A uniform ladder of weight $W$ and length $2L$ rests against rough wall(coefficient of friction $2/3$) and has the lower end on a rough floor(coefficient of friction $1/4$) the ladder is inclined at an angle $X(\theta)$ with the horizontal where $\tan X=5/3,$ calculate how far a person weighing $3W$ can ascend the ladder.

Now I've resolved the forces acting vertically where we have $R$(the normal force acting vertically from the bottom of the ladder) is equal to $W$(the weight of the ladder) + $2/3S$(our friction force working vertically at the top end of the ladder with $S$ being the normal force reacting at that point) This gives us $R=W+\frac23S$

Resolving Horizontally I get $S=\frac14R$ as thats the friction force at the bottom of the ladder equal to the normal force working the other way at the top of the ladder.

Now for taking moments I decided to put my point of axis at the bottom of the ladder so having clockwise = anticlockwise I get this equation $W\cdot2L\cos X=S\cdot2L\sin X+\frac14S\cdot L\cos X$

Unfortunately this is where I am stuck I don't know how to apply the person weighing $3W$ into the question at this point and how to answer how far they could ascend the ladder I have watched so many separate torque videos online and none of them go into how far a person who isn't already on the ladder could ascend it so any help would be greatly appreciated on what i need to do next