Force and motion problem with two tensions and 3 masses and an ideal pulley

[runphysicsrun]

Homework Statement

Two clay pots (m1 = 6 kg, m2 = 3 kg) joined together by a light string rest on a table. The frictional coefficient between the pots and the table is 0.35. The pots are also joined to a suspended mass (m3 = 4kg) by a string of negligible mass passed over an ideal pulley as shown. Calculate the tensions in strings and the acceleration of the system when the suspended mass is released.

Homework Equations

Ffr=uN
SumF=ma
w=9.81*m

The Attempt at a Solution

sumFm3=T2 + Wm3, m*a=T2- (4kg*9.81), ------> 4a=T2-39.2 (eq1)
sumFm1+m2 y= FNy+FGy, N-w=m*a , N-w=0, w=N, 9.81*9kg= N, 88.29=N
Ffriction=uN, Ffr=0.35 * 88.29N, Ffr=30.9015
sumFm1+m2 x= T2-Ffr, 9*a=T2-30.9015 (eq2.)

Subtracting eq1 from eq 2 gives me 5a=8.2985, a =1.6597
plugging a back into equation 1 gives me 45.839 for T2
sumFm2=T1-Ffr, m*a= T1- N, N=(.35 * 6 * 9.81), 6(1.6597)=T1-20.601, T1=30.6692 N

The answers are: .64m/sˆ2 for a and T1=24N and T2=37N.

 

[Doc Al]

sumFm3=T2 + Wm3, m*a=T2- (4kg*9.81), ------> 4a=T2-39.2 (eq1)
sumFm1+m2 y= FNy+FGy, N-w=m*a , N-w=0, w=N, 9.81*9kg= N, 88.29=N
Ffriction=uN, Ffr=0.35 * 88.29N, Ffr=30.9015
sumFm1+m2 x= T2-Ffr, 9*a=T2-30.9015 (eq2.)

You have a sign problem in your equations. If the acceleration of m1 & m2 is 'a' to the right, then the acceleration of m3 is 'a' downward. Give those accelerations the proper signs.

 

[runphysicsrun]

[PLAIN]http://dl.dropbox.com/u/14756133/physics%20two%20clay%20pots.pngSorry. I accidentally typed plus. If you look at the following equation, though, I have it as a minus. I'm subtracting the force due to gravity from the force due to tension in the downward motion of mass 3. and my positive direction is right for m1 and m2. There's an image of the problem!

 

[runphysicsrun]

Wait. I just understood what you were talking about. So it should read -4a = T2-39.2! Thanks!