What is the acceleration of block B?

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The discussion revolves around calculating the acceleration of block B in a system of three blocks connected by cords over frictionless pulleys. The masses are given as mA=6kg, mB=6kg, and mC=18kg, with gravity set at g. The user initially attempts to separate the problem into two parts but realizes that since all blocks are interconnected, they share the same acceleration. The confusion arises regarding the signs in the equations, particularly when considering the forces acting on each block. Ultimately, the acceleration of block A, and consequently block B, is determined to be +9.8 m/s², as they all experience the same acceleration due to the heavier block C.
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[SOLVED] Friction and blocks

Homework Statement


Three blocks are attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA=6kg, mB=6kg, and mC=18kg. The accleration due to gravity is g.


Homework Equations


F=ma


The Attempt at a Solution



I thought that it would be wise to treat these as two separate problems. I thought I would solve one side by the following. Block B T=ma Block C F(g)-T=ma and thus mg-T=ma
I thought that to figure out the acceleration of block B (since we know C's is -9.8m/s^2) that i would use the equation a(B)=(m(C)g - (m(C)g))/m(B) But when I do that i would get 0/6. That can't be right because the block C would move because it is heavier than block A
 

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A acts in the opposite direction to C so you could think of it as a negative weight on the same side as C. That will simplify it into a single falling block pulling on B.
 
Oops, i forgot to mention, i have to find the acceleration of block a. So would i still be on the right track? I feel like my signs are messed up in that equation i had
 
Since they are all on the same wire the accelaration of all the blocks is the same.
 
so the accleration of block A is +9.8 m/s^2?
 
or are you saying that once i figure out the acceleration of one block, then that would be the same for all of them
 
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