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Question about net force and acceleration

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data

    Picture related

    1. There is a system of two blocks tied to each other. There is a force F on the small block causing the system to move at a certain acceleration.

    Find the acceleration.

    2. (This is the one I'm not sure about) Instead of force F, we add a pulley with a weight that's equal to force F. Will the acceleration become smaller/bigger/remain the same?

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    1. Net force = ma
    a = Net force (F) / total mass (m+M) = F/(m+M)

    2. Logically I feel like it doesn't matter "which force" pulls the system, to the right. But if we use the equation, then a = F/ total mass and in this case there are three masses instead of two, so acceleration is smaller?

    Is this correct? I am confused...
     
  2. jcsd
  3. Dec 29, 2015 #2
    Hi , i think all the block will be moving with the same acceleration and velocity. PF~ = ma projected on the x-axis . if they are lying on ground , but pulley problems , u must split the vectors i guess.
     
  4. Dec 29, 2015 #3
    Became smaller, the acceleration that move M and m is given by the tension of the string that connect m and mx : T= (m + M)*a not more by F, because F acts also to the other mass that is falling, ( mx )

    { T= (M+m)*a
    {F-T= mx * a

    Solve this system for T and a
     
    Last edited: Dec 29, 2015
  5. Dec 29, 2015 #4

    PeroK

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    I'm not sure why you are confused. You have explained that in the second case the same force has more mass to accelerate, hence there will be less acceleration.

    What makes you doubt your own analysis?
     
  6. Dec 29, 2015 #5
    I was not sure that it is the same force. I thought maybe in the second case we need to add a force to the net force, like tension, that I missed.

    This was a question on a test today and some students' answer was different from mine.

    Okay, I see! The third mass is not the same as F force.

    (M+m)a=F-mxa
    (M+m+mx)a=F
    a=F(M+m+mx)

    Nice! So I got the answer right!

    Thanks everyone. :)
     
  7. Dec 29, 2015 #6
    If anything, an easier way you could look at that second situation is to just treat all 3 blocks as an entire system that has the same magnitude of force F that will cause all three blocks to accelerate. Since the weight of the mass hanging from the pulley is equal to the magnitude of force F, we know that the second system will have a lower acceleration because now it accounts for an extra mass, not just m and M.
     
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