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!

 

[rwisz]

I would first commend the student for their attempt at solving the problem and for using the correct equations and units. However, I would also point out that there are a few errors in their calculations.

Firstly, in equation 1, the student has used the weight of the suspended mass (m3) as the force acting on the system. This is incorrect, as the weight of m3 is already included in the sum of forces acting on m1 and m2. The correct equation should be: 4a = T2 - Ffr.

Secondly, in equation 2, the student has used the total mass (m1 + m2) instead of just m1 in the calculation of the frictional force (Ffr). The correct equation should be: 9a = T2 - (0.35 * 6 * 9.81).

Finally, the student has made a calculation error in solving for T1. The correct equation should be: 6(1.6597) = T1 - 20.601, which gives T1 = 30.6692 N.

With these corrections, the final answers should be: a = 0.64 m/s^2, T1 = 30.67 N, T2 = 45.84 N.

Overall, the student has shown a good understanding of the concepts of force and motion, and with some attention to detail, they can improve their problem-solving skills.