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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: enter image description here

My working: FBD of 100kg:enter image description here

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): enter image description here

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!

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In reality static equilibrium would be maintained over a range of values of M2. Near the minimum value of M2, frictional force would be directed up the incline and would be near its maximum value of coefficient * N. Whereas near the Maximum value of M2, it would be directed down the incline but still near its Maximum value. It makes sense if you sit down and think about it and makes you realise the concept that friction force opposes relative motion. Hope this helps

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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.

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If you take the friction in the direction you took, then you will have the maximum value of M_2 and in these kind of problems you will get a range of value of the second mass because frictional force can adjust its direction, so that the "relative motion of the block and the wedge" stops/tends to stop.

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