maths - m1 edexcel Light Inextensible String Smooth Peg

[mufc802]

There are two questions that I am unsure on.
1)

A light inextensible string passes over a smooth peg, and is attached at one end to a particle of mass m kg and at the other end to a ring also of mass m kg. The ring is threaded on a rough vertical wire as shown in the diagram. The system is in limiting equilibrium with the part of the string between the ring and the peg making an angle of 60° with the vertical wire.
Calculate the coefficient of friction between the ring and the wire, giving your answer to three significant figures.

2)

A light inextensible string passes over a smooth peg, and is attached at one end to a particle of mass 3m kg and at the other end to a ring of mass 2m kg. The ring is threaded on a rough vertical wire as shown in the diagram. The system is in limiting equilibrium with the part of the string between the ring and the peg making an angle of 30° with the vertical wire.
Calculate the coefficient of friction between the ring and the wire, giving your answer to three significant figures.

The wording seems almost exact except for the angle and mass yet in the solution it shows the first answer with the force pointing up, but on the second one it shows the force pointing down. I get the same answer anyway it is just negative but I want to know how you know whether the force is pointing up or down. Here are the diagrams for both..
1)

2)

 

[mufc802]

Yeah I had to calculate the tension which was 3mg, then find out what the frictional force was which is Tcos30 + 2mg which is 3mgcos30 + 2mg which is greater than the downwards force 2mg, so the friction must point down to keep the particle from going up. On the first one the force is less than mg, so it must point up to prevent the particle moving down. They are two questions from the book and aren't exam questions lucky, I just wanted to know how to solve it anyway.

 

[mufc802]

If you tell me which year and which paper it was from I can find the mark scheme for you.

Thanks but it's from the book, that diagram is part of the solution. It's not in any exam papers.

 

[mufc802]

OP, I don't understand. Surely you answered your own question when you worked out the answer. You must have assumed the actual direction of the frictional forces when you worked out the resulting vertical component force of the ring?

Yeah I ended up with a negative coefficient of friction so I had the frictional force pointing the wrong way. I was just asking how you can find out the direction before working out the final answer.