Force and tension

1. Sep 27, 2007

aligass2004

1. The problem statement, all variables and given/known data
The figure below shows two 1.8kg block connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 350g. The entire assembly is accelerated upward at 4.4m/s^2 by force F. a.) What is F? b.) What is the tension at the top end of rope 1? c.) What is the tension at the bottom end of rope 1? d.) What is the tension at the top end of rope 2.

http://i241.photobucket.com/albums/ff4/alg5045/p8-26.gif

2. Relevant equations

F=ma

3. The attempt at a solution

I (again) confused as to what goes into the free body diagrams for each block.

2. Sep 28, 2007

learningphysics

As with your other problem, the trick is to first examine the system as a whole... that will let you get F.

Examining block A alone will give you the tension at the top of rope 1... examining block B along with rope 2 as a system will give you the tension at the bottom of rope 1.

examining block A, rope 1 and block B as a system (leaving out rope 2)... will give you the tension at the top of rope 2.

There are other choices for freebody diagrams... this is just one way.

3. Sep 29, 2007

aligass2004

I still don't understand examining the system as a whole.

4. Sep 29, 2007

learningphysics

Draw a circle around the whole system (draw a circle around the 2 blocks and the two ropes)... treat it as one body... what are the external forces acting on this body? in other words... what are the forces acting on the inside of the circle by something outside (the tension forces won't count because it is inside the circle... it is exerted by one part of the body on another... it isn't an external force)

5. Sep 29, 2007

aligass2004

The forces acting on the whole system is just the gravitational force right?

6. Sep 29, 2007

learningphysics

And F.

7. Sep 29, 2007

aligass2004

So F=ma. Is the mass all of the masses added together times 4.4 m/s^2?

8. Sep 29, 2007

learningphysics

yes, but that's not the same "F" as in the diagram...

$$\Sigma{F} = ma$$

substitute into this equation...

9. Sep 29, 2007

aligass2004

I understand the concept. I'm just not sure how to put it on paper. F - (sum of the masses) a = ma?

10. Sep 29, 2007

learningphysics

no. (sum of masses)a is wrong for the left side. Like you said before, the external forces are gravity, and F... the external forces go in the left side of the equation...

11. Sep 29, 2007

aligass2004

F - g = (sum of the masses)a

12. Sep 29, 2007

learningphysics

yes... F - (sum of masses)g = (sum of masses)a.

13. Sep 29, 2007

aligass2004

I got 61.103N. Now how do I start part b?

14. Sep 29, 2007

learningphysics

look at post #2 in the thread.

15. Sep 29, 2007

aligass2004

I got part b. I used T - (sum of the mass for B and the two ropes)g = (sum of the same masses) a. The answer was 35.525N

16. Sep 29, 2007

aligass2004

I tried doing the same thing for part c, but it didn't work.

17. Sep 29, 2007

learningphysics

yes. looks right.

18. Sep 29, 2007

learningphysics

what exactly did you do for part c?

19. Sep 29, 2007

aligass2004

T - (sum of the mass for B and the bottom rope)g = (sum of the same masses)a.

20. Sep 29, 2007