Atwoods - applied force - reduced acceleration

In summary, two masses, 6.45 kg and 11.90 kg, are connected by a string with a constant tension T. An additional force f is applied to the bottom mass, causing the acceleration of the system to decrease by 59%. Using the equations Fnet = ma and Fnet = T - m1g - f = m1(.59a), the resulting tension is found to be 83.4N and the applied force f is 9.7N. However, when considering the second mass, it is determined that the tension T should not include the applied force f. This results in a different value for the tension, 61.9N, and it is uncertain if this is the correct
  • #1

Homework Statement

Suppose that in the same Atwood setup another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1. If m1 = 6.45 kg and m2 = 11.90 kg, what value of f will reduce the acceleration of the system by 59%?

Homework Equations

Fnet = ma

The Attempt at a Solution

So I thought about this problem for quite some time, and then all of a sudden it was BAM I got it! But sadly, my answer was not correct and now I'm here quite disappointed that all my hard work was wrong. Anyway, you probably didn't care to read all that and now I'll just tell you what I did.

I first drew my free body diagrams like a good little Honors Physics student. Then I set up two Fnet equations that went something like this:

T = tension
m = mass (subscript denoting mass1 or mass2)
g = gravitational force (9.81)
f = applied force

Fnet = T - m1g - f = m1(.59a)
which reduced to
T - f = 93.1N

Fnet = T + f - m2g = m2(.59a)
which reduced to
T + f = 73.7N

Then I added the two reduced equations together to eliminate the f and got T = 83.4N.
Plugging that back into one of the equations, I got f = 9.7N.

If someone could help me out I would really appreciate it! Thanks!
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  • #2
dandy9 said:
If m1 = 6.45 kg and m2 = 11.90 kg, what value of f will reduce the acceleration of the system by 59%?

shouldn't you have taken acc. as 0.41a ??
  • #3
Still didn't work, but thanks anyway.
  • #4
well i solved it but can't post the answer. so...

and OH! for 2nd eqn ... for m2 block ... why are you using f?

only T acts on it.
  • #5
Does T include f?
The way I thought about it was that T is the original tension between the two masses, and then there is an additional force, f, that has to be taken into account. So I'm adding T and f for the "upward force" and subtracting m2g for the "downward force."

Hmm. I guess not then? Let me try your way.
Thanks again for the reply.

Alright, I got 61.9N... Let's hope for the best!
  • #6
Just kidding. I didn't realize that I got locked out of the question.
Perhaps you could tell me if my answer is reasonable just so I know for the future if I did it right in the end?
  • #7
If 61.9 is the force then I'm afraid that its wrong ...
what's your value of a (not a = .49a but a in initial case)?

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