# Acceleration of system

1. Dec 22, 2011

### canicon25

1. The problem statement, all variables and given/known data

determine the acceleration of the system. there is no friction on the surface

2. Relevant equations
Fnet=ma

3. The attempt at a solution

Left block
Ft=9.81+a1

Middle block

Ft=Fs-7a2
Right block
Fs=5a3+49.1

Not sure how to proceed with this one. What is equal to what?

2. Dec 22, 2011

### technician

The first thing to do is draw the forces acxting on each object on the diagram.
The 1kg mass has weight acting down and a tension (T1) acting up. The 5kg mass has a weight acting down and a tension (T2) acting up.
The 7kg mass has the 2 tensions acting on it.
You should be in a better position to write down the resultant force equations with this diagram

3. Dec 22, 2011

### canicon25

i did make diagrams, i just didn't post them.

Ft = T1
Fs = T2

the equations i wrote are from the diagrams

left
T1-mg=ma
T1-9.81=a
T1=9.81+a
right
T2-mg=ma
T2-49.1=5a
T2=49.1+5a

The middle block would be

-T1-(-T2)=ma
-T1+T2=ma
-T1+T2=7a

i am assuming the acceleration is the same for all? can it be solved for acceleration this way? please help.

4. Dec 23, 2011

### Delphi51

Yes, your method is correct. The individual equations look okay, though it can be tricky to get all the signs right. Check your work by thinking of the whole system as one 13 kg mass with forces 5g and -1g so net of 4g to the right. Use F = ma on that.

5. Dec 24, 2011

### cupid.callin

Would it be T2 - mg ?