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Calculating normal force for wedge

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that normal force between a block and wedge is: N = m*sqrt(g^2-a^2)


    2. Relevant equations
    I used the pythagorean theorem to attempt this.


    3. The attempt at a solution
    See attached.

    Am I doing this right, or am I making up math here to fit the answer?
     

    Attached Files:

  2. jcsd
  3. Jun 28, 2011 #2

    rock.freak667

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    Homework Helper

    I think that is correct but that is only for the given case when a=gsinθ. (Which is correct as your free body diagram is showing)
     
  4. Jun 28, 2011 #3
    Thanks for your help!
     
  5. Jun 28, 2011 #4
    [itex]F_{N}=F_{g}sin\theta[/itex]
     
  6. Jun 28, 2011 #5

    ElijahRockers

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    Gold Member

    Looks right. I assume you are wondering about pulling the m from the square root. It may be the long way around, but try expanding.

    [tex]N^2 = (mg)^2-(ma)^2[/tex]

    can be expanded into:

    [tex]N^2 = mmgg-mmaa[/tex]

    factor out the m squared:

    [tex]N^2 = mm(gg-aa)[/tex]
    [tex]N^2 = m^2(g^2-a^2)[/tex]

    Take the square root of both sides (raise both sides to the half power):

    [tex]N = \sqrt{m^2(g^2-a^2)}[/tex]
    [tex]N = m\sqrt{g^2-a^2}[/tex]

    I know this is a long winded response to a relatively simple question, but I am new to helping on the forums and trying to figure out how to use LaTeX. Sorry I am just a beginner myself, but I hope it may give you some insight into the math involved.
     
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