Why is the downward force not equal to the weight of the components ?(Physics question)(Image below)?

That's a great question. The short answer is: If gravity is not acting directly on the pulleys (I assume the pulleys are postulated to be massless), then only those things that directly touch the pulley can exert a force on it. In this case (say for Pulley "B"), that means an upward tension (from the string in the middle of "B") and two downward tensions (from the strings connected to m1 and m2).

It may intuitively seem that the the two downward tensions ought to equal the weights of m1 and m2, but that's not true either. In particular, if a mass (suspended by a string) is accelerating downward, then the tension must be LESS than the mass's weight (otherwise it wouldn't accelerate downward!), and likewise, if a suspended mass is accelerating upward, the tension must be GREATER than the weight.

The only time the tension is EQUAL to the weight is when the mass has zero vertical acceleration.

Given these observations, now we can begin to solve the problem.

Let:

m1, m2, m3, m4 = the masses of the blocks;

Ta = tension of string wrapped around "A"

Tb = tension of string wrapped around "B"

Tc = tension of string wrapped around "C"

a1, a2, a3, a4 = accelerations of the blocks;

aB, aC = accelerations of Pulley "B" and Pulley "C"

(We are also assuming the strings are massless and there is no mass or friction in the pulleys. Under those assumptions, the tension in any given string is constant throughout the string's length.)

For consistency, choose the "up" direction to be positive.

First, notice some relationships between the accelerations:

Pulley "B" accelerates at the same rate (but in the opposite direction) as Pulley "C". So:

aB = -aC

m1 and m2 accelerate at the same rate (but in opposite directions) _relative_ to Pulley B. The blocks' accelerations relative to B are "a1-aB" and "a2-aB" respectively. So since they are equal and opposite:

a1 - aB = -(a2 - aB)

And by the same reasoning:

a3 - aC = -(a4 - aC)

Next, write four "F=ma" equations, one for each block. Notice that, on each block, only two forces act: (1) the tension of the string connected to the block; and (2) the block's weight. Therefore, the net upward force on each block looks like: "T-mg". The four equations are therefore:

Tb - (m1)g = (m1)a1

Tb - (m2)g = (m2)a2

Tc - (m3)g = (m3)a3

Tc - (m4)g = (m4)a4

Now we have SEVEN equations, in EIGHT unknowns (aB, aC, a1, a2, a3, a4, Tb, Tc). That's a problem, since you generally can't solve when the number of unknowns exceeds the number of equations. But there's one more thing we can do:

We said the pulleys are massless, but let's pretend for a moment that B and C have (small) masses mB and mC. That means we can write "F=ma" equations for those two pulleys. Each pulley has an upward force consisting of the tension "Ta" of the upward pulling string, and downward forces consisting of the pulley's weight plus two downward pulling strings.

So for B:

Ta - (mB)g - 2Tb = (mB)aB

And for C:

Ta - (mC)g - 2Tc = (mC)aC

Now if we take the limit as mB and mC go to zero, we have:

Ta - 2Tb = 0

and

Ta - 2Tc = 0

Combine those to get:

Tb = Tc

And that is our eighth equation. NOW, use the algebra of simultaneous equations to solve for the eight unknowns. (Or as many of them as the problem is asking for.)