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I Integrating force

  1. May 30, 2016 #1
    If force isn't constant and you want to find the force of something at a specific time, you'd integrate right?
    For a specific question I came up with this
    Integral (m * da)
    Integral m * dv/t
    (m/t) * dv
    (m/t) * v
    F = (m/t) * v

    Does this seem right? Or I'm I making illegal moves. Help.
  2. jcsd
  3. May 30, 2016 #2
    The question is not clear. F = ma is not a relation between one force and the acceleration. It is a statement of a law which says that if a number of forces act on a body, then their sum is equal to the mass of the object times its acceleration. If there is only one force acting on the body, then that force F = ma. Now, if that force is not constant, then certainly, both force and acceleration are functions of time, F(t) = m a(t). So if you know the acceleration as a function of time, you know the force as a function of time. There is no integration to be done.
    So I do not know what is meant by the second line: integral (m * da).
    If you don't know the acceleration as a function of time, but know the velocity as a function of time, then the acceleration can be obtained by differentiating the velocity, since a = dv/dt. So the third line, da = dv/t is not correct, and integrating that has no significance.
  4. May 30, 2016 #3
    Sorry I wasn't clear on the question. Thanks btw. So if I want force at a specific time and velocity, all I have to do is substitute dv/dt with a number?
  5. May 30, 2016 #4
    Sorry I'm not very good at this.
  6. May 31, 2016 #5


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    The manipulations you are making are both invalid and poorly motivated. Instead of writing a sequence of formulas down, you need to write a sequence of equations down and explain how each one follows from the ones.
    This is a good starting point. An equation that is valid for constant acceleration.
    This expression comes out of the blue and is incorrect. What is it that you are trying to write down here? Do you, for instance, want to write down the formula for the momentum transferred by a non-constant force acting over a period of time? If so, where is the time variable that should appear in the formula?
  7. May 31, 2016 #6
    That will give you the acceleration. Multiplying that with mass gives the sum of all the forces acting on the body.
  8. May 31, 2016 #7
    I wrote the da to indicate the quantity that is changing.
  9. May 31, 2016 #8


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    But that's not how the "d" symbol is conventionally used. Normally you would consider acceleration as a function of some other parameter such as time. Then you would denote the integral of "acceleration over time" as ##\int a(t) \, dt##
  10. May 31, 2016 #9
    So writing acceleration as that tell me that acceleration is changing over time? If I want to integrate something which is not constant I need to consider that something as a function of another thing?
    Exammple, if I want to find the kinetic energy of something (a spinning disk), would we write velocity in terms of distance from the center of the disk? Since eery small area on the disk has a different velocity.
    Lol sorry I'm new to calculus infused physics
  11. May 31, 2016 #10
    Yes, you will need the velocity as a function of radius and then integrate over the radius.
  12. May 31, 2016 #11
    Integration is not exactly the same as summation. There is also the variable of integration. That's why we write ##\int F(x)dx## for example and not ##\int F(x)##.When you integrate a function you have to specify with respect to which variable the integration is done. So in your post #1 when you just say "integrate force" it just doesn't say anything..
  13. May 31, 2016 #12
    So when you specify the variable of integration, you can then integrate. Thank You . Last question, just so I understand this, let's say I want to find the area of a disk, I would integrate area with respect to radius? A(r) ?
  14. Jun 1, 2016 #13
    In that case the variable of integration is implied, but for the force, a force can be a function of time , or a function of spatial coordinates (as for example ##F=-kx##) or force is always function of acceleration, so you have to say if you want to integrate force w.r.t to acceleration to get the integral ##\int mada=\frac{1}{2}ma^2##, though not sure if this last integral has some physical meaning.
  15. Jun 1, 2016 #14
    Something else to clarify my post , post #11, integral is a summation of infinitely many, infinitesimal small products ##f(x)dx##. If we were just to summate f(x) and write ##\int f(x)## we would just get infinite as a result in all the cases regardless of what f(x) is.
  16. Jun 1, 2016 #15


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    I disagree with this. The variable of integration is not implied. There is no rule that requires us to find the area of the disk as the sum of the infinitesimal areas of a bunch of concentric rings. We could equally well find it as the sum of the infinitesimal areas of a bunch of side-by-side isosceles triangles with each of their tips at the center of the disk. Or we could find it as the sum of the area of a bunch of side by side vertical strips running from top to bottom of the disk.

    Let us work the problem. We have a disk of radius r and we want to compute its total area.

    First the concentric ring approach. We want to add up the areas of a bunch of thin rings. Each ring is at a radius x from the center of the disk and is of infinitesimal width, "dx". If we unroll one of these rings to a rectangle, its area is its length (##2 \pi x##) times its width (##dx##). So our formula is

    [tex]A = \int_0^r 2 \pi x \, dx[/tex]
    The ##2pi## term is a constant. We can take that out of the integrand and get
    [tex]A = 2 \pi \int_0^r x \, dx[/tex]
    That integral is easy to evaluate. ##\int x \, dx = \frac{x^2}{2}##. We are evaluating a definite integral (the bottom and top of the evaluation range are specified). The limits of integration are 0 and r, so the formula becomes:
    [tex]A = 2 \pi ( \frac{r^2}{2} - \frac{0^2}{2} )[/tex]
    This obviously reduces to
    [tex]A = \pi r^2[/tex]

    Now the triangle approach. We want to add up the areas of a bunch of narrow isosceles triangles. Each triangle has one vertex at the center of the disk and its other two vertices on the rim of the disk separated by an infinitesimal distance dx and at a distance x from an arbitrary starting point. [Think about measuring x along the rim of the disk with a flexible measuring tape]. Each of these triangles has an area of one half base (##dx##) times height (r). We want to add up triangles until the distance measured on the rim gets back to the starting point, a distance ##2 \pi r## away. So our formula is
    [tex]A = \int_0^{2 \pi r}\frac{r}{2} \, dx[/tex]
    Our variable of integration is x. ##\frac{r}{2}## does not depend on x. It behaves like a constant. We can pull it out of the integrand and get
    [tex]A = \frac{r}{2} \int_0^{2 \pi r}1 \, dx[/tex]
    That's a trivial integral. It reduces to the difference between the upper limit of integration and the lower. So our formula becomes
    [tex]A = \frac{r}{2} ( 2 \pi r - 0 )[/tex]
    Which simplifies to
    [tex]A = \pi r^2[/tex]
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