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Tension Between Objects in 1 Dimension

  • Thread starter Versaiteis
  • Start date
  • #1

Homework Statement



Here is the problem verbatim (values have been slightly changed, also assume a frictionless environment):

"A 4.3 kg block A and 6.0 kg block are connected by a string of negligable mass. Force FA = (15 N) acts on block A; force FB = (24 N) acts on block B. What is the tension in the string? "

The diagram can be found here:
http://img713.imageshack.us/i/w0085n.jpg/" [Broken]

Homework Equations



F=ma
[tex]\vec{F}[/tex]A = 15 N [tex]\hat{i}[/tex]
[tex]\vec{F}[/tex]B = 24 N [tex]\hat{i}[/tex]
Fn - Fg = 0 (forces on y-axis are balanced)

The Attempt at a Solution



I'm not exactly looking for an answer, I just need a place to start. What I'm having trouble with is drawing the free-body diagram, and if I am doing that correctly then I need help interpreting the values that I have.

Heres pretty much what I have been able to come up with as a free-body diagram:
http://img153.imageshack.us/img153/3972/freebodydiagram.jpg" [Broken]

I have a feeling that my diagram is missing some vital forces (I believe this problem requires me to apply Newtons Third Law, but I'm unsure how to display that graphically)
 
Last edited by a moderator:

Answers and Replies

  • #2
1,137
0
find acc. of blocks when no string is attached
if aA > aB then T is ?

if a B> aA use fact that T is same for both blocks and also acc.
 
  • #3
alright so...

acc. of A = 3.4884 m/s^2

acc. of B = 4 m/s^2

but I'm not sure I understand exactly what that means. That the 0.512 m/s^2 difference is somehow related to T? I mean F=ma, but if I were to apply this formula what would be the mass? A? B? the sum of the system?
 
  • #4
1,137
0
now you got that aB > aA
and they are connected with string ... so will B move faster than A? ...NO!!!

Now consider their net acc. as "a"

use newton's laws to find a --- using forces acting on abjects and tension
 
  • #5
Alright so to find the net acceleration "a" I set it up like this:
("F" = Net Force, "m" = Net Mass)

F=ma

39N = (10.3kg)*a

39N/10.3kg=a

a=3.786m/s

But this would be the acceleration of the entire system would it not? Doesn't there need to me forces in the opposite direction (-i) for there to be any Tension?
 
  • #6
1,137
0
Alright so to find the net acceleration "a" I set it up like this:
("F" = Net Force, "m" = Net Mass)

F=ma

39N = (10.3kg)*a

39N/10.3kg=a

a=3.786m/s

But this would be the acceleration of the entire system would it not? Doesn't there need to me forces in the opposite direction (-i) for there to be any Tension?
As the blocks are connected by string and individual acc. of B>A ... do you think they would have same acc. or different???

In this case you have taken masses+string as system. tension becomes an internel force in this case and therefore do not have any effect on acceleration
your answer is perfectly correct
 
  • #7
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998

Homework Statement



Here is the problem verbatim (values have been slightly changed, also assume a frictionless environment):

"A 4.3 kg block A and 6.0 kg block are connected by a string of negligable mass. Force FA = (15 N) acts on block A; force FB = (24 N) acts on block B. What is the tension in the string? "

The diagram can be found here:
http://img713.imageshack.us/i/w0085n.jpg/" [Broken]

Homework Equations



F=ma
[tex]\vec{F}[/tex]A = 15 N [tex]\hat{i}[/tex]
[tex]\vec{F}[/tex]B = 24 N [tex]\hat{i}[/tex]
Fn - Fg = 0 (forces on y-axis are balanced)

The Attempt at a Solution



I'm not exactly looking for an answer, I just need a place to start. What I'm having trouble with is drawing the free-body diagram, and if I am doing that correctly then I need help interpreting the values that I have.

Here's pretty much what I have been able to come up with as a free-body diagram:
"[PLAIN]http://img153.imageshack.us/img153/3972/freebodydiagram.jpg [Broken] [Broken]

I have a feeling that my diagram is missing some vital forces (I believe this problem requires me to apply Newtons Third Law, but I'm unsure how to display that graphically)
Hi Versaiteis.
[PLAIN]http://img153.imageshack.us/img153/3972/freebodydiagram.jpg [Broken] [Broken]

I re-posted your free body diagram(s). Actually, you have two of them in your figure, one for each block. You're missing the force exerted on each block by the tension, T, in the string.

As long as [tex]\frac{\left|\vec{F}_A\right|}{m_A}<\frac{\left|\vec{F}\right|}{m_B}\,,[/tex] you can do the problem the way you and cupid have done it.

To find the tension in the string using that method, realize that the net force required to accelerate block A at a=3.786 m/s2 is greater than the 15 N supplied by [tex]\vec{F}_A[/tex]. The extra force needed for that acceleration is the tension in the string. Similarly, the tension is opposite [tex]\vec{F}_B[/tex] and is the the amount needed to accelerate block B with acceleration of a=3.786 m/s2.
 
Last edited by a moderator:
  • #8
Oh I think I get it so

Acc. of A is 3.4883 m/s^2
Acc. of B is 4 m/s^2

F(net) = FA + FB = 24N + 15N = 39N

m(net)=mA + mB = 6kg + 4.3kg = 10.3 kg

a(net)= F(net) / m(net) = 39 N / 10.3 kg = 3.786 m/s^2

Now I need the force necessary to move A at 3.786 m/s^2

FA2 = mA * a(net) = 4.3 kg * 3.786 m/s^2 = 16.2798 N

Now I know that FA=15 N is contributed by A so the tension T must be the difference

T = FA2 - FA = 16.2798 N - 15 N = 1.2798 N or 1.28 N

And after repeating this problem with my actual problem (mA = 3.2, FA = 10) and entering into WebAssign it says the answers correct, so the methods right.

Thank you very much Cupid and Sammy, you guys have been a big help, I'll be sure to come post on this site again if I have any more questions.
 

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