- #1

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.

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.

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