area!>> It is, hence my comments that the problem would rather be connected to the "shear strength" of the mud rather than "friction".

Posted by: Mats on December 14, 2003 11:43 PM

Timothy Klein, observe that your story, as well as the one in Brad's 3. is misleading. Frictional force is independent of contact area. Overcoming frictional force -sliding - is independent of the supporting force from e.g. the hillside to the shoe (explaining among other things the annoying 'stuck drawer' effect)

big al, friction is higher between non-sliding than between sliding surfaces, as you point out. Observe how much in some cases ("waxed wood on dry snow") in the table I was linking to above.

Dan, David, dynamics is of course important here. Note that it is conceptually much more difficult than static mechanics, and should preferably be left outside a basic pedagogic hillside problem.

Posted by: Mats on December 15, 2003 12:15 AM

Oops, friction is not dependent on surface area. I was thinking of baseballs in a fluid, where surface area matters. I always confuse that, because it only works out that friction is not proportional to the surface area by a neat accident, and it is only an approximation -- like all of frictional formulas. In the purest sense, the surface area does matter, but as the downward force increases, it happens to neatly increase the surface are in contact (microscopically speaking) by the right amount to make the approximation work.

There, thats a nice little dodge to cover up being wrong :-)

Posted by: Timothy Klein on December 15, 2003 01:21 AM

I may be wrong, but, in reference to your depiction of F, I seem to remember my esteemed high skool physics teacher pounding into our heads that vectors never push, only pull.

Posted by: The Squire on December 15, 2003 02:35 AM

I may be wrong, but, in reference to your depiction of F, I seem to remember my esteemed high skool physics teacher pounding into our heads that vectors never push, only pull.

Posted by: The Squire on December 15, 2003 02:38 AM

Pounds? Why not kilos, and g os approx equal to 9.81 m/s(squared)? Don't you think it is high time that the next generation of americans is taught the S.I unit system?

Posted by: MRM on December 15, 2003 02:55 AM

i) Follow South Beach Diet or similar

iii) Weigh self every day

iv) Plot weight over time, plus N-day average after N days (where N is 5, 10, 20 etc)

v) Lose 20lbs in 8 weeks at a steady rate without feeling particularly hungry or particularly denied.

Nasi "now 195lbs" Lemak

Posted by: Nasi Lemak on December 15, 2003 03:09 AM

Actually, if the hill's slope really was 45 degrees, there's the possibility that you would have lost contact with the ground when the dog jumped on you (Lesson 2: Conservation of Linear Momentum for Inelastic Collisions)-- with consequences that one hesitates to contemplate.

Posted by: Matt on December 15, 2003 06:34 AM

Oh thank you God I haven't developed Alzheihmers yet -- was it Alzheimers!?

Posted by: Bulent Sayin on December 15, 2003 06:39 AM

The dog sensibly assumes normal creatures have four legs. The static paw pressure is probably higher than Brad's; but a trotting dog has two feet on the ground at all times, and a lower centre of gravity. So it can correct for slips better than Brad can. The trouble with bipedalism is instability, not low friction.
(Is a dog's fastest gait galloping or bounding? Did Muybridge stop-photograph dogs?)

Posted by: James on December 15, 2003 08:00 AM

Well, as someone at 6'1" and 168 lb (lots of cycling), clumsiness is the marginal force determining whether or not I fall down a muddy hill. Especially if it gives me an excuse to use cosecants. Is there a physical model of clumsiness out there?

Posted by: Chris on December 15, 2003 01:03 PM

This is a very good example for introducing some concepts of mechanics, evoking the real-world experiences of thirteen-year-olds; and I agree that going into the detail mentioned here would place a good deal of stress on thirteen-year-old heads, possibly resulting in explosive strain. However, characterizing the shear-stress as "friction" will give the student a deeply misleading picture of friction that some high-school or college physics teacher will have to labor mightily to overcome.

(Strunk and White would argue that following a semicolon with "and", as I just did, would also set a bad example for a thirteen-year-old, but somehow it seemed right.... :-)

Posted by: Jonathan on December 17, 2003 05:34 PM

This is a very good example for introducing some concepts of mechanics, evoking the real-world experiences of thirteen-year-olds; and I agree that going into the detail mentioned here would place a good deal of stress on thirteen-year-old heads, possibly resulting in explosive strain. However, characterizing the shear-stress as "friction" will give the student a deeply misleading picture of friction that some high-school or college physics teacher will have to labor mightily to overcome.

(Strunk and White would argue that following a semicolon with "and", as I just did, would also set a bad example for a thirteen-year-old, but somehow it seemed right.... :-)

Posted by: Jonathan on December 17, 2003 05:39 PM

This is a very good example for introducing some concepts of mechanics, evoking the real-world experiences of thirteen-year-olds; and I agree that going into the detail mentioned here would place a good deal of stress on thirteen-year-old heads, possibly resulting in explosive strain. However, characterizing the shear-stress as "friction" will give the student a deeply misleading picture of friction that some high-school or college physics teacher will have to labor mightily to overcome.

(Strunk and White would argue that following a semicolon with "and", as I just did, would also set a bad example for a thirteen-year-old, but somehow it seemed right.... :-)

Posted by: Jonathan on December 17, 2003 05:41 PM

Oh, man, I've *got* to learn to hit "reload" before assuming that "site not found" means that the post didn't go through... :-(

Posted by: Jonathan on December 17, 2003 06:31 PM

What are these things: pounds, feet? I thought this was the 21st century.