Tension acting on ropes with several pulleys [1]

Find the force $F$ that makes the mass go down with constant velocity, given that the mass is $300kg$ and neglect the mass of the pulleys.

My question is not about the problem given in the homework, it is rather conceptual, that's why I thought of putting it here. In this photo, I know $F=T$ where $T$ is tension, but apparently the second Newton's law applied on the mass is: $$mg-6F=0$$ my only issue is with the 6F, because I thought that the forces would divide by 2 each time it passes through one of those 3 inferior pulleys, hence I initially thought that it should've been $$mg-(\frac{F}{2}+\frac{F}{4}+\frac{F}{8})=0$$ as the force is splitted in 2 each time it passes through the inferior pulleys, but it's not, why?
[EDIT:*This was what my thoughts were regarding what I said about "forces would divide by 2 each time it passes through one of those 3 inferior pulleys"

 A: The mechanical advantage of a pulley system is equal to the number of ropes supporting the movable load, which in this case is six. Then the tension in the rope, which is the same all along the rope given massless and frictionless pulleys, is the load divided by the mechanical advantage, in this case $mg/6$.
This can readily be seen from the free body diagram of the block/movable pulley system below where the net force on the system is zero (so it will either remain stationary or move with constant velocity).

yes, I've understood your diagram, but I'd like to know what's wrong
about my thoughts, I've just edited my post and posted the diagram
that made me think like this, I was taught that each time a rope would
pass through a pulley like the ones at the bottom, the tension, in
this case the force, would be divided by 2, as I thought I did right
in the diagram, but why is this not the case?

Now that I see your edit, I understand what you were trying to say. But as @Biophysicist pointed out ideal ropes can only experience tension, not compression. Any attempt to push down or up on a pulley with a rope would make the rope collapse. The direction of the forces that the ropes apply to the pulleys have to be away from the pulleys.
See my second diagram below

Why disconnected? I've never seen that "disconnection" to be used.

My undergraduate mechanics professor used to constantly tell us: DRAW A FREE BODY DIAGRAM!. And yes, he yelled it, because the first instinct of his students (me included) was to try and use "intuition" to solve a problem, and get into trouble. When you "disconnect" a body from everything else you set it "free", thus the term "free body diagram" (FBD). An FBD enables you to identify all the external forces acting on a body, i.e., the forces that determine the net force, if any on a body. When you draw an FBD you replace the external connections with the forces that may be acting at those connections. I have replaced my original figures with those below to further elaborate.

Also why do we add the 6 forces at the top pulleys instead of the 3
forces at the bottom pulleys? It'd be more intuitive to me to add the forces at the bottom.

Keep in mind you need to identify the external forces that act on the body. See FIG 1 for the FBD of the combination of the block and inferior pulleys.
In this case, the block and the inferior pulleys act as a single body moving down together with a constant velocity. The only external forces acting on this combination is the force of gravity acting downward on the block (pulleys considered massless) and the six tension forces acting upward on the ropes connecting the inferior pulleys to the superior pulleys. The three tension forces acting on the ropes connecting the block to the inferior pulleys are internal forces. This can be seen in the FBD.
On the other hand, FIG 2 is a FBD of the block alone. You would draw this if you were interested in the tension in the ropes connecting the block to the inferior pulleys. Now the three forces at the bottom are external forces acting on the block.  Summing the vertical forces we get $T=mg/3$.
Hope this helps.


A: Tension can only pull, it can't push. So the ropes always pull on the pulleys. This means that the some of the vectors in your diagram need to be reversed.
A: The problem is that you're mixing ideas. Pulleys make the total effort easier, but that does not mean that each individual rope gets reduced by a factor 2 respect to the previous one. It is the net effect what you observe.
But, in general, you should not presume how a system works before you solve it. That's why you want to sovle it; because you want to check what actually happens, instead of what you think it happens...
First of all, you should not presume anything different than the most basic ideas. What you really have is

*

*Tension force appears in pairs: the one in the first end of the rope section is the same as the one in the second end of the rope section. This is due to Newton's 3rd law: tension appears in pairs.

So, if you are rigurous, you write something like

As you're sure that each  rope section has two equal and opposite tension forces. But you know nothing else...
Then, we add more information. If a pulley has no mass and no friction, then you can be sure that the tension at both sides is the same. This is due to the torque equations.
If you¡re sure you've got massless pulleys, then you can be sure that
$$T_1=T_2=T_3=T_4=T_5=T6$$
But this is not neccesarily true if pulleys have mass.
Then, you keep calculating. Proceed by drawing the free body diagram.
For the mass itself, you have
$$T_A+T_B+T_C-mg=ma$$
If the weight is static, then $a=0$ and the forces are compensated.
$$T_A+T_B+T_C=mg$$
That's what you can deduce from Newton's laws. You know nothing else from them.
This equation talks about translation. Of course the weight doesn't move if upward forces compensate the downwards weight...
But if you want to know more things from $T_A, T_B,T_C$, you'd need information about rotations. If the body does not rotate, then you'll need $T_A=T_C, and so on.
In summary, never presume what happens. Be rigurous when adding information. State as many unknowns as you need, you'll reduce them as soon as you find new equations
A: Imagine we pulled on the left rope with enough force to make the mass go up with constant speed.
If we pull 60cm down on the left rope with the force $F$.  Only 10cm is 'removed' from each of the six ropes on the right, (so the amount of rope moved through the system is consistent), and the mass is raised by 10cm.
from
$Work  = Force \times distance$
$$ F\times0.6 = mg\times0.1$$
$F=\frac{1}{6}mg$
The same is true for lowering a weight.
Perhaps you thought that the force would halve each time, as that would be true for one pulley, with two ropes, but the $6F$ comes from the reason above.
A: A way to understand this type of problem is to suppose that the weight is being supported by 6 separated strings. It is clear in this case that the force is divided by 6 for each string if they are equally spaced.
Adding massless pulleys and joining all strings to form only one doesn't change the division of the weight.
