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F = MA 2010 #10 (Two blocks on top of each other, one is being pulled)

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    See:
    http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf
    #10

    EDIT: There are several constants.


    2. Relevant equations
    F = ma
    Ff = μN


    3. The attempt at a solution
    First: Draw a free-body diagram for block 2 (the bottom block)
    Up: Normal force from floor pushing on block 2
    Down: weight of block 1 on block 2
    Down: weight of block 2 down
    Right: Force F
    Left: Force of Friction Ff

    Up forces = Down forces
    N = m1g + m2g
    N = g(m1 + m2)
    Ff = μg(m1+m2)

    F_y = 0
    F_x = F_net
    so
    Fnet = F - Ff
    m2a = F - ug(m1+m2)
    a = (F - ug(m1+m2)/ m2
    This is the wrong answer.
    The answer says it should be 2m1 + m2 not m1 + m2.
    Why is this?
     
    Last edited: Jan 28, 2013
  2. jcsd
  3. Jan 28, 2013 #2
    First let me point out this is not true. We know F, mu, m1 and m2. They are not variables, they are constants. We don't know their numerical value, so we will manipulate them symbolically. But they are known. The unknown is a2.

    I think your mistake is simple, look at your calculation for the normal forces. Make sure you get the masses right, mass 1 is pushing down on mass 2. So at the point where mass 2 contacts table, how much mass is pressing down? What normal force must be needed to counteract this?
     
  4. Jan 28, 2013 #3
    I don't see what you're saying. Mass 1 is pushing down on mass 2 - Yes, so there is a force m1g down. Mass 2 is also pushing on the table, so there is a force m2g down. Shouldn't that be it?
    So: F_down = m1g + m2g?
    Where does the 2nd m1g come from? Could you please explain?
     
  5. Jan 28, 2013 #4
    If what you said was true, I could put my hand under mass 2 and have you increase mass 1 ten times and not feel a difference. If I am trying to feel the force under mass 2, what do I feel? What if you climbed on top of the whole thing and became mass 3, would that change the force at the table?
     
  6. Jan 28, 2013 #5
    So is what your saying that the table has to overcome each of the three following forces:
    1) mass 2 gravity force
    2) mass 1 push on mass 2
    3) mass 1 gravity force

    This gives 2m1 + m2 and thus the right answer.
    But I still can't picture why you need to account for both mass 1 push on mass 2 and mass 1 gravity force both.
     
  7. Jan 28, 2013 #6
    You need to think about the surfaces where friction is taking place. At the mass 1 / mass 2 interface you only have the weight of mass 1 pushing down. So there the frictional force will be less than at the mass 2 / table interface. There you have the weight of mass 2 and mass 1 each pushing down on the table. So with more force you will have more friction.

    Consider some extreme cases to get an intuition about it. Replace mass 1 with a massive block from stonehenge. Clearly if the force at the table / mass 2 interface will increase. It would crush you if you were under it, not matter how small mass 2 is.

    Two surfaces, two normal forces, each depend on ALL the mass above them (because of gravity's direction)
     
  8. Jan 28, 2013 #7
    Ok, so you can kind of view it as one big system of mass m1 + m2. And then include the force between the m1 and m2 interface. That seems much more logical to me. Thanks.
     
  9. Jan 28, 2013 #8

    SammyS

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    (I see that you've worked this out while I was composing the following, but here's my 2 cents worth anyway.)

    For block 2, there are two different normal forces; one on the upper surface, and one on the lower surface.

    The first thing I would do is draw a free-body-diagram for Block 1. It's a fairly simple diagram: gravity, m1g, points down, normal force, N1 points up, and friction, f1 points to the right.

    The resulting equations are straight forward:
    N1=m1g

    m1a1 = f1 = μm1g .​

    The frictional force, f1, which gives Block 1 its acceleration, is exerted on Block 1 by Block 2. Therefore, according tp Newton's 3rd Law, Block 1 exerts a force on Block 2, equal to f1 and to the left. This is the force you neglected and it's equal to μm1g .
     
  10. Jan 28, 2013 #9
    Oh thanks so much -- I didn't see that before.
    So on block 2 we have:
    Friction from block 1 equal to μm1g on the left
    Friction from the ground equal to μ(m1+m2)g on the left
    Force F going to the right
    F - g(2m1 + m2)

    Thanks. I finally get it. Essentially all I was missing was that the friction force between one and two needs to be accounted for. THIS is where the extra m1 term comes from.
     
  11. Jan 28, 2013 #10

    SammyS

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    Yes.

    Well, I glad I went ahead and replied.
     
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