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Dynamics problem.

  • Thread starter deliveryman
  • Start date

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.
 
Last edited by a moderator:

enigma

Staff Emeritus
Science Advisor
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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.
 

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