# Confused about direction of friction

Question : The system of two blocks and fixed pulley are joined by a cable as shown. Determine the mass $$\textbf{M_2}$$ required for static equilibrium. All wheels and pulleys have negligible friction:

My working: FBD of 100kg:

Then for static equilibrium (using a rotated coordinate system)we require

$$\sum F_y=0$$

$$N=m_{100}g\cos(25^{\circ})$$

$$\sum F_x=0$$

$$2T=F_f+m_{100}g\sin(25^{\circ}) \Leftrightarrow T=\frac{1}{2}m_{100}g (\mu \cos(25^{\circ}+\sin(25^{\circ}))$$

And now consider a free body diagram of the other block (M_2):

Then $$\sum F_y=0$$

$$T=m_2g\cos(10^{\circ})$$

$$\frac{1}{2}m_{100}g (\mu \cos(25^{\circ}+\sin(25^{\circ})) = m_2 g\cos(10^{\circ})$$

$$m_2 = 36.6kg$$

However my answer is wrong the correct $$m_2=6.06kg$$, this is because they took friction in the opposite direction to me.

But I was confused, why do they do this? Why does friction point to the right for the big block and not to the left? I think the friction points to the left because $$M_2$$ is on a bigger slope compared to $$M_1$$ so I think the whole system SHOULD move to the right.

I also think the question is underspecified and there should be two answers. Maybe I am overthinking this, maybe I am wrong. This is why I have came here to clear my doubts. How do I figure out the /correct/ way of friction acting in a question like this.

I am aware of the non-homework posting rule but I think this is different because I think I have written enough for a discussion :)

Thank you!

Both your answers are equally correct and I think the question meant the minimun/smaller mass of $M_1$ as taking fixed Tension in the string of you take friction in the opposite direction of tension you get the minimum mass $M_1$ while taking friction in same direction of tension would yield the bigger mass.
In reality both of these masses would result in a static equilibrium as the friction can have a range of values and thus $M_1$ can thus have a range of values.