Still working this problem of intersection of line and circle. Trying to simplify. Here is initial calculation:

Code 1:

ClearAll["Global`*"]
ClearAll[x, w, b, m, R, xo, phiw, y, J, solns1]
{x, w, b, m, R, xo, phiw} \[Element] Reals

Solve[{x, w} \[Element] 
   InfiniteLine[{{0, b}, {-b/m, 0}}] && {x, w} \[Element] 
   Circle[{xo, 0}, R], {x, w}]
FullSimplify[%, 
 0 < phiw < Pi/2 && R >= 0 && m >= 0 && b >= 0 && xo >= 0]

% /. m -> Tan[phiw]
solns1 = 
 FullSimplify[%, Reals, 
  Assumptions -> 0 < phiw < Pi/2 && b >= 0 && J >= 0 && xo >= 0 ]
x /. solns1[[2]]

This gives two solution for x and w. The algebraic is easier to follow, but the trig gives a simpler solution as last step. After some manipulation, I can clean these up. For example, for one solution, if I do by hand, I get:

Code 2:

m = Tan[phiw]

w = (b + m (xo + Sqrt[(1 + m^2) R^2 - (b + m xo)^2]))/(
  1 + m^2) = (b + m xo + m Sqrt[(1 + m^2) R^2 - (b + m xo)^2])/(
   1 + m^2) = (b + m xo)/(Sqrt[(1 + m^2)] Sqrt[(1 + m^2)]) + (
    m Sqrt[(1 + m^2) R^2 - (b + m xo)^2])/(
    Sqrt[(1 + m^2)] Sqrt[(1 + m^2)])
= (b + m xo)/(Sqrt[(1 + m^2)] Sqrt[(1 + m^2)]) + (
   m Sqrt[(1 + m^2) R^2 - (b + m xo)^2])/(
   Sqrt[(1 + m^2)] Sqrt[(1 + m^2)]) = (b + m xo)/(
    Sqrt[(1 + m^2)] Sqrt[(1 + m^2)]) + (
    m Sqrt[R^2 - (b + m xo)^2/(1 + m^2)])/Sqrt[(1 + m^2)] = 
   Ro/Sqrt[(1 + m^2)] + (m Sqrt[R^2 - Ro^2])/Sqrt[(1 + m^2)]

w = Ro Cos[phiw] + Sqrt[R^2 - Ro^2] Sin[phiw]

(b + m xo)^2/(1 + m^2) = Ro^2

x = (-b m + xo + Sqrt[(1 + m^2) R^2 - (b + m xo)^2])/(
  1 + m^2) = (-b m + xo )/(1 + m^2) + Sqrt[
    R^2 - (b + m xo)^2/(1 + m^2)]/Sqrt[(1 + m^2)] = (-b m + xo )/(
    1 + m^2) + Sqrt[R^2 - Ro^2]/Sqrt[(1 + m^2)]

x = -b Cos[phiw] Sin[phiw]  +  xo Cos[phiw]^2  + 
  Sqrt[R^2 - Ro^2] Sin[phiw]

Note m is the slope of the line. And you get a little simpler after using Tan[phi]. Ro is the specific circle radius that gives one solution, or the tangent line.

First question, how do I do these manipulations in Mathematica. Doing by defeats the purpose.

Second, when I try and equate the initial solution set with my manipulated set, they are clearly equal. But I can't get == or === or in combination with Expand or Reduce, can't get a True. This was the code for this comparison next. What is the secret to showing these are equal?

Workbook is attached.

Code 3:

ClearAll["Global`*"]
ClearAll[x, w, b, m, R, xo, phiw, y, J, solns1, x1, x2, s1, s2]
{x, w, b, m, R, xo, phiw} \[Element] Reals

m = Tan[phiw]
x1 = (-b m + xo + Sqrt[(1 + m^2) R^2 - (b + m xo)^2])/(1 + m^2)
x2 = -b Cos[phiw] Sin[phiw]  +  xo Cos[phiw]^2  + 
  Sqrt[R^2 - (b + m xo)^2/(1 + m^2)] Sin[phiw]
x1 == x2
s1 = FullSimplify[x1, Reals, 
  Assumptions -> 0 < phiw < Pi/2 && b >= 0 && J >= 0 && xo >= 0 ]
s2 = FullSimplify[x2, Reals, 
  Assumptions -> 0 < phiw < Pi/2 && b >= 0 && J >= 0 && xo >= 0 ]

s1 == s2

Attachments:

Wall_Friction_Ca...nb

Have you seen http://reference.wolfram.com/language/tutorial/ManipulatingEquationsAndInequalities.html? If so, what's missing?

(I found it by going to the documentation page for Solve[] and clicking on one of the linked tutorials. About half the links are linked to some subsection of this tutorial.)

I have not seen a tutorial, in written or video format, that takes me through all the obvious steps in setting up simultaneous equations and simplification and manipulations of results. Is there one??? I little disappointed in Mathematica here. Lots of and lots of tutorials, but seems to be a whole lot less on this, which seems weird. Most are either over simplistic or focused on a random set of topics, and not core math and engineering calculations.

thanks, that worked on full simply and can have && for multiple assumptions, yes?

How can I restrict all calculations and variables to real domain, or eg. positive reals and zero. Is that possible for whole workbook? and does it get passed down to future expressions.

Or at least the results of trig functions.

also trying to assign the result of a solution back to one of the variables, see last post

You should check the syntax of FullSimplify:

Solve[-b^2 m^2 + m^2 R^2 + m^4 R^2 - 2 b m^3 xo - m^4 xo^2 == 0, R]
% /. m -> Tan[phiw]
FullSimplify[%, 0 <= phiw < Pi/2]

ClearAll does not clear all... for example, if x=sin(b), it will remember that.

In your code snippets above, you have just the raw head ClearAll. You need to pass it an argument(s). So, in your x example:

ClearAll[x]
(* or *) 
ClearAll["x"]

If you want to clear everything in the Global namespace:

ClearAll["Global`*"]

First, while Solve does accept complicated expressions as variables, it seems to rule out ones with certain heads, like Plus, Times, and Power. I surmise the reason is that these have internal simplification rules that are applied automatically. It seems difficult to disentangle b when b^2 m^2 becomes J^2 m^4 (or worse, since m has a complicated value). You could try solving for J instead of J*m and then recover b from J * m /. solution.

Second, the desired answer indicated in one of posts suggests you want to simplify or reduce, rather than solve for b. However, doing so indicates a problem in your setup. Maybe there's an error in one of the formulas or assumptions?

Simplify[-b^2 m^2 + m^2 R^2 + m^4 R^2 - 2 b m^3 xo - m^4 xo^2 == 0, 
 Assumptions -> R > 0 && b > 0 && J > 0 && m > 0 && y > 0 && sig1 > 0]
(*  False  *)

Reduce[-b^2 m^2 + m^2 R^2 + m^4 R^2 - 2 b m^3 xo - m^4 xo^2 == 0 && 
  R > 0 && b > 0 && J > 0 && m > 0 && y > 0 && sig1 > 0, Reals]
(*  False  *)

For reference, here are the definitions I copied:

y = Sin[phiw];
xo = (sig1 + J)/(1 + y);
R = y*xo;
m = Sqrt[y/(1 - y^2)];
b = J*m;

Above text not clear, not sure why post runs it all together. Repeated againenter code here:

   ClearAll
y = Sin[phiw]
xo = (sig1 + J)/(1 + y)
R = y*xo
m = Sqrt[y/(1 - y^2)]
b = J*m

Solve[-b^2 m^2 + m^2 R^2 + m^4 R^2 - 2 b m^3 xo - m^4 xo^2 == 
  0, b, Reals,
 Assumptions -> R > 0 && b > 0 && J > 0 && m > 0 && y > 0 && sig1 > 0]

The final answer should look something like :

   Sin[w/x] = -(((J + sig1) Sin[phiw])/(-sig1 + J Sin[phiw]))

btw m = tan[phiw], or tangent to the circle given by xo and R . Which are related as given above, with y = sin[phiw] .