Two Bad Mice ponder Two Bad Questions |

Most of this month's problem sets I wrote while sitting at the Starbucks on the Pearl Street Mall in Boulder, Colorado. I do highly recommend such a retreat for summer planning. All I did was to copy the problems that I've used successfully for many years, adding explicit prompts such as "Explain why the tension is greater than the weight of the stoplight," or, "Describe, as to a non-physicist, how much force the string exerts by comparison to a force with which you are familiar. Justify your comparison." This approach has worked well.

However, I discovered -- too late -- a couple of minor mistakes that turned into ill-posed problems. As a

1. This one is based on an old Giancoli problem, but I copied it wrong:

*mea culpa*to my students, and*pour l'encouragement d'les autres*, I present to you these questions that you should not ask.1. This one is based on an old Giancoli problem, but I copied it wrong:

In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to push grocery carts along the ramps, as shown above. A grocery cart has mass 30 kg; the coefficient of kinetic friction ism= 0.10. Determine the maximum angle of the ramp such that the customer will not have to exert a force in excess of 50 N.

Sounds like a great question... but, as my students pointed out to me, it requires either numerical analysis on a graphing calculator or maximizing a function using calculus in order to solve. I promise the class that they will never have to use any math beyond algebra I and the definitions of the trig functions, so they were pretty confused.

The original question that I should have transcribed correctly asks whether a 5 degree angle would be too steep, knowing that customers should not be asked to exert more than 50 N. That's a straightforward problem mathematically, which allows the rest of the question to focus on a description of just how much force is 50 N, anyway. And the whole point was to provide an easy, confidence boosting problem for the class, who is finally getting the hang of equilibrium. Grr.

2. I just didn't think this one through at Starbucks:

A 150 N block sits on an inclined plane, as shown. The coefficient of static friction between the block and incline is 0.30. Calculate the angle of the incline and the force of friction on the block.

Simple enough so far... it's straightforward if you recognize that sin/cos = tan. Since that's the one trig identity that shows up in introductory algebra-based physics, I use this problem sort of as a reminder to know that identity. But then I screwed up:

Now imagine that the same 150 N block slides down the same plane at constant speed. Is the force of friction greater than, less than, or equal to the value you calculated previously?

Well, this question requires understand extremely subtle issues about friction, far more subtle than the AP exam or I cares about. The coefficient of static friction can take on any value up to a MAXIMUM of

*Ff/Fn*. The question as stated is confusing to students who have had ten days of physics. On one hand, they see that equilibrium requires*Ff*and*Fn*to be*mg*sin*θ*and*mg*cos*θ*, respectively; so*Ff*should be unchanged with the same block on the same angled incline. But they also understand that the coefficient of kinetic friction is less than the [maximum] coefficient of static friction. So the friction force should be smaller. Which is right?The correct answer is,

*don't ask this question to begin with.*I don't care how cool the underlying physics is -- and really, static friction is pretty awesome when considered from an expert, dispassionate, and deeply intellectual perspective -- a justification question this early in the course without a definitively correct answer can spawn not just confusion but hopelessness.Now, you might suggest that a college physics course can and should include problems at this level of difficulty. The course you took back in college sure did, and everyone got it wrong, and we were better for it, right? Well, no. We're not teaching physics majors here, we're teaching high school students at the college level. Their confidence is a fragile thing, and must be preciously guarded. In the first week, I have already "torn down" a number of students who struggled for a week just to draw a free body diagram and understand what a normal force means. I need to build back these students' confidence, show them that if they follow the problem solving procedure I've described in class, they can and will do well. But they don't see physics as "cool," they see it as "hard." Let's first convince the class that physics is "doable," so that at the end of the year they will be ready for "cool."

And, oh yeah, since I'm assigning rewritten problems, I'm going to write a solution to every problem set this year, starting now. I can't make this kind of mistake again.

*Mea culpa.*

GCJ

Maybe something else is shown in the figure for problem #2, but unless motion is pending, the angle of the incline could have any value from zero to arctan(0.30). The friction coefficient doesn't change, but the normal force changes and therefore the friction force changes as the angle changes.

ReplyDeleteHi Greg,

ReplyDeleteIn your first question, are the shopping carts on wheels? If so, can't the coefficient of sliding friction can be ignored (no slippage, just rolling), and wouldn't the solution be ~ 10 deg?

Jay Fogleman

To Anonymous above -- you're right, that's why this was such a bad problem. The coefficient of static friction is a maximum, and the problem wasn't clear whether or not the incline was as high as it could go without slippage.

ReplyDeleteTo JayF, the "coefficient of friction" here is more precisely a drag coefficient, taking into account primarily friction in the axes of the wheels. You're right that there's no sliding, but the situation is not without a force opposing motion, which acts essentially identically to sliding friction. (At least at the introductory level, which is what I'm teaching.)