Kinda depressing that it’s taken me two years to get from chapter 4 to chapter 8.
Much of the trig can be avoided, at the expense of more effort in figuring out the various angles in the diagram, by working with the force vector, R, acting at point B, rather than its components normal and tangent to the surface (N_B and F_B).
Force balance along direction of R immediately gives magnitude of R:
R=W cos(alpha+phi)
since the force at A has no component in the R direction.
Moments about point A gives
2hR sin(theta-phi) = hW sin(alpha+theta).
Substituing for R results in
2 cos(alpha+phi) sin(theta-phi) = sin(alpha+theta)
This is the point you had almost reached after all the trig work above. Using the identity for 2cos(x)sin(y) results in
sin(theta-alpha-2 phi)=0
theta=alpha+2 phi, and if friction acts in the direction opposite from assumed, use negative phi, for theta=alpha-2 phi
I think you might be missing a term. I get:
\begin{equation}
\phi=tan^{-1}\mu_{s}\\
tan \phi=\mu_{s}\\
\end{equation}
\begin{equation}
1+\mu_{s}^2\\
=1+\tan^2\phi\\
=\sec^s\phi\\
\end{equation}
\begin{equation}
-W\sin\lgroup\alpha+\theta\rgroup+\frac{2W\lgroup-\mu_{s}\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\mu_{s}\cos\theta\rgroup}{1+\mu_{s}^2}=0\\
-W\sin\lgroup\alpha+\theta\rgroup+\frac{2W\lgroup-\mu_{s}\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\mu_{s}\cos\theta\rgroup}{\sec^2\phi}=0\\
-W\sin\lgroup\alpha+\theta\rgroup+2W\lgroup-\mu_{s}\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\mu_{s}\cos\theta\rgroup\cos^2\phi=0\\
-W\sin\lgroup\alpha+\theta\rgroup+2W\lgroup-\tan\phi\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\tan\phi\cos\theta\rgroup\cos^2\phi=0\\
-W\sin\lgroup\alpha+\theta\rgroup+2W\cos\phi\lgroup-\sin\phi\sin\alpha+\cos\phi\cos\alpha\rgroup\lgroup\cos\phi\sin\theta-\sin\phi\cos\theta\rgroup\\
\end{equation}
Eh, in for a penny. I’ve come this far, I would like to see if I can get this to resolve to something usable.
ETA: Forgot to factor out W.
\begin{equation}
-\sin\lgroup\alpha+\theta\rgroup+2\cos\phi\lgroup-\sin\phi\sin\alpha+\cos\phi\cos\alpha\rgroup\lgroup\sin\phi\sin\theta-\sin\phi\cos\theta\rgroup\\
\end{equation}
You have an extra cos(phi). After you multiplied each of the 2 terms in parentheses by cos(phi), you kept an extra factor of cos(phi) in the next line.
That’s why I posted; I feel like you left the cos(phi) term out.
Look at the line before that.
\begin{equation}
-W\sin\lgroup\alpha+\theta\rgroup+2W\lgroup-\tan\phi\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\tan\phi\cos\theta\rgroup\cos^2\phi=0\\
\end{equation}
The last term is cos^2(phi). Multiply the preceding terms by cos(phi):
\begin{equation}
\lgroup-\tan\phi\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\tan\phi\cos\theta\rgroup\cos\phi\\
\lgroup-\sin\phi\sin\alpha+\cos\phi\cos\alpha\rgroup\lgroup\cos\phi\sin\theta-\sin\phi\cos\theta\rgroup
\end{equation}
and there’s still another cos(phi) term left, isn’t there?