Force and Tension Homework: F, Rope 1 & 2 Tension

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The discussion revolves around calculating force and tension in a system of blocks connected by ropes, with specific mass values and an upward acceleration. Participants clarify the use of free body diagrams to analyze the system, emphasizing the importance of considering external forces like gravity and the applied force F. The correct approach involves summing the masses and applying Newton's second law, leading to the calculation of force F as 61.103N. Tension values are derived for different points in the system, with results of 35.525N for the top of rope 1, 30.552N for the bottom of rope 1, and 4.974N for the top of rope 2. The discussion highlights the iterative nature of problem-solving in physics, with participants refining their calculations and understanding.
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Homework Statement


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


Homework Equations



F=ma

The Attempt at a Solution



I (again) confused as to what goes into the free body diagrams for each block.
 
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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.
 
I still don't understand examining the system as a whole.
 
aligass2004 said:
I still don't understand examining the system as a whole.

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)
 
The forces acting on the whole system is just the gravitational force right?
 
aligass2004 said:
The forces acting on the whole system is just the gravitational force right?

And F.
 
So F=ma. Is the mass all of the masses added together times 4.4 m/s^2?
 
aligass2004 said:
So F=ma. Is the mass all of the masses added together times 4.4 m/s^2?

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

\Sigma{F} = ma

substitute into this equation...
 
I understand the concept. I'm just not sure how to put it on paper. F - (sum of the masses) a = ma?
 
  • #10
aligass2004 said:
I understand the concept. I'm just not sure how to put it on paper. F - (sum of the masses) a = ma?

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
F - g = (sum of the masses)a
 
  • #12
aligass2004 said:
F - g = (sum of the masses)a

yes... F - (sum of masses)g = (sum of masses)a.
 
  • #13
I got 61.103N. Now how do I start part b?
 
  • #14
aligass2004 said:
I got 61.103N. Now how do I start part b?

look at post #2 in the thread.
 
  • #15
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
I tried doing the same thing for part c, but it didn't work.
 
  • #17
aligass2004 said:
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

yes. looks right.
 
  • #18
aligass2004 said:
I tried doing the same thing for part c, but it didn't work.

what exactly did you do for part c?
 
  • #19
T - (sum of the mass for B and the bottom rope)g = (sum of the same masses)a.
 
  • #20
aligass2004 said:
T - (sum of the mass for B and the bottom rope)g = (sum of the same masses)a.

what answer did you get?
 
  • #21
I just redid the math, and what I did was right. Somehow I put the entirely wrong answer. I got 30.552N. I did the same thing for part D, and I got 4.974N.
 
  • #22
aligass2004 said:
I just redid the math, and what I did was right. Somehow I put the entirely wrong answer. I got 30.552N. I did the same thing for part D, and I got 4.974N.

cool. good job.
 

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