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Dumb normal force question

  1. Oct 5, 2011 #1
    I'm feeling really stupid right now.

    But let's say a block is resting on a floor with friction. If someone were to pull to the right of the box at some angle theta above the horizontal, there would be components in the y (perp. to horizontal) and x (parallel to horizontal) of the force you pull at.

    Now, when you draw the free body diagram, mg points straight down (the force of the block) and the Normal force points up (also equal to mg).

    Now, doesn't that mean when you add up all the vectors the normal force and the weight of the block will cancel, and the only vector left in the y direction will be the y component of the force vector which was enacted at some angle theta above the horizontal.

    Thus, since there's a net force in the y direction the block will lift from the ground. In fact, following this reasoning, any old arbitrary (non-zero) force at some non-zero angle should lift the block.

    Obviously something's missing here from my free body diagram description?
     
  2. jcsd
  3. Oct 5, 2011 #2
    You are nearly there.

    It is a standard question/procedure to determine whether the block will slide or tip under the action of your force.

    What happens if you consider moment equilibrium as well as horizontal and vertical?
     
  4. Oct 5, 2011 #3

    berkeman

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    Staff: Mentor

    When you unweight the box a bit by pulling up at an angle, you reduce the normal force. Since the box is not accelerating up or down, the sum of the vertical forces must be zero. That includes the weight of the box, the upward force from the rope, and the normal force pushing up from the ground...
     
  5. Oct 5, 2011 #4

    berkeman

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    Staff: Mentor

    Dagnabit. Beat out by Studiot again....
     
  6. Oct 5, 2011 #5
    Hello mike

    o:)
     
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