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Conservative forces

  1. Oct 10, 2004 #1
    Hello everyone, I have a question about conservative forces.
    I am given a function F = -kx^2 + ax^2 + bx^4, where a, b, and k are constants. I am asked to determine if this force is conservative or not, I don't know know how to prove this mathematically or theoretically. Please help!

    In this particular example the force is a spring force, spring forces aren't always conservative are they?
  2. jcsd
  3. Oct 10, 2004 #2
    You can use the fact that Work down by a conservative force in a closed path is zero.
    I guess spring forces are always conservative if the spring is ideal.
  4. Oct 10, 2004 #3

    Doc Al

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    Staff: Mentor

    There are several ways to mathematically test that a force is conservative. Here are a few:
    (1) You can check that the line integral is zero along any closed path (equivalent to what rhia suggested).
    (2) Or you can see if you can find a potential function whose -gradient equals the force function (hint: integrate).
    (3) Or you can see if the curl of the force is zero.​
    The last two checks should be easy. :smile:
  5. Oct 10, 2004 #4
    Hi Doc AI,
    Can you please explain the physical significance of the last two checks?
    I don't understand things if I can't relate them to physical world.:(
    Thanks a lot!
  6. Oct 10, 2004 #5
    Okay I though that if the force was conservative, then the gradient could be used, I didn't know that it proves the force is conservative.

    A correction in the force equation

    F = -kx + ax^3 + bx^4

    After integrating a found the potential equation to be
    U = (kx^2)/2 - (ax^4)/4 - (bx^5)/5

    using F = -dU/dx.

    I found the gradient of U to be

    grad(U) = (kx - ax^3 - bx^4)i

    Is there a reason for the sign difference? I just learned gradients, in my Calc III class that I am taking right now, two weeks ago so I am still not comfortable with all of its properties.

    Okay what if I have a force function of the type:

    F = -kx^2y +(y^2)(z^2) + xyz ( I made this one up)

    So took three integrals, with respect to x, y, and z seperately, to get the potential function in each direction. Then I found the gradient of the potential function to be

    grad(U) = (kyx^2 - (y^2)(z^2) - xyz)i + (k(x^2)y - (y^2)(x^2) - xyz)j + (k(x^2)y - (y^2)(z^2) - xyz)k

    To find if the gradient equals the force function, do I find the absolute value of grad(U) , to make it a scalar function??? I am not sure how to find if these two are equivalent.

    Okay I took a guess to convert the force function into a vector function. First I found the partial derivatives of F with respect to x, y, and z. Then I took the integral with respect to x, y, and z. I found

    F = -(k(x^2)y - xyz)i - (k(x^2)y + (y^2)(z^2) + xyz)j + ((z^2)(y^2) +xyz)k

    When I took the integral with respect to each I didn't add the Integral constant, so maybe that could account for the difference. According to the scalar function F, when x = O, F = (y^2)(z^2), when z = 0, F = -k(x^2)y . .so these could be the missing constants in my vector function F??
    Thanks for all your help.
  7. Oct 10, 2004 #6
    I think the simplest way, is the curl.

    If [tex]\vec{F} = M(x,y,z)\vec{i} + N(x,y,z)\vec{j} + Q(x,y,z)\vec{k}[/tex], then

    [tex] \vec{\nabla} \times \vec{F} = \left| \begin{array}{ccc}
    \vec{i} & \vec{j} & \vec{k} \\
    \partial/\partial x & \partial/\partial y & \partial/\partial z \\
    M & N & Q \end{array} \right| = 0

    The reason for this, is the Stokes theorem, which says:

    Let be S a surface oriented with a normal vector [tex]\vec{n}[/tex], limited by a closed simple curve C. If [tex]\vec{F}[/tex] is a vectorial field and its partial derivatives are continuous in that region, then:

    [tex] \oint_{C} \vec{F}d\vec{r} = \iint_{S} (\vec{\nabla} \times \vec{F})\vec{n}dS[/tex]


    If the curl is equal to zero, then the right term is zero as well, and the definition of conservative field is that the left integral is equal to zero.
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