Masses & Pulley: Balancing Forces with mA, mB, mC

[KMjuniormint5]

In the figure below, three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 22.0 kg, mB = 40.0 kg, mC = 30.0 kg.

figure:

Pulley-----------mA
|
|
|
mB
|
mC

Now I apply Fnet = ma so. . .

the massA in the x direction would be T1 (tension) = (mA)(ax) and in the Y direction n-(mA)(g) = (mA)(ay).
the massB in the y direction would be T1-T2 - (mB)(g) = (mB)(ay)
the massC in the y direction would be T2 - (mC)(g) = (mC)(ay)

but in the equations above wouldn't acceleration in the x direction (ax) = 0? and accel. in the y (ay)direction be (9.8 m/s^2 aka gravity (g))

 

[learningphysics]

No ax isn't 0. The acceleration of mA towards the left equals the acceleration of the hanging masses downwards...

using your conventions ax = -ay.

so you have 3 equations with 3 unknowns... T1, T2 and ax.

One "trick" you can use to simplify calculating T1 and ax, is to take mB and mC together as one system. that way you get 2 equations with 2 unknowns, T1 and ax.

But the 3 equations you have seem fine.