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How long does it take mass m to fall distance h?

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://rawrspace.com/Capture.JPG [Broken]

    So when you draw out a force diagram, we find these two equations
    mg-T=ma
    rewriting in terms of T
    T=mg-ma
    and
    T-MgSin[tex]\vartheta[/tex]-f=Ma
    rewriting in terms of T
    T=Ma+MgSin[tex]\vartheta[/tex]+f

    I set the two equations equal to each other so I could solve for the acceleration since it is as rest when it starts (I figured that means that initial velocity is 0 )
    setting them equal I get:
    mg-ma=Ma+MgSin[tex]\vartheta[/tex]+f
    Then I moved them so that I could start factoring out terms:
    g(m-MSin[tex]\vartheta[/tex])-f=a(M+m)
    finally, solving for a:
    [g(m-MSin[tex]\vartheta[/tex])-f]/(M+m)=a

    I also know that y-y_0=V_0t-(1/2)at^2
    Since y final is 0 and y_0 is h, I get this equation:
    -h=-(1/2)at^2
    So this means that
    (2h/a)^(1/2)=t

    Pluging in a that I solved for above, I get

    {(2h)/[g(m-MSin[tex]\vartheta[/tex])-f]/(M+m)}^(1/2)=t

    That was my answer of the time it takes for the mass m to fall a distance h. Does that seem correct to you, or am I overlooking something ?
     
    Last edited by a moderator: May 4, 2017
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  3. Sep 16, 2010 #2

    kuruman

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    It looks right to me except your answer has f in it and you are specifically required to write the answer in terms of m, M, h, θ and μ. So how can you get rid of f and introduce μ?
     
  4. Sep 16, 2010 #3
    Thank you for taking a look. I don't know why I wrote it like that on here, on my paper I have μN for the frictional force instead of just the f for frictional force..

    Thank's again for taking a look.
     
  5. Sep 16, 2010 #4

    kuruman

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    You're still not there because μN won't do either. What is N in terms of the given quantities m, M, h, θ?
     
  6. Sep 16, 2010 #5
    Well N is the normal force so it is the opposite component the gravitational force which is Mg so it would be -Mgcos[tex]\vartheta[/tex] wouldn't it (assuming I take my g to be positive)? That would still leave me with a g though which isn't in those terms. Is it okay to have g since it is a constant ? If so then I would replace -f with μMgcos[tex]\vartheta[/tex] (it would be positive correct, as -f = -μN = -μ(-Mgcos[tex]\vartheta[/tex]) )

    So I would like to write my answer as

    {(2h)/[g(m-Msin[tex]\vartheta[/tex])+μMgcos[tex]\vartheta[/tex]]/(M+m)}^(1/2)=t
     
  7. Sep 16, 2010 #6

    kuruman

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    Don't forget that, in your original expression, f stands for the magnitude of the frictional force. That is a positive number. So f = μN = Mgcosθ. You already took care of the direction when you wrote down T-Mgsinθ-f=Ma.
     
  8. Sep 16, 2010 #7
    So It would be μMgcosθ correct since N = Mgcosθ , and f = μN

    So that means that it would be -μMgcosθ in the problem instead of -f

    t={[(2h)(M+m)]/[g(m-Msinθ)-μMgcosθ]}^1/2

    I believe that would be correct as it is now in the correct terms, I moved the M+m to the numerator. Does that seem correct now, If it was -f which took care of the magnitude since it subtracted in the original force equation. Then replacing it with the information I derived above should be correct.
     
  9. Sep 16, 2010 #8

    kuruman

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    Right.
     
  10. Sep 16, 2010 #9
    Thank you for helping me understand where I went wrong.
     
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