Finding tension between connected 4 kg and 5 kg blocks pulled by 30 N

[SS2006]

there's two blocks

a 4 KG one and a 5 KG one right next to each other, connected by a c ord, the 5 KG is on the right and its conected to another cord pulling 30 N to the right.

so

Box1----Box2----30 N
get it
so theyre asking for the tension in the cord connecting them
it slipped my mind how to do this :(
i got acceleration of the system to be 3.3 m/s tho

 

[Pyrrhus]

Did you draw the FBDs?

 

[SS2006]

yes i did
but I am confused really
i know its a simple questoin
test is 2morrow

 

[Pyrrhus]

Ok then let me see your equations for each body

 

[SS2006]

well for acceleration

i did

9a = 30
and got 3.33 m/s, which was right
now they ask for tension of cord connecting them, i just hit a blank spot here for some reason.
lets say i make a fbd for the 5 kg box, so it got 30 N going one way, and the other way? is it 4 x10? no right. and there's no friction btw
i can't think, relaly, just help me out, I am not trying to get u guyst od o my homework or antyhign honeslty, this is just a review Q for a test tmorrow!

 

[hotvette]

Explain how you got a. Also, consider the following:

1. What force is acting on the left block?
2. What can you say about the relationship between the acceleration of block 1 and the acceleration of block 2
3. What equations of motion are you using?

 

[Pyrrhus]

basically you considered your system the whole 2 boxes, that's ok, but then the tension force will be an internal force instead of an external force, analyze for each body in order to find the tension.

 

[SS2006]

k got the asnwer
sorry for the stupidness
i had some brain crmap or something :D

 

[Alpha2005]

Further thought to the above question by SS2006, can I know how come everytime we draw a FBD, the rope has force in both opposite direction

box 1 -- box 2---
for the -- between the box1 and 2, the direction of force is box1--<-->--box2--30N
*the arrow shown is for the tension of the rope only
hope you get what I mean:)