# atwood machine

• Feb 5th 2009, 01:17 PM
checkmarks
atwood machine
A 3.0 kg counterweight is connected to a 4.5 kg window that freely slides vertically in its frame. How much force must you exert to start the window opening with an acceleration of 0.25 m/s^2?

It seemed like an easy problem to me but I've gone about it so many ways and still can't get it right! :( Please help.
• Feb 5th 2009, 09:21 PM
ixo
force = mass x acceleration
Plug in (window mass - counterweight mass) then multiply by the given acceleration in the problem and you will get your force. Force is (kg x m/s^2)
• Feb 7th 2009, 09:45 AM
checkmarks
no that can't be right.
the correct answer is 17 N.
• Feb 7th 2009, 10:22 AM
ixo
Sorry I couldn't solve it for you but this looks like a physics question? Might try Physics help forum?
Since a Newton is http://upload.wikimedia.org/math/c/d...f481c8f9d2.png if you look at the equation you have 17(kg*m)/(s^2) = xkg * .25 m/s^2 which means there is a mass of 68 kg? Idk.. sorry have you double checked the given numbers to make sure you wrote them all down right, as well as the answer? The only other thing i can think of.
• Feb 7th 2009, 10:49 AM
skeeter
Quote:

Originally Posted by checkmarks
A 3.0 kg counterweight is connected to a 4.5 kg window that freely slides vertically in its frame. How much force must you exert to start the window opening with an acceleration of 0.25 m/s^2?

Quote:

the correct answer is 17 N.
I do not believe 17N is correct.

let F = force required

T = tension in the cable connecting the window to the counterweight

net force acting on the 3kg counterweight ...

\$\displaystyle 3g - T = 3a\$

net force acting on the 4kg window ...

\$\displaystyle F + T - 4g = 4a\$

combining the equations ...

\$\displaystyle F - g = 7a\$

\$\displaystyle F = 7a + g\$

\$\displaystyle F = 7(.25) + 9.8 \approx 11.6 \, N\$