Tricky Physics Problem - Picture Included -

[Elthamer]

Homework Statement

The problems asks:
According to the above sketch, find accelartion (a1) of mass1? Note that the cables and pulleys are ideal (massless and frictionless). Everything started at rest until the pulleys lock were realsed. m1, m2, m3 have all different masses.

Homework Equations

a1 = a2 + a3
F = ma

The Attempt at a Solution

Here's what i have set up so far, correct me if i am wrong:
Known: m1, m2, m3, g
Unknown: a1, a2, a3, T1, T2, T3 (T=Tension, a= acceleration, m=mass)
System 1: m1
x: 0
y: T1 + W = m1a1
System 2: m2
x: 0
y: T2 + W = m2a2
System 3: m3
x: 0
y: T3 + W = m3a3
a1 =a2 + a3
I don't know what to do from here, i guess its because my set up is wrong. I was told by a friend that just 2 systems are required, and that system 2 is m2 and m3 combined. Thus Tension 1 would be T1 = T2+3. But i am not sure if that's correct either.
Any help solving or setting up the problem would be greatly appreciated. Thanks in advance.
Thamer H.

 

[Elthamer]

Edit: Fixed the link for the picture. It should be working now. I was wondring why no one could answer the question lol.

 

[DarylMBCP]

I'm quite sure that u just have to take (net force acting on m_{1/SUB]) / m1 to find its acceleration. Since taking that m1 is accelerating downwards instead of upwards, find the gravitaitonal force acting on m1 and minus the force acting on m2 + m3 and you will get the net force.}

 

[Elthamer]

Hmm i am not sure what you are trying to say DarylMBCP, but this is what i came up with after trying to rework this problem this morning:

Known: m1, m2, m3, g
Unknown: a1, a2, a3, T1, T2, T3 (T=Tension, a= acceleration, m=mass)
System 1: m1
x: 0
y: T1 + W = m1a1
System 2: m2
x: 0
y: T2 + W = m2a2
System 3: m3
x: 0
y: T3 + W = m3a3

a1 =a2 + a3

Based on the forces on the y direction, I assume that:
a1 = (T1-m1g) / m1
a2 = (T2-m2g) / m2
a3 = (T3-m3g) / m3

So, based on a1 =a2 + a3 I assume:
(T1-m1g) / m1 = (T2-m2g) / m2 + (T3-m3g) / m3

We also know that a2 = -a3, so:
(T2-m2g) / m2 = - [(T3-m3g) / m3]

With this, I can go ahead and solve for T3 and T2 to find T1.
T2 = -m2 [((T3-m3g) / m3) - g] = -m2[(T3/m3)-g-g)

T3 = -m3 [((T2-m2g) / m2) - g]

Now that i know what T3 equals to, i plugged it in here T2=-m2[(T3/m3)-g-g). To find what T2 equals to. But when i do that, i end up getting:
T2 = -T2 which doesn't help me solve for T3.


I don't know exactly where in my steps i messed up. Any suggestions?


EDIT: Could it be that this problem would require momentum principle and velocity?

 

[Elthamer]

Okay, i think i see what i did wrong. After working on it for the past few hours i realized that the tension (T2) of mass 2 and the tension (T3) of mass 3 and their acceleration is irrelevant to this problem. I realized we do not need it to find a1.

Here's what i came up with as my final solution to this problem:

From system one (system of m1), we know that m1g -T1 = m1a1 by using Newton's second law (F=ma).

I then set:
T1= m2g + m3g
Because Tension1 on mass 1 is equal to the tension on the string that is holding m2 and m3 right?

So, from this formula m1g -T1 = m1a1 we can say that:
a1 = (m1g1 - T1)/m1

Now that we know that T1 equals T1= m2g + m3g, we can plug it in the equation a1 = (m1g1 - T1)/m1. So we get:

a1 = g(m1 - m2 - m3)/ ma

Could someone please verify if this is correct? I would greatly appreciate it.

 

[Elthamer]

bump~

Anyone? Its due tomorrow morning i just want to make sure my answer is correct before i turn it in >.<

 

[rl.bhat]

In the system, one string (S1) connects m1 and the movable pulley. Second string (S2) connects m2 and m3 masses. When the masses move net length of the strings remain constant. But length of their segments change. When m1 goes up through a distance x1, movable pulley goes down by x1. When m3 goes down by a distance x3, the net change in that segment is x3 - x1. When m2 goes up by a distance x2,net decrease in the segment is -x2-x1. So the net change in the length in S2 = x3 - x1 - x2 - x1 = 0
Or x3 - x2 = 2x1.
So 2a1 = a3 - a2.
Now you can proceed.

 

[DarylMBCP]

Sorry there. I was really busy this past week. I think rl.bhat is correct though. Rlly sry again.