yawtsok said:
Lol, I don't think he wants me to dig too deep into it. Perhaps saying that the Normal force balances the mass of the block out so mass is not a factor in determining Mu of K is enough.
I'm sorry, but what you've said here is not correct, and doesn't make much sense. You should re-read my explanation of why mass doesn't matter in post #7. More massive objects, in addition to having more weight, also experience more friction. The latter is due to the higher normal force between them and the surface they are sitting on.
yawtsok said:
No, it really isn't. To "derive" a result means
to mathematically show that that result is true. In other words, you should be able to arrive at that result algebraically, as a natural consequence of what you already know (the established principles that you start out with). In this case that major starting point is Newton's second law.
If you set up the equations in the way I suggested, then after a
very small number of algebraic manipulations of those equations, the result that μ
k = tanθ WILL come out.
EDIT: Here is an idea of what I'm talking about:
Condition 1 - sum of forces acting parallel to the plane = 0
[1] (component of weight pulling down the plane) + (friction) = 0
Of course, one of these forces will be negative, since there are two directions "along the plane". For simplicity, you could say that the "down the plane" direction is positive, and the "up the plane" direction (in which friction points) is negative.
Condition 2 - sum of forces acting perpendicular to plane = 0
[2] (component of weight pushing into the plane) + (normal force) = 0
Again, these two forces have opposite sign.
[1] & [2] are the two equations that you get by considering the force balancing. I've expressed them in words, rather than symbolically. Now, you need to figure out these two components of the weight -- the parallel and perpendicular components. The full weight is mg, and the components will be related to this by trigonometric ratios involving the inclination angle. If you draw a picture (a right triangle) of the weight vector resolved into parallel and perpendicular components, it should be obvious what the components are.
That's it. Now I've really held your hand and guided you through it.