I think I know what you are talking about. For example, there is a famous problem called the 4 colour problem. It states that any region that has been decomposed into smaller regions can be coloured with at most 4 colours such that no two adjacent ones has the same color. When mathemations where working (or worked) on this problem they completely changed the problem. Because the original problem was not mathematical enough, I am myself not sure how you even begin to define something like that but that is what they did.Actually, I've noted Mathematicians using this one too. They simply redefine the problem into one they know how to solve.
I know that joke too.Physicists, on the other hand, tend to use "spherical horses," as the joke goes.)
the angle wouldn't be zero when the force is minimixed. If the force shall be minimixed, should be minimized. The equation abowe describes a line sloping downwards, crossing the axis when is . Since the force should be in a right angle to this line, the tangent of the angle should equal if I have got this rigt.
No time still to post the soulution---I type really slow, because I am a one-finger typist---I will do it in maybe 2 hours from now. [Household chores first.]
I am very sorry I just found out now that I really made a mistake in my fast calculations this morning. I said I found an answer where x is not zero, but I had a nagging question in my mind almost all day long---because my reasoning without computations said for minimum P, the angle x cannot be more than zero.
Without P, the block will slide down the incline by a force of [60sin(30deg) -(0.25)(60cos(30deg)] = [30 -13] = 17 lbs.
So with P, and angle x is zero, P = 17 lbs going up the incline to stay the block.
If the P is raised at an angle x with the incline, then the total normal force on the incline will be decreased by P*sin(x). Hence the friction will be lesser than 13 lbs. So the sliding force will increase, will be more than 17 lbs. So the component of P parallel to the incline will be more than 17 lbs also to stay the block. Thus, the P, which is be larger than this parallel component, will be even more than 17 lbs. Therefore there is no way that P will be less than 17 lbs if angle x is more than zero. That is why for minimum P, the angle x must be zero.
That was what I had in mind while driving to the jobsite. But to be sure, when I had a chance, I did a quick calculation, and I found a P less than 17 lbs with an angle x greater than zero. I said, "Wow, really?" No time to check the computations, I was a little bothered by the written answer the whole day. Until a while ago before I finally had time to sit down before this laptop.
Then I again quickly solved the problem on a sratch paper. And there I saw what went wrong this morning.
Along the incline,
(0.25)[51.96 -P*sin(x)] +P*cos(x) = 30
-(0.25)P*sin(x) +P*cos(x) = 30 -12.99
P[(0.25)sin(x) +cos(x)] = 17 --------------there was the mistake.
The (0.25)sin(x) should have been negative!
Of course, if it were negative, then
P[-(0.25)sin(x) +cos(x)] = 17 ----this will lead to an optimization:
P =  / [sqrt(1 -sin^2(x)) -(0.25)sin(x)] -----(i)
Get the derivative of P with respect to sin(x), then equate that to zero.
To avoid confusion, let s = sin(x),
P = 17 / [sqrt(1 -s^2) -(0.25)s]
dP/ds = 0 will lead to
-s/[sqrt(1 -s^2) -0.25 = 0
-s = (0.25)sqrt(1 -s^2) -------------and here I could have found the mistake. And I should have not written in this thread here a while ago as I arrived home that I found a P and an x.
Oh, well, sorry again.
It always pay to review your computations first before stopping in your calculations. , .