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I'll have a go... someone correct me if I make a mistake.

basically we're looking at something like this...

You'll notice in the diagram that the angle between the gravitational force acting straight down and the normal force on the block (force due to gravity acting perpendicular to the surface of the ramp) is equal to theta, where theta is the angle of the block with the horizontal.

Another picture to help explain that (if you didn't know already)

So... using these relations (note the first picture) and summing all the forces, we can say...

-mu*M*g*cos(theta)+M*g*sin(theta)=a

where 'mu' is the coefficient of friction, 'M' is the mass, 'g' is the gravitational constant, and 'theta' is the angle between the ramp and horizontal

the M cancels out of the equation to leave

g*sin(theta) - mu*g*cos(theta) = M*a [eqn 1]

Looking at the constant acceleration formulas, the best one to use would be

V^2=V_0^2+2*a*(x-x_0) [eqn 2]

where 'V' is the final velocity, 'V_0' is the initial velocity, 'a' is acceleration, 'x' is final location, 'x_0' is initial location

You can see that both 'V_0' and x_0 are zero (starts from rest, and we'll call the starting location zero. You also know 'V' to be 5.25 m/s and 'x' to be 8.35 m from the problem statement. You've solved for 'a' using [eqn 1] and then you can plug 'a' into [eqn 2] to obtain a value for 'mu'.

That's about it... like I said, if I messed something up, someone jump in and save me!

~Derek

For the first problem, I think the one I was missing was

[0,1:0,0:0,0]

Anyone have thoughts on the second problem?

~Derek

Sorry for the long delay... I haven't got a great reference right off hand. The book that I used back when I took Diff.Eq. was

A First Course in Differential Equations

The Classic Fifth Edition

Dennis G. Zill

ISBN: 0-534-37388-7

~$40 on Amazon... I'm sure that are cheaper alternatives, but this is the only one that I've used.

~Derek

So far as having the RHS not equal to a constant....

There are general methods for solving the equations depending on the type of function on the right hand side. For instance... if you've got

left hand side = cos(t)

your particular solution would look something like

A*cos(t) + B*sin(t)

I don't know if you have access to a Diff.Eq. book, but there should be a list of the different things that you can end up with depending on the right hand side. Another common one would be something like

left hand side = A*t^n + ... +B*t + C

then the particular solution would be in the form

D*t^(n-1) + ... + E*t + F

fill in the '...' with the descending powers of t

as you can see, the more complicated the right hand side, the more constants you have to solve for in the end (using the initial conditions).

And college, well... I'm in my senior (4th) year at the University of Michigan majoring in Mechanical Engineering. Anyone out there looking for an intern for next summer???

~Derek

It's actually not that difficult after a while... for simple linear differential equations (like this one) it's mostly just a matter of following a sort of algorithm.

Find roots of characteristic equation

Determine "class" of roots (complex, distinct real, repeated real)

Form your "base" equation (C*exp(r1*t)+D*exp(r2*t)+etc...) depending on the types of roots

Solve for constants using initial conditions

Have you solved diffeq's before? If not, I can see how it would seem a bit daunting They are quite useful, as you could probably imagine. I'm taking an Automatic Controls class, and what it basically boils down to is using the different properties of differential equations to choose an equation that will model a system to gain a desired output (like maximum value, settling time, etc...)

Anyways, that was a bit off topic. Like I said, if there's something that's unclear, let me know!

Alright... I can show how I would solve this w/o using Laplace transforms. There might be a more streamlined way to do it, but if there is, someone else will have to show you

Here goes...

First we'll create what I think is called the "characteristic polynomial" for the differential equation. Basically what that means is that we're going to neglect the right hand side (300) for now, and work on the left hand side. By neglecting the right hand side, we're dealing with something called a homogenous differential equation.

Now, to start, we'll replace all d^n(q)/dt^n with m^n... so the equation becomes

m^2+8m+25 = 0

Now, solve for m, and you get -4 ± 3j (use quadratic formula, or calculator, or whatever...)

What this means is that we've got complex roots of the homogenous equation. The solution to a differential equation is usually in the form

C*exp(r1*t) + D*exp(r2*t)

where C and D are constants, and r1 and r2 are the roots of the homogenous equation. Because we have complex roots, we have to apply Euler's formula, which states something like

exp(j*K) = cos(K) + j*sin(K)

You can go through the algebra, but what this ends up boiling down to for our purposes is something that looks like this

C*exp(r*t)*cos(k*t) + D*exp(r*t)*sin(k*t)

where "r" is the real part of the complex root, and "k" is the imaginary part, so right now we've got

q(t)_homogenous = exp(-4*t)*[D*cos(3*t)+E*sin(3*t)]

Now let's deal with the right hand side of the equation that we neglected earlier. Because it is just a number, there's a bit of a trick that you can use. Essentially the right hand side is describing the "input" to the system... which means that after the transient behavior of the system dies off (the differential terms) you're left with the "input" and the linear terms of q(t). What this means in terms of our equation is that

25*q(t) = 300

and if you solve for q(t) you get q(t) = 12, which we will call our "particular" solution... because it is particular to the "input" or something like that. To get q(t) we'll add the homogenous and particular solutions

Ok, now we're basically done, we've got

q(t) = exp(-4*t)*[D*cos(3*t)+E*sin(3*t)] + 12

Now it's just a matter of plugging in initial conditions (you need (order of differential equation - 1) initial conditions) and solving for the constants, D and E.

If the initial conditions were

q'(0) = 0 and q(0) = 0

You should end up with

q(t) = exp(-4*t)*[-12*cos(3*t) - 16*sin(3*t)] + 12

I'll spare you the algebra, I'm sure you can do it (assuming you can take the derivative of q(t).

So I think that's it, if something's unclear, which I'm sure there's at least a few things, post back and I'll see if I can try to clear it up. Enjoy!

~Derek

in my opinion, the easiest way to solve this would be to use laplace transforms. do you know how to use them? if not, we can solve it using another method... reply if you want to know more!

~Derek

p.s. Matlab or Maple or a TI-89 calculator can be used to solve differential equations as well if you're not interested in the method

**kybasche**- Replies: 3

Hey all, lovely forum! I've got 2 problems, both seem to be pretty straight forward, but I'm getting stuck. I've used the notation for matrices that my calculator accepts... a comma seperates values in the same column, and a colon denotes the start of a new row, I also seperated rows with spaces to make it a bit easier to read

1) (this is paraphrased from Linear Algebra w/ Applications by Otto Bretscher)

two nxm matrices in reduced row-echelon form are the the same type if they have the same number of leading 1's in the same locations... i.e. [*1,2,0 : 0,0,*1] and [*1,3,0 : 0,0,*1] --- stars indicate corresponding leading 1's

How many types of 3x2 matrices in rref are there?

I get 3... [1,a : 0,0 : 0,0] [1,0 : 0,1 : 0,0] and the [0,0 : 0,0 : 0,0]

However, the book's answer is that there are 4 types... am I missing something??

2) (again paraphrased from the same book)

consider matrix A = [a,b : b,-a] where a^2+b^2=1

Find two nonzero perpendicular vectors 'v' and 'w' such that A*v=v and A*w=-w

Solve for the vectors in terms of 'a' and 'b'

For this one, the matrix 'A' is a reflection transformation, and in order for the reflection of 'v' to be equal to 'v,' I would imagine that 'v' has to be parallel to the line about which the reflection is taking place. I can't seem to find a way to get to the answer though.

The answer is given in the book as

v = [b : (1-a)]

w = [-b : (1+a)]

Sorry that this was so long Any help on either of these would be very much appreciated. Thanks a bunch!

~Derek

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