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Lebeling forces correctly

  1. Oct 8, 2011 #1
    Problem: Two blocks are pushed along a frictionless horizontal surface by a constant 6.00 N force.
    Create a diagram with all forces... I'm not going to include that weight and normal forces.
    PS. My vectors are not drawn to proportion....

    My explanation to my Diagram:

    Given: Push force on blocks 6.00 N Towards blocks.

    *** I added a "reaction Push force by small block in opposite direction" This one is not given in the book.

    2. Action force: Small block pushed on big block.. reaction force big block pushes back on small block

    3. The two red arrows are equal and in opp direction, they cancel and all that is left is the black arrow and purple arrow... purple arrow being smaller in magnitude to black arrow... hence the acceleration in direction of black arrow....

    Is this incorrect?????
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2011 #2

    Doc Al

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    It's a bit confusing. I assume one block is on top of the other with friction between them and on a frictionless table. And the 'push force' is applied to the top block.

    Do this: Just list all the forces acting on the top block (include vertical forces) and on the bottom block separately. (We can worry about the diagram later.)
     
  4. Oct 8, 2011 #3
    No, they are actually side by side as such.....


    Sorry for the confusion
     

    Attached Files:

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  5. Oct 8, 2011 #4

    Doc Al

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    OK. So list the forces on each.
     
  6. Oct 8, 2011 #5
    Forces acting on the top block is the constant push 6.00 N,

    Is there also a force by the small block pushing in the opposite direction?
     
  7. Oct 8, 2011 #6

    Doc Al

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    You mean the right block? (Assume they are being pushed to the left.)
    Definitely. There will be a 'normal' force between the two blocks as they push against each other.
     
  8. Oct 8, 2011 #7
    Here is the original picture...

    Why are you saying they are being pushed to the left when the acceleration is to the rights?
     

    Attached Files:

  9. Oct 8, 2011 #8

    robphy

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    Some points to keep in mind....

    In general, whenever two objects, A and B, are in contact,
    there is a "vector-force-on-A due to B" and
    an equal-magnitude, oppositely-directed "vector-force-on-B due to A". (Newton III)
    Note these forces are on different free-body diagrams.

    Each of those forces can broken up into
    a part parallel to the surface of contact [usually called friction, if any] and
    a part perpendicular to the surface of contact [usually called a normal force].

    For a given object (say A), the sizes and direction of friction and
    the size of the normal force on A
    depend on the acceleration and mass of object A,
    as well as the vector-sum of all of the forces on object A
    [as shown on A's free-body diagram].
    (The relation among these is Newton II for object A.)
     
  10. Oct 8, 2011 #9
    Ok this is what I have....

    Force 1:Action force: Hand pushing with force 6.00 N on small block.
    Force 2: Reaction force: Force by small block pushing back on hand.

    Force3: Action force: Small block pushing on big block
    Force4: Reaction force: Big block pushing back on small block.....

    .... However the book does not have "Force 2" in the diagram... and I am asking why not?
     
  11. Oct 8, 2011 #10

    Doc Al

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    I had to guess based on your description. (I didn't see if you mentioned the acceleration before.)
     
  12. Oct 8, 2011 #11
    Sorry about that...
    But my only confusion comes with why force two doesn't play a role?
     
  13. Oct 8, 2011 #12

    Doc Al

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    The reaction force acts on the hand, so we don't care about that. Just list the forces on the blocks.

    OK.

    Because force 2 acts on the hand, not on one of the blocks. All they want are the forces on the blocks. And it's a good idea to list the forces on each block separately, just so it's clear where the forces act.
     
  14. Oct 8, 2011 #13
    Ok thank you... and as I thought about it, it wouldn't make sense to add force on hand..
    Last questions... I believe that forces 3 and 4 are equal in mag and opp in direction... do these forces cancel one another?????
     
  15. Oct 8, 2011 #14

    Doc Al

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    Depends on how you define your system. If you look at each block separately, then they do not cancel since they act on different bodies. (In that sense, Newton's 3rd law pairs of forces never cancel.)

    If you look at the two blocks as a single composite system, then the forces between the blocks are internal forces and sum to zero. The net force on the two block system is just the applied force (force #1).
     
  16. Oct 8, 2011 #15

    robphy

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    Are these forces on the same free body diagram? [...seconds too late :rofl:]
     
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