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Free Body Diagram and Newton's 2nd

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    3. The attempt at a solution
    My attempt is in the picture. I am trying to get the diagram done before I continue on because if I mess up in the diagram, my work is guaranteed to be wrong. So basically what I am asking is is my diagram correct?

    NOTE: the diagram in the bottom right was drawn by the professor.
     
  2. jcsd
  3. Oct 30, 2007 #2

    PhanthomJay

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    When you draw a FBD of the system as a whole (both blocks together), you must identify only the weight of the system and the forces external to your FBD. The friction forces between the two blocks or normal force between the 2 blocks do not enter into the FBD of the system.
     
  4. Oct 30, 2007 #3
    [​IMG]
    Is that right?
     
  5. Oct 30, 2007 #4

    PhanthomJay

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    Yes, more or less. You have the horizontal and vertical components of the applied force F, and the weight.....which you should identify as the weight of BOTH blocks....and the Normal force of the floor on the bottom block. Your FBD should look more like your professor's diagram with those forces identified on it.
     
  6. Oct 30, 2007 #5
    Ok, I drew it on my paper. So for B:
    [tex]\sum F = Fcos\theta = (m_{1} + m_{2})a[/tex]
    [tex]a = \frac{Fcos\theta}{m_{1} + m_{2}}[/tex]
     
  7. Oct 30, 2007 #6
    And C:
    [tex]\sum F_{m1} = Fcos\theta - \mu_{s}m_{1}g = m_{1}a[/tex]
    [tex]\sum F_{m2} = \mu_{s}m_{1}g = m_{2}a[/tex]
    [tex]a = \frac{\mu_{s}m_{1}g}{m_{2}}[/tex]
    [tex]Fcos\theta - \mu_{s}m_{1}g = \frac{m_{1}^2\mu_{s}g}{m_{2}}[/tex]
    [tex]Fcos\theta = \frac{m_{1}^2g}{m_{2}}\mu_{s} + m_{1}g\mu_{s}[/tex]
    [tex]\mu_{s} = \frac{Fcos\theta}{\frac{m_{1}^2g}{m_{2}} + m_{1}g}[/tex]
     
  8. Oct 30, 2007 #7

    PhanthomJay

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    Looks good!
     
  9. Oct 30, 2007 #8

    PhanthomJay

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    A very nice attempt with your FBD's, but you fell a bit short. The friction force is mu_s times the normal force. You have not correcty identified the normal force between the 2 blocks. And don't lose sight of what the problem is asking for. It gets a tad confusing with so many letters and oh so few numbers.
     
  10. Oct 30, 2007 #9
    Had class about 3 hours ago (turned this in with the wrong answer). At least it wasn't lonely, I had a whole paper with wrong answers to keep it company.

    [tex]Normal_{m2/m1} = m_{1}g + Fsin\theta[/tex]
    [tex]F_{friction} = \mu_{s}n = \mu_{s}(m_{1}g + Fsin\theta)[/tex]

    [tex]\sum F_{m2} = F_{friction} = m_{2}a[/tex]
    [tex]a = \frac{F_{friction}}{m_{2}}[/tex]


    [tex]\sum F_{m1} = Fcos\theta - F_{friction} = m_{1}{a}[/tex]
    [tex]Fcos\theta - F_{friction} = m_{1}\frac{F_{friction}}{m_{2}}[/tex]
    [tex]F = \frac{\frac{m_{1}F_{friction}}{m_{2}} + F_{friction}}{cos\theta}[/tex]

    It still looks wrong, and I'm still laughing at me solving for [tex]\mu_{s}[/tex] and turning that in.
     
  11. Oct 30, 2007 #10

    PhanthomJay

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    Your ability to draw FBD's is exceeded only by your abilty to use LaTex, good job at both. I don't see anything wrong with what you've done; i'm wondering if you used the result of the acceleration from the 'system' FBD if the result for F would be simpler. If F exceeds a certain value, then the block on top will slide. Too many letters for me to figure it out.
     
  12. Oct 30, 2007 #11
    Thanks for the help. I just realized [tex]F_{friction}[/tex] has [tex]Fsin\theta[/tex] in it. Even using the system acceleration, I get this:
    [tex]F = \frac{\mu_{s}m_{1}g}{\frac{m_{2}cos\theta}{m_{1} + m_{2}} - \mu_{s}sin\theta}[/tex]
    That is after quite a bit of simplification... I'm surprised this question was so complicated. The other questions did not take nearly as long to complete.
     
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