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Homework Help: Forces exerted by two rods with uniform mass.

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the gravitational force betwee two uniform(1-D) rods as shown below.(hint you have to integrate the force over the mass elements of both rods . we did the first integral in class. As before you only have to worry about the componet in the x-direction by symmetry. Demonstrate that , in the limit that the seperation of the rods becomes large compared to their lengths, the force the correct limiting form.

    Here is a drawing of the two rods on my course webpage. You might have to click on the homework tab of the main page and then proceed to click on assignment three if link directly
    fails to link you to the homework page containing the problem

    http://courses.ncsu.edu/py411/lec/001/ [Broken]

    2. Relevant equations

    principle of superposition for Forces


    3. The attempt at a solution

    General information about the forces between two or more rods
    F=lim(as delta(x)==> 0) sigma(-G*m0*lambda*delta(xi) (x-hat)/(x0-xi)^2

    F0=\lmoustache{-L/2 , L/2}(-G*m0*lambda(x)*dx/((x0-x)^2)*(x-hat)

    since object is uniform, all lambdas are constant and therefore all lambda's equal lambda(0).

    F0=integral(from -L/2 to L/2)(-G*lambda(0)*m(0)*(x-hat)/(b-x)^2=-G*m0*lambda(0)(x-hat) *integral(from -L/2 to L/2) dx/(b-x)^2. From u-substitution, I can say that :



    F0=-G*m0*lambda(0)*(x-hat)*integral(from b-L/2 to b+L/2)(-du/u^2)=-G*m0*lambda(0)*x-hat[-1/(b-L/2)+1/(b+L/2)]

    1/(b+L/2)-((1)/(b-L/2))= -L/(b^2-(L/2)^2)


    in limiting case, as b>>L, then F0 = G*m0*M/b^2.

    My solution
    let M/L=lambda

    with cos(alpha)=b/r and r=sqrt(b^2+y^2)

    by symmetry, the x component is the only left since force is only x direction.

    F=integral(from -L/2 to L/2)dy(-G*m(particle)*lambda*dy/(r^2)*cos(alpha))x-hat=-G*m(particle)*lambda*integral(from -L/2 to L/2)b*dy/r^3=-G*m(particle)*lambda*b*integral( from -L/2 to L/2)

    ==>G*m(particle)*lambda*b^-1*integral(from -L/2 to L/2)*(dy/b)/(1+(y/b)^2)^(1.5)

    using u-substitution , I find that y=b*tan(theta)

    F=-G*m(particle)*lambda*b^-1*integral(from arctan(L/2b) to arctan(-L/2b)*(sec^2(theta)*dtheta/(1+tan^2(theta)= G*m(particle)*lambda*b^-1 integral(arctan(-L/b) to arctan(L/b)) dtheta/sec(theta)=-G*m(particle)*lambda*cos(theta)*dtheta=-G*m(particle)*lambda*b^-1*sin(theta)*integral(arctan)

    F=-G*m(particle)*lambda*b^-1 * [(2L/b)/(sqrt(1+(L/b)^2)

    Force between two rods is r=sqrt(b^2+(y'-y)^2)

    I'm not sure where to go from here :(
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 14, 2008 #2


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    oh come on, Benzoate!

    that's virtually unreadable … :frown:

    either learn how to do LaTeX …

    http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000 [Broken]

    http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000 [Broken]

    or use unicode characters and the X2 and X2 tags.
    Last edited by a moderator: May 3, 2017
  4. Sep 14, 2008 #3
    I am having difficulty using latex. I will paste one of the equations from my OP and you can show me how I can transformed text into latex:

    F0=integral(from -L/2 to L/2)(-G*lambda(0)*m(0)*(x-hat)/(b-x)^2=-G*m0*lambda(0)(x-hat) *integral(from -L/2 to L/2) dx/(b-x)^2

    Sorry , I should have asked you. Can you show me how to transformed the above equation into latex?
    Last edited by a moderator: May 3, 2017
  5. Sep 14, 2008 #4


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    I'll start you off … you get:

    [tex]F_0 = \int_{-L/2}^{L/2}\frac{-G \lambda_0 m_0 \hat{x}}{(b - x)^2}[/tex]

    if you type

    [noparse][tex]F_0 = \int_{-L/2}^{L/2}\frac{-G \lambda_0 m_0 \hat{x}}{(b - x)^2}[/tex][/noparse]

    Please look at those links I gave you!
  6. Sep 14, 2008 #5
    I was unable to edit my OP , so I will have to repost my solutions.

    [tex]F=\begin{displaymath}\lim_(n \rightarrow 0)\-int_{-L/2}^{L/2}\frac{-G \lambda_0 m_0 \hat{x}{( b-x)^2}[/tex]
    [tex]F_0 = \int_{-L/2}^{L/2}\frac{-G \lambda_0 m_0 \hat{x}}{(b - x)^2}[/tex]


    [tex]F_0 =frac{-G \lambda_0 m_0 \hat{x}} \int_{b-L/2}^{b+L/2}\frac{{-du}{u}^2}[/tex]

    just a test! still converting all my text into latex
    still a test

    I give up! I've worked on the same equation for latex for 30 min I've still failed to convert my equation into latex
    Last edited: Sep 14, 2008
  7. Sep 14, 2008 #6


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    Hi Benzoate! :smile:

    Some corrections:

    i] no need to type "begin displaymath" … tex will atuomatically assume it's maths, unless you type \text{}

    ii] instructions like lim and ^ and _ must be followed by {}, not ()

    iii] you typed \-int instead of \int, and left out some { and }s … ou must count thsoe brackets carefully :wink:

    This, I think, is what you wanted …

    [tex]F=\lim_{n \rightarrow 0}\int_{-L/2}^{L/2}\frac{-G \lambda_0 m_0 \hat{x}}{( b-x)^2}[/tex]
    and you get it if you type:

    [noparse][tex]F=\lim_{n \rightarrow 0}\int_{-L/2}^{L/2}\frac{-G \lambda_0 m_0 \hat{x}}{( b-x)^2}[/tex][/noparse]

    Try again! :smile:
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