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Integrating the rational fractions

  1. Sep 9, 2008 #1
    please........... i need a help in integrating the partial fractions
    i cant proceed to the integration part if i don't understand the patter in finding the constant....
    that is....

    if the given is:

    ʃ ( (x^5+1) / ((x^3)(x+1)) )dx

    then;

    ʃ ( x-2 + ( 4x^3+1 ) / ( x^4 + 2x^3) )
    ʃ ( x-2 + ( 4x^3 + 1 ) / ( x^3 (x + 2 ) )
    ʃ ( x-2 + A/x + B/x^2 + C/x^3 + D/x+2)

    Ax^2(x+2) + Bx(x+2) + C(x+2) + D(x^3) = 4x^3 + 1 ----- i dont know what going on
    in this part
     
  2. jcsd
  3. Sep 9, 2008 #2

    tiny-tim

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    Hi MadPhysics08;! :smile:

    (I assume you mean ʃ ( (x^5+1) / ((x^3)(x+2)) )dx)

    Ax^2(x+2) + Bx(x+2) + C(x+2) + D(x^3) = 4x^3 + 1

    is simply the previous two lines (except for both the x - 2 s), multiplied by x^3(x + 2)

    A B C and D are unknowns, and you find them by solving that equation :smile:

    (though I think you can find D using the usual substitution method :wink:)
     
  4. Sep 9, 2008 #3
    Prove that F=mgtanθ

    Can everyone help me to solve this problem....

    A small sphere of mass hangs from a string of L as in below. A variable horizontal force F is
    applied to the sphere in such a way that it moves slowly from the vertical position until the string makes angle θ with the vertical. Assuming the sphere is always in equilibrium

    1. show that F=mgtanθ
    2.show that work done by F is mgL(1-cosθ)

    __________________________
    |\............................
    |..\..........................
    |.θ.\.L......................
    |......\.......................
    |........\.....................
    |.........0------F..........
    0.........|....................
    ...........|....................
    ...........|....................
    ..........W...................
     
  5. Sep 9, 2008 #4
    Prove that F=mgtanθ

    Can everyone help me to solve this problem....

    A small sphere of mass hangs from a string of L as in below. A variable horizontal force F is
    applied to the sphere in such a way that it moves slowly from the vertical position until the string makes angle θ with the vertical. Assuming the sphere is always in equilibrium

    1. show that F=mgtanθ
    2.show that work done by F is mgL(1-cosθ)

    __________________________
    |\
    | \
    | θ \ L
    | \
    | \
    | 0------F
    0 |
    |
    |
    W
     
  6. Sep 9, 2008 #5

    tiny-tim

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    Hi MadPhysics08! :smile:

    ("Can anyone help me to solve this problem...." would be more realistic! :wink:)

    "moves slowly" is examiners' code for saying that, at any particular time, the acceleration is so small that you can treat it as zero.

    Does that help? :smile:
     
  7. Sep 10, 2008 #6
    so, let d be the displacement W(work)=F*dcosθ
    ∑F=ma=0 then,
    ∑F=W+T+F
    W+T+F=0
    -W*dcosθ + T*dcosθ + F*dcos(θ+90)

    [T*dcosθ=0 because they are perpendicular]

    W*dcosθ=F*dcos(θ+90)

    but cos(θ+90)=sinθ

    w*dcosθ=F*dsinθ
    F=Wtanθ
    F=mgtanθ

    is my solution correct?
     
  8. Sep 10, 2008 #7
    thnks tim....

    Can i asked one favor...
    plz... i need a hint for the problem number 2.....
     
  9. Sep 10, 2008 #8

    tiny-tim

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    Hi MadPhysics08! :smile:

    Let's do part 1 first:
    You should say which direction you are taking components in, and why.

    Anyway, if w*dcosθ=F*dsinθ, then F = w cotθ. :frown:
    Whatever is this supposed to be?? :confused:

    Hint: work done = energy gained.

    (this is the only reason we bother about work done)
     
  10. Sep 10, 2008 #9
    okay....

    There three forces acted on the sphere:
    tension(T), applied force F and gravitational force(mg)

    so..
    W=T*Scosθ=0
    the displacement of the sphere along the arc length is perpendicular to the tension
    While..
    W=-mg*d cos(θ+90)=-mg*dsinθ (trigonometric identity of cos(θ+90)=sinθ)
    because the angle between the displacement of the sphere is more than 90 degrees..
    thus...work done by weight is negative
    however..the displacement of the sphere started from the vertical position to the slanting position until the two position made an angle θ
    and,
    W=F*dcosθ (angle between F and d is θ)
    so...
    equate W to W:
    F*dcosθ= -mg*dsinθ (cancel d)
    F*cosθ=-mg*sinθ
    F=-mgtanθ--------this is my problem...i just end up with negative sign...
     
  11. Sep 10, 2008 #10
    work done by gravitational force(W=-mgcos(θ+90)) is negative because it opposed the direction of W=F*dcosθ
     
  12. Sep 10, 2008 #11

    tiny-tim

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    Hi MadPhysics08! :smile:

    I'm sorry, I can't follow any of this …

    you seem to be using work done to find the value of F. :confused:

    1. to find F, just take components perpendicular to the unknown force, whch is … ? :smile:

    2. to find the work done, either:

    i] integrate the force "dot" the displacement (which works, but is really complicated :frown:), or:

    ii] just use the work-energy theorem: work done equals energy gained …

    so what is the energy gained? :smile:
     
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