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Help with Newtons

  1. Aug 16, 2004 #1

    abc

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    can anyone help me in this problem ?????
    see the attachment plz
     

    Attached Files:

  2. jcsd
  3. Aug 16, 2004 #2

    abc

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    a = 5.4 newton >>>> i think it is not clear in the attachment ???
     
  4. Aug 16, 2004 #3
    This problem sounds a bit tricky, but it really isn't. What they aren't telling you is that first you'll pull the box with whatever force you need to get it moving. Once you've done that it wants to know what force keeps it at a constant velocity.

    Using newton's laws, what force on an object produces a constant velocity?

    If you will answer that I'll help with the rest of the problem.
     
  5. Aug 16, 2004 #4

    HallsofIvy

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    The total weight of the objects is 9.0 Newtons. Since the coefficient of friction is 0.35, the friction force will be (0.35)(9.0) Newtons. The force necessary to keep them moving at constant velocity is exactly the force necessary to offset that.
     
  6. Aug 16, 2004 #5

    abc

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    thanx HallsofIvy ..... but the answer written in the book is 4.05 ( but i don't know the steps to bring this answer )
    dear locrian..... do u mean this >>>> sum F= 0 ????
    thanx guys ..... but plz continue to solve this problem
     
  7. Aug 16, 2004 #6

    arildno

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    Either you've been given insufficient information, or your book has an incorrect solution.
    Let's look at the forces working on each object A, B in the horizontal direction
    Object A:
    1. The frictional force from the ground, [tex]-f_{G}[/tex]
    2. The frictional force between A and B:[tex]-f_{AB}[/tex]
    3. the rope tension: -T
    4. The force you drag the along system with: f
    Object B:
    1. The frictional force between A and B:[tex]f_{AB}[/tex]
    2. the rope tension:-T
    (I've added signs for assumed directions)

    Are you with me on this?
     
  8. Aug 16, 2004 #7

    Gokul43201

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    HallsofIvy, there's also friction between the blocks (I would imagine - maybe with the same coefficient).

    But assuming that, I get F= 4.4N

    abc, draw the free-body force diagram, and take it from there.
     
  9. Aug 16, 2004 #8

    arildno

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    We'll proceed:
    The weight of object B is [tex]w_{B}[/tex], and we have therefore:
    [tex]f_{AB}=\mu_{AB}w_{B}[/tex]
    where [tex]\mu_{AB}[/tex] is the friction coefficient between A and B

    That A moves with constant velocity implies that B also moves with constant velocity, since otherwise, the rope length can't remain constant.

    Hence, the rope tension must satisfy (by Newton's 2.law applied to object B in the horizontal direction):
    [tex]T=\mu_{AB}w_{B}[/tex]

    Clearly, the normal force from the ground on object A must balance the weights of both objects; hence we have for the frictional force:
    [tex]f_{G}=\mu_{G}(w_{A}+w_{B})[/tex]

    Newton's 2.law in the horizontal direction for object A must read:
    [tex]f-f_{AB}-T-f_{G}=0[/tex]
    Or, solving for f:
    [tex]f=\mu_{G}w_{A}+(\mu_{G}+2\mu_{AB})w_{B}[/tex]
    By setting [tex]\mu_{G}=\mu_{AB}=0.35[/tex] we gain:
    [tex]f=0.35*(5.4+3*3.6)=0.35*16.2=5.67[/tex]

    your book's solution is wrong
     
    Last edited: Aug 16, 2004
  10. Aug 17, 2004 #9

    abc

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    thanx all for replying .....
    regard
    abc
     
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