How Do Newton's Laws Apply to a Frictionless Pulley System?

[Swigs]

The problem was pretty simple i thought, but according to the book i get the wrong answer. I think i might be approching the problem wrong.

"In the drawing, the weight of the block on the table(block1) is 422N and that of the hanging block(block2) is 185N. Ignoring all frictional effects and assuming the pulley to be massless, find (a) the acceleration of the two blocks and (b) the tension in the cord."

o-----------Block1
I
I
I
Block2

(excuse my poor excuse for trying to draw with letters)

What I did was since we know that block2 is 185N, I thought that would be the only force acting on Block1, science it is frictionless. And they would have the same accelration since they are connected so i set up an equation like this Fx=185=m1a = a=4.3m/s^2. But this is wrong, and I know i set it up wrong.......

 

[HallsofIvy]

The force on the system- that is both blocks 1 and 2- is the gravitational pull which is 185 N. The hanging block cannot move without pulling the other block after it- that's why I call it a "system". F= ma with F= 185 N and m= the mass of both blocks: (185+422)/g kg. (notice the "/g" to change weight to mass!).
F= ma becomes 185= (607/g)a so a= 0.305g.

Now look at the two blocks separately. Block 2 has mass 185/g kg. Since it has accelerating at 0.305g, the total force must be
(185/g)(0.305g)= 56.4 N. That's a lot less than it's weight! What happened? Block 2 is pulling back on it with force 185- 56.4= 126.6N.
Block 1 has mass 422/g kg. Since it is accelerating at 0.305g, the force on it must be (422/g)(0.305g)= 128.6 N. That should be no surprise: every action has an equal and opposite reaction. Block 1 is pulling back on block 2 and block 2 is pulling forward on block 1 with exactly the same force. And that 128.6 N is, of course, the tension in the rope.

 

[Swigs]

Thankyou very much, HallsofIvy. that helped a lot.