Finding acceleration of a three-object pulley system

[kathyt.25]

Homework Statement

"Three blocks are connected on a table as shown in the figure below.
http://i4.photobucket.com/albums/y111/kathy_felldown/sb-pic0550.png

The table is rough and has a coefficient of kinetic friction of 0.360. The three masses are m1 = 3.74 kg, m2 =1.39 kg, and m3 = 1.86 kg, and the pulleys are frictionless. Determine the magnitude of the acceleration of each block."

I think I've figured out how to do it, but I keep on getting the wrong answer. I denoted left to be the POSITIVE x-direction, and down to be the POSITIVE y-direction, since that's the direction of movement. Can someone check my equations to see if I have them right?

Homework Equations

Fnet=ma
f(k)=u*n (where u=coefficient of friction)

The Attempt at a Solution

For m1, the 3.74kg mass:
Fnet(y) = m1*g - T1 = ma
Therefore, T1 = m1*g - m1*a --> EQUATION #1

For m2, the 1.39kg mas on the table:
Fnet(y) = m2*g - n = 0
Therefore, n = m2*g

Fnet(x) = T1 - T3 - f = ma
Therefore, T1 - T3 - u*n = ma --> EQUATION #2

For m3, the 1.86kg mass:
Fnet(y) = -T3 +m3*g = ma
Therefore, T3 = m3*g - m3*a --> EQUATION #3


I then substituted equations #1 and #3, into equations #2, and isolated for acceleration "a".

 

[Doc Al]

For m3, the 1.86kg mass:
Fnet(y) = -T3 +m3*g = ma
Therefore, T3 = m3*g - m3*a --> EQUATION #3

You've made a sign error. The acceleration of m3 is upward and must have the appropriate sign.

 

[davieddy]

If m1 has acceleration a downwards, the acceleration of m3
is a upwards (or -a downwards if you prefer)