Dynamics Problem: Calculating Tension in a Pulley System with Two Masses

[deliveryman]

Ok, it's 3am and I can't seem to get the answer that concurs with the answer in the back of the book, I went over my work 10 times and I don't see any mistakes in my work!

10. Two masses of 6 kg and 9 kg respectively are fastened to the pull ends of a cord passing over a light, frictionless pulley supported by a hook. Find the pull on the hook while the masses are in motion.

Here's my work:

Fnet = (M1 + M2)(A)
Fnet = (-6 kg + 9 kg)(10 m/s^2) The 6 is negative because it's going upward (opposite to the 9, which is going downward)
Fnet = 3(10m/s/s)
Fnet = 30 N

Now to find the acceleration of the system:

As = Fnet / M1 + M2
As = 30 / 6 + 9
As = 30 / 15
As = 2m/s^2

We have the acceleration, now we just find the Tension between the first or second mass and the hook.

Fnet = M1(As)
Fg1 + FT = M1(As)
Ft = M1(As) - Fg1
Ft = 6(2m/s^2) - (-60)
Ft = 12 + 60
Ft = 72N

It says the Answer should be 144N. Maybe I'm missing something because it's 3am, but please someone help? I'm dead tired and I have no idea what went wrong.

 

[enigma]

Tension between the first mass and the hook is 72N. What about the second mass and the hook?

The hook is holding both of them up, after all.

 

[deliveryman]

Lol! Wow how could've I missed that, the hook is holding them both up, i forgot about the other tension. Silly me, so I take the tension of one end and mutiply it by 2, since both the tensions are equal, just in opposite directions, which would give me 144N. Thanks enigma, can't believe I missed that.