Solve Pulley System: Find Acceleration of m3, B, m1, m2 & Tension in A & C

[awelex]

Homework Statement

See picture: http://session.masteringphysics.com/problemAsset/1038667/6/YF-05-85.jpg

All pulleys are weight- and frictionless, as are the ropes.

Find the acceleration of block m3, of pulley B, of block m1 and m2, the tension in String A, and the tension in string C.

Homework Equations

The Attempt at a Solution

To find the acceleration of m3, I thought of m1 and m2 as a single mass, which should turn the problem into a simple Atwood problem. The two equations that I got out of this are:

(m1+m2)g - T2 = (m1+m2)*a
T2 - m3*g = m3*a

Solving for a yields g*(m1 + m2 - m3)/(m1 + m2 + 2m3), which is not even close to the solution.

What am I doing wrong?

 

[zorro]

Your logic won't lead you to the answer. You cannot think 2 masses as a single mass.
Hint: Use Constrained relations

 

[awelex]

Hi Abdul,
thanks for your answer. I know that it is somehow possible to express one tension in terms of the other (and I remember reading somewhere that one is twice the other), but I don't understand at all how to do that.
If pulley B was in equilibrium, then it would make sense: the tension in C must equal twice the tension at point A. But since the system is not in equilibrium, my reasoning is wrong, isn't it?

 

[zorro]

Whether the system is in equilibrium or not, from Newton's third law T1+T2 = T3,
where T1 and T2 are the tensions in the string connecting m1 and m2 respectively and T3 is the tension at C.
Since it is mentioned in the problem that pulleys and ropes are massless, T2=T1=T and as you said, the tension in C must equal twice the tension at point A (T3=2T).

If you are smart, you can use simple logic to get the relation between accelerations of different masses (instead of working on time taking constrained relationships).

Assuming you are ,
Consider this- if m1 moves down with a1= 1m/s^2
and m2 moves up with a2= 1m/s^2, accleration of m3, a3=0

if m1 moves down with a1=1m/s^2
and m2 moves down with a2=1m/s^2, a3=1m/s^2 UPWARDS

It easily follows that a1 + a2=2a3

Now use Newton's second law for each mass, get 3 equations, solve them with the equation above, you will get your answer

 

[awelex]

Hi Abdul,

thanks again. It's all making sense now, except that I still can't figure out why T3 = 2T.

Again, in my earlier example I assumed that pulley B was in equilibrium. In that case, the net force must be zero, so

Fnet = 2T - T3 = 0

In that case, it is evident that T3 = 2T. But the system is obviously not in equilibrium, so instead we have

Fnet = 2T - T3 = aB*mB

How can you deduce from this that T3 = 2T? Is it because mB = 0?

 

[zorro]

How can you deduce from this that T3 = 2T? Is it because mB = 0?

Yes.
Consider the motion of the pulley B,
The forces on this light pulley are
1) T3 upwards by the upper string
b) 2T downwards by the lower string

As the mass of the pulley is negligible
2T-T3=0