• Support PF! Buy your school textbooks, materials and every day products Here!

Atwoods - applied force - reduced acceleration

  • Thread starter dandy9
  • Start date
  • #1
28
0

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 in to one of the equations, I got f = 9.7N.

If someone could help me out I would really appreciate it! Thanks!
 

Answers and Replies

  • #2
1,137
0
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
28
0
Still didn't work, but thanks anyway.
 
  • #4
1,137
0
well i solved it but cant post the answer. so...

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

only T acts on it.
 
  • #5
28
0
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
28
0
Ahh.
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
1,137
0
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)?
 

Related Threads for: Atwoods - applied force - reduced acceleration

Replies
2
Views
4K
Replies
3
Views
887
  • Last Post
2
Replies
26
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
12
Views
40K
  • Last Post
Replies
9
Views
6K
Top