Solving Pulley Block System Acceleration & Tension

[Elena14]

1. Homework Statement
FInd acceleration and tension. Take g=10m/s^2

2. The attempt at a solution
By drawing the free body diagrams of every block and simultaneously solving all the equations, I got the answer acceleration= 30/7 m/s^2, which is also correct.
But when I try to treat all the pulleys and blocks as one system to get the acceleration, I get a different answer.
f=ma

⇒ 40g+10g= 70 * a (since weight of 40 kg block and 10 kg block are the only forces acting downwards) ⇒ a=500/70 m/s^2
Somebody told me that I should have rather subtracted 40g and 30 g, so that would be 40g-10g=70a ⇒a=30/7 which happens to be the right answer.
But I don't understand why will we subtract the force vectors if they are all acting in the same direction ;ie, downwards. Where am I wrong with the second approach?

 

[BvU]

Imagine one single pulley with two weights, 10 kg and 10 kg. Will the acceleration be (10+10) g / 20 or will it be (10-10) g / 20 ?

 

[Elena14]

20 g according to my logic. Since, weight( a force vector) of the bodies is in the downward direction, we will add the vectors together to get the total weight and hence the acceleration=mg=(10+10)g.

 

[Elena14]

Imagine one single pulley with two weights, 10 kg and 10 kg. Will the acceleration be (10+10) g / 20 or will it be (10-10) g / 20 ?

Alright I get it now, when gravity acts on those blocks ,both will have acceleration in opposite directions; the 40 kg block will move downward being heavier and the lighter block will be forced to move up and hence we subtract their weight.(mg)
Thank you so much.

 

[BvU]

(Especially for later exercises) it is important to clearly identify the physical role of the rope: you should have seen that already when "drawing the free body diagrams of every block": the tensions at the two ends of a straight rope are equal and opposite forces. When the chord is run over a pulley, in the no friction case the magnitudes remain the same but the directions change.
But for the straight sections - again - the tensions at each of the two ends are equal and opposite.

As a simple trick you can (imagine to) cut each straight section halfway and apply two equal and opposite forces to keep the loose ends in place.