Help me prove my physics teacher wrong

I’m working on an insanely complicated problem for my AP physics class. I’ve reduced it to 6 equations and 6 unknowns. According to my physics teacher, you just have to hack the algebra out by hand. I decided to try to solve this problem using Cramer’s Rule, but he told me that it probably wouldn’t work. I understand roughly how Cramer’s works with 2 equations and 2 unknowns, but what happens when it is 6 and 6? Is it feasible to solve the problem using this method?

Note: he wants it done by hand, so even if I use Cramer’s, it should be something that doesn’t require a calculator.

I see no reason why you shouldn’t be able to hack out a 6x6 matrix using Cramer’s Rule. This is, essentially, second-year algebra. I would not want to try it by hand (the calculations would be horrendous) but there is no reason, in principle at least, why it should not be doable, even by hand.

Okay, now a few questions. What kind of physics problem is this that requires six variables, and what kind of sadist is a teacher that makes you do it by hand? I’d have said ‘Cramer’s Rule isn’t rocket science’, but that might be exactly what you’re doing. :slight_smile: Good luck with the math (and your grades).

well, in case you’re curious: it’s a double Atwood’s machine. There is a pulley - on one end we have a mass, and on the other end we have another pulley with two masses on it. I have to find all of the accelerations and tensions in terms of the masses and g (it’s all symbolic). Hardly a nice problem for a high school class.

As far as I recollect, Cramer’s rule isn’t limited to 2 unknowns. This site shows it done for 3 (and later on, for N) - seems that upgrading it to 6 should be straightforward.

Straightforward, but awfully computationally intensive by hand. You’ll need to calculate the determinants of seven different 6x6 matrices - while do-able, that sounds pretty damn tedious. Depending on just how the unknowns appear, simple Gaussian elimination might be less work.

Cramer’s rule only works if the equations are all linear in the unknowns, BTW. If your unknowns are a,b,c,d,e and f, there had better not be any ab terms or a[sup]2[/sup], or anything like that. There may be some kind of nonlinear extension of Cramer’s rule that I’m not aware of, of course…in fact, I’d say it’s likely, but it’s probably horrid. :slight_smile:

Have fun with it…

Okay, there are several ways to do large, simple simultaneous equations w/o a calculator. All are pains in the butt. You can do row reduction, Cramer’s rule, Laplace development, and probably some other stuff too. Here’s how to use Cramer’s rule for matrices w/ n>2:

The determinant “D” for the 2x2 matrix [[a,b],[c,d]] is ad-bc. Simple, right? For the 3x3 matrix [[a,b,c],[d,e,f],[g,h,i]], D is determined using minors and cofactors. To do this, imagine crossing out one row and one column of the 3x3 matrix, leaving a simple 2x2 matrix. Now take the determinant of this 2x2 matrix (which you already know how to do). Now multiply this answer (D1) by the number which got crossed out both vertically and horizontally. This number is called the minor. Now multiply by the cofactor. The cofactor is not a number, but either a + or - sign. Start with a + sign in the upper left hand corner, and then just checkerboard the rest of the matrix to see what the cofactor of every other square is. Then repeat this process for an entire row or column, and add 'em all up.

Perhaps an actual example will help here, no?

Take the following matrix:

1 5 2
3 4 9
7 8 6

We can pick any row or column we want to expand upon. Let’s pick the second row. Start with the first number in the second row, 3. Cross out the row and column that it is in. This leaves us with the matrix [[5,2],[8,6]]. The determinant D1 is 14. Thus the first number we get is 3 * 14 * -1=-42. Remember, -1 is the cofactor. Here’s how the cofactors look for the matrix:

Now let’s do this with the second number in the second row, 4. D2= -8. Our second number is -8 * 4 * 1 = -32.

Our third number is -27 * 9 * -1 = 243.

So the determinant is 243±32±42=169. A quick check on Mathematica reveals this to be correct. You could have used any row or column and arrived at the same result. Naturally, to simplify things, you’ll wish to use a row/column that has the most 0’s in it.

Now, for a 4x4 matrix, simply expand the process. Pick a row or column. Cross off a number to get a 3x3 matrix. Find its determinant, multiply by the minor and cofactor, and then continue down the row or column. Keep expanding for a 5x5 and 6x6 matrix. This is already a lot of work, and you’re still just finding the determinant, not using Cramer’s rule yet. Here come’s Cramer’s rule:

The value of “x” in some set of simultaneous equations is found as follows. Take the determinant of the matrix, using the coeffecients next to the variables as the elements (already done). To solve for any single variable, replace its coefs in the matrix with those of the coefs with no variables next to them, and find its determinant (D2). Take D2/D1=answer. Do this six times for all six variables, and you have yourself an answer. I’ll do a simple example with a 3x3 matrix:

3x + 2y + z = 14
x - 4y + 3z = 4
4x + y - z = 2

We find the determinant of:

3 2 1
1 -4 3
4 1 -1

It is 46.

Now, we plug in the numbers with no variables next to them for the x values, giving us the matrix:

14 2 1
4 -4 3
2 1 -1

Take the determinant. It is also 46. 46/46 = 1, so x is 1.

Now do y:

3 14 1
1 4 3
4 2 -1

Now our determinant is 138. 138/46 = 3, so y = 3. You can now solve for z the easy way.

You can usually use Gaussian row reduction to simplify your matrices (and hence your determinant-finding) a little bit, but that’s a bit complicated for high school. It’s definitely too complicated for the SDMB! Good luck.

My advice, for 6x6 matricies, Cramer’s rule would be too hard and error-prone, although it would work. Look up row reduction or Gaussian elimination in an Algebra book- these techniques are tedious, but it’s easier to spot and reverse a mistake.

I’m having a hard time seeing six equations from a double Atwood’s machine: Last time I did that problem, I only ended up with two, one for each pulley. Maybe you’re on the wrong track?

I’m surprised to see Cramer’s rule being taught, but maybe I shouldn’t be. I have an old, old numerical methods text which has a funny section in the middle entitled “Interlude - What Not to Compute”. He mentions how many people will solve a set of linear equations by using the library inversion routine, and multiplying the target vector, which is substantially more calculation than Gaussian elimination. He then has this footnote:

He also suggests (concerning the student who inverts and multiplies, rather then using elimination):

Solving a 6x6 by hand for an academic exercise is, well, to borrow another thread, beyond the pale by any method.

To reiterate:

We have a pulley. On one end of the string running through the pulley, we have mass 3. on the other end, we have a massless pulley on which we have mass 1 and mass 2 hanging from opposite sides. We are asked to find all of the accelerations and tensions in terms of m1,m2,m3 and g.
The six variables in my case are:

  1. acceleration of mass 1
  2. acceleration of mass 2
  3. acceleration of mass 3
  4. tension of string connecting m1 and m2 in the sub-Atwoods machine
  5. tension of string connecting the sub-pulley to m3
  6. the acceleration of the sub-pulley

I actually did manage to solve this problem, with manual algebraic manipulation to isolate one variable at a time, and then with a TI-89 to simplify in order to solve for that variable after I’ve isolated it. I was really not supposed to use any sort of computer thing, but I don’t have the time, skill, or effort to simplify the nasty algebra involved.

I was able to verify the solution by using a special case where mass 1 = mass 2 = 1/2 mass 3. This would result in equilibrium, causing all of the accelerations to be equal to 0.

Thanks for all the input on Cramer’s rule, I decided that it wouldn’t save me any time in this problem, and would probably result in lots of careless errors. I hate to admit defeat though, you can imagine how I feel about this physics teacher - a man who says “this is not a math class” and proceeds to give us three hours of algebra homework.

I agree with Chronos; having 6 equations for this simple system sounds rather suspicious. That aside, I recommend using simple Gaussian (or Guass-Jordan) elimination like everyone else has to get a solution. It shouldn’t take you more than 5 minutes if you get familiar with the process. You basically subtract rows by multiples of other rows to reduce it. On the other hand, if you’re working with acceleration, you’re likely to have some second order terms, and that will require differnt techniques altogether.

I never quite understood the use of Cramer’s Rule except for computers…

Ah, I see the difficulty. You do, indeed, have six variables there, but some of the equations relating them are trivial. For instance, the acceleration of the separate mass is equal to the acceleration of the sub-pulley (assuming that the sring doesn’t stretch), which is in turn equal to the average of the accelerations of the two sub-masses.

I always did wonder as to the usefulness of Cramer’s rule; thank you, yabob, for justifying my suspiscion.

I agree with Chronos. Some of the equations do seem trivial and might not apply. I remember doing tons of pulley problems in college and I don’t remember most of them being solved with a 6x6 matrix. On a second note, I’m glad to see that you are challenging your teacher a bit. I’ve found out that in the professional world, as well as in school, everybody doesn’t know as much as you think. It could be on this particular problem or subject that your teacher only has limited knowledge. He or she may only know what is listed to the problem in the teachers guide to the book and drawing there only conclusions. Or… your teacher may a genius amd I’m an idiot.

The use of Matricies for solving multiple equations with multiple unknowns is helpful only when using a computer to solve the equations. It is my understanding that it is the sole reason it is taught. As far as the size of the matrix, it doesnt matter, as long as there are as many distinct equations as there are unknowns. I rarely have seen more than 4 unknowns/equations for any given problem, however, and usually a simple substitution of terms is all that is required to solve more problems.

Of course, I am only an engineer and have only had college level math through multi-dimensional calculus. I am sure there are plenty of other methods to solve such equations.

It would seem to me that the student made the problem too complicated. At a glance, massless Pulley 2 (with masses M2 and M1) could be solved independent of the rest of the system. Then Pulley 1 (with the mass of the Pulley 2 system and M3) could be solved. Accelerations of M2 and M1 then could be translated for the motion/acceleration of Pulley 2). Another way to think of it is there are 4 Forces involved (F=ma), with F4 equal to F2 plus F1. I would think 4 equations could do this.

Yes, two of the equations are trivial because:

  1. The acceleration of the massless pulley equals -1 * the acceleration of mass 3 (the one that isn’t attached to the sub-pulley)

  2. The tension on the string connecting the sub-pulley and mass 3 is just two times the tension of the string running through the sub-pulley.

So it does actually reduce to 4 equations, for some reason it didn’t occur to me to perform this substitution before I attempted to use matrices in order to solve the problem.