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Finding minimum force

  1. Jul 19, 2007 #1
    The problem Im having is that in all the figures (in the image) I have 3 variables with only 2 equations. In figure 3 The only way I could solve for the variables would be to set the normal force equal to the weight, however if the tension is also pulling up it would decrease the normal force. What am I missing?

    Any help would be greatly appreciated.

    [​IMG]
     
  2. jcsd
  3. Jul 19, 2007 #2

    berkeman

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    Staff: Mentor

    The sketches are a little hard to decipher. But if I understand them correctly, the block will start to move only when the tension is enough to counteract the weight of the block. Which means that the tension in the rope attached to the block will have to have a vertical component equal to the weight of the block (and a horizontal component of ?). That's if the initial motion of the block is vertical. Seems like you also need to check to see if slightly less tension would lighten the block enough so that the horizontal component of the tension could slide the block sideways.....
     
  4. Jul 19, 2007 #3
    That makes sense if the initial movement is vertical. But how could I find the tension if the initial movement is horizontal. Would'nt that bring us back to the 3 variable 2 equation thing?

    If it would help I could scan the problem from the book also.
     
  5. Jul 19, 2007 #4

    berkeman

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    Staff: Mentor

    I'm not sure about the number of equations thing, but the first movement will be horizontal if the horizontal component of the tension required to overcome the diminishing horizontal static friction force (F=mu*N) becomes big enough to slide the block, before the vertical component is enough to lift the block. Are you given the mu for the block on the surface? If not, then they must just be asking for the lifting event.

    And if it's the lifting event, then yes, the vertical component of the tension would equal the weight of the block. Does that give you an equal number of unknowns and equations now? (I don't know -- I haven't looked in detail at the problem)
     
    Last edited: Jul 19, 2007
  6. Jul 19, 2007 #5
    They do give mu as .3. I think I might be able to solve the problem now. If I cant Ill post back. thanks for the help.
     
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