Mechanics Maths Year 2 Edexcel Question -Need help

Please can someone talk me through the mechanics question below. I have the mark scheme but I don’t understand part b and kind off understand part a
thank you
the answer to mew for part b is 18/49 or 0.37 (2.s.f.)

Please can someone talk me through the mechanics question below. I have the mark scheme but I don’t understand part b and kind off understand part a
Thank you
This question should only require content from chapters 4/5/7
the answer to mew for part b is 18/49 or 0.37 (2.s.f.)

For part a) I assume they took moments about A, so you balance the clockwise moment due to the weight with the anticlockwise moment due to the normal reaction at C? If you dont understand that, post what youre unsure about.
For b) resolve vertically to get the normal (vertical reaction) at A, then resolve horizontally to get the necessary friction force and hence mu (using the limiting equilibrium/vertical reaction). Again, post what youve tried / unsure about?

Upload your diagram and what you've tried. We don't give solutions but will guide you through it

Could the confusion be from not splitting the normal reaction at C into its horizontal and vertical components? And then considering them when resolving ?

Agree for a).
For b) Agree about RA. Id have drawn a clearer diagram for the peg/normal reaction as the normal reaction makes an angle of 90-alpha with the horizontal or alpha with the vertical and that should be clear. For the horizontal friction must be equal but opposite to the horizontal component of the reaction at the peg. Then use Ra to get mu. You seem to drop the cos(alpha) on line 2 of your second pic working.
It would be better to do the simplification of each resolved horizontal force before the division to get mu.
To come back to the diagram at C, draw the horizontal at C, then the angle the rod makes with that must be alpha as its a "z" shape (parallel lines). Then the angle the normal reaaction makes with the horizontal must be 90-alpha, so it makes an angle of alpha with the vertical. So you have a 90 : alpha : 90-alpha right triangle to resolve the normal reaction. So sin(alpha) gets the horizontal component and cos(alpha) gets the vertical component.

Thank You!!
is it possible if you could draw what you mean for the component forces at C (sorry i’m more of a visual learner)

Sort of, will do it if you cant annotate something like (right diagram)SoSee if you can make sense of that and upload a sketch if necessary. If its not clear, Ill do a diagram.

is this correct? (if the image quality is rlly bad then lmk and i’ll reupload it, idk why my device is lowering the quality)

That looks spot on. The key thing is to do alternate angles to get theta up to C, then its the usual resolving forces.
A simple sanity check is to pick something like theta=10 degress (so its almost horizontal), then the normal reaction at C is 90-10=80 with the horizontal, which is about right/plausible.