How Do I Determine the Correct Direction of the Friction Force?

[LCSphysicist]

Homework Statement

nothing

Relevant Equations

nothing

Exercise like this:

I think that the friction force can assume two direction, up and down, but how can i know the direction of the force?
I want to say, i don't even know which mass is greater to say i the block will upward or downward.
that is the doubt.

That is the correct forces directions:

[how can i know that is the right direction to friction force?]

The only answer i wonder is, if i choose the wrong direction, eventually i will fall into absurdity, that is, the magnitude force will be negative, but the magnitude is only positive, so that is the absurdity

 

[haruspex]

You do not need to know the direction of the force in advance. Just choose one. If you guess wrong you'll get the value as negative.
Why do you consider it absurd? Scalars can be negative. Don't think of it as a magnitude.

 

[LCSphysicist]

You do not need to know the direction of the force in advance. Just choose one. If you guess wrong you'll get the value as negative.
Why do you consider it absurd? Scalars can be negative. Don't think of it as a magnitude.

Hmmm wel this make sense, but if i can't know if the value is negative? Well, i will use this example.

The question is, find the aceleration of the system.
Its not hard, calcule torque and... well... go on, but i choose torque as r(T-T'), when the "question" in the resolution use r(T'-T), the results?
answer right: m'g/(m' + m + M/2)
answer i came: m'g/(m' + m - M/2)

Thats because i choose "the wrong" direction to torque, but how will i know that is the wrong?

 

[haruspex]

Hmmm wel this make sense, but if i can't know if the value is negative? Well, i will use this example.
View attachment 260211
The question is, find the aceleration of the system.
Its not hard, calcule torque and... well... go on, but i choose torque as r(T-T'), when the "question" in the resolution use r(T'-T), the results?
answer right: m'g/(m' + m + M/2)
answer i came: m'g/(m' + m - M/2)

Thats because i choose "the wrong" direction to torque, but how will i know that is the wrong?

Your error was not that you chose the wrong direction for torque but that you were inconsistent about it. I can't pinpoint your error without seeing all your working.
Here's how it should have gone:
Taking anticlockwise as positive, torque is R(T-T').
$\frac 12MR^2\alpha=R(T-T')$
Taking right and down as positive for linear acceleration, a,
m'a = m'g - T'
ma = T
And here is the critical one:
$a=-R\alpha$
Do you see why?

 

[LCSphysicist]

Your error was not that you chose the wrong direction for torque but that you were inconsistent about it. I can't pinpoint your error without seeing all your working.
Here's how it should have gone:
Taking anticlockwise as positive, torque is R(T-T').
$\frac 12MR^2\alpha=R(T-T')$
Taking right and down as positive for linear acceleration, a,
m'a = m'g - T'
ma = T
And here is the critical one:
$a=-R\alpha$
Do you see why?

Oh that's good, Now i know that i made a mistake in this end. Thank you, i will be more attentive with my choose.