Hard physics problem!

I just have no idea how to solve this!


Link of the picture : http://tinypic.com/r/16k2hvt/7


The suspended 2.4 kg mass on the right is
moving up, the 1.4 kg mass slides down the
ramp, and the suspended 7.5 kg mass on the
left is moving down. The coefficient of friction
between the block and the ramp is 0.11.
The acceleration of gravity is 9.8 m/s2 . The
pulleys are massless and frictionless.

What is the acceleration of the three block
system?
Answer in units of m/s2.

Call the tension in the left rope T1, the tension in the right rope T2, and acceleration a.

The masses will move in a direction such that the 7.5kg mass moves down. Treat motion in this sense as the +ve direction..

For the 7.5kg mass, use F=ma, where F = resultant force = mg-T1:
(7.5x9.8) - T1= 7.5a (equation 1)

For the 2.4kg mass, use F=ma, where F = resultant force = T2-mg:
T2-(2.4x9.8) =2.4a (equation 2)

The 1.4kg mass has 4 forces in the direction of motion:
T1 down
T2 up
Component of weight down =mgsin(theta) = 1.4x9.8xsin(24degrees)
Friction up = (mu) x normal reaction = mu x mgcos(theta) = 0.11x1.4x9.8xcos(24degrees)
Use F=ma, where resultant force F = T1 - T2 + 1.4x9.8xsin(24degrees) - 0.11x 1.4x9.8xcos(24degrees)
T1 - T2 + 1.4x9.8xsin(24degrees) - 0.11x 1.4x9.8xcos(24degrees) = 1.4a (equation 3)

The rest is messy algebra and arithmetic. Three simultaneous equations and 3 unknowns (T1, T2 and a);
solve for 'a'.
Note. Luigi (previous answer) is on th eright track but has forgotten to allow for the component of the 1.4kg mass parallel to the the slope (mgsin(theta))

Let m1 = 7.5 kg; m2 = 1.4 kg; m3 = 2.4 kg.
Newton's 2nd law:
m1 g - T1 = m1 a
T1 - T2 - μ m2 g cos24° = m2 a
T2 - m3 g = m3 a
Sum
g*(m1 - μ m2 cos24° - m3) = (m1+m2+m3) a
Solving
a = g*(m1 - μ m2 cos24° - m3)/(m1+m2+m3) = 4.3 m/s²