Maths question help please

Hi, please could I have help on question 14a of this paper? So I’m really confused and trying to resolve around all the particles. What I’m confused on is what would the acceleration of a and b be? C has an acceleration of a but a is attached to b with a different string and b is attached to two strings?
Question: https://www.ocr.org.uk/Images/667257-question-paper-pure-mathematics-and-mechanics.pdf
Thanks!!

The strings are all inextensible and under tension so all particles must have the same acceleration "a" as the string lengths do not change. To determine it, you could think of it as a single system ABC (so forget about any internal tensions) where the resolved gravity acts in opposite directions on A and C and friction acts on B.

oh ok i see, for part b i'm confused because i thought that the magnitude of the contact force would be the resultant of the normal contact force and then the horizontal resultant force? thanks!

Just had a look at the mark scheme and for part a) splitting it up into seperate parts so AB and BC and then combining isnt wrong but its unnecessary (you should be able to do it though). Arguing that the strings remain under tension so you can treat it as a single system with a common acceleration you can write down
(2+3+4)a = 4gsin(60)-2gsin(30)-3g*mu
which eliminates the first half of the working. Its also a good thing to practice as, for some reason, people seem to model things independently and then combine, even when you dont need to (as here).
But for B, the forces on it from the horizontal plane are the normal reaction (vertical) and friction (horizontal), so pythagoras to get the magnitude. I guess youre including acceleration or tension or ... in the horizontal? If so, thats not a contact force between B and the horizontal plane as a diagram would make clear?

Oh so is tension never included as a contact force, like if I drew a single diagram of the forces on b, there would be no tension,just friction weight and normal contact force?

If you treat it as a single body, the strings (under tension) are simply part of the body and newton 3 says each string acts in an equal but opposite way at each end, so sums to zero. It would be "no" different if A, B and C were glued together. So here youd have the single system ABC (mass) being accelerated with acceleration "a" by the resolved gravitation force to the right of 4gsin(60) and being re tarded by 3gmu and 2gsin(30) to the left. The actual slopes are irrelevant apart from thinking about the magnitude of the gravitational forces to the right and left.

So hypothetically if there was only one string pulling on a block, would the tension be included in the force diagram for the block because it won’t be cancelled out?