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Finding force vectors of gravity along a curve

  1. Nov 8, 2015 #1
    hello.

    I'm looking for some general guidance as to which methods to use where.

    i'm wanting to find the force vectors on an object as it moves along a general curve in a space (imagine a roller coaster). for example the parametric curve.

    z=t
    x=cos(t)
    y=sin(t)


    I know i need to find the partial derivatives of the curve to find the gradient vector and the directional derivative. finding that isn't an issue.

    i'm stuck now where i keep wanting to jam the vector R= (R+VT+1/2AT^2)<i,j,k>
    head on with the R(t) vector for the curve.

    and because of that i can visualize vector components of gravity. i think its one of those things that's right in front of me and i'm just not seeing it. most likely the changing acceleration is what's throwing me off.

    once i have F(g)(x,y,z) and the direction, that will give me F(n) and f(k). and from that all i need is (S) to find the work done. after that there's F(c) which i'm expecting to also be a pain.


    i'm in the process of coding this and all that's left to finish is this bit.

    thanks.

    chris
     
  2. jcsd
  3. Nov 8, 2015 #2

    mfb

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    If acceleration is not constant, that equation does not work. You'll have to replace it with an integral.
     
  4. Nov 8, 2015 #3
    I think you might be overthinking this - you've got R(t) = <cos(t),sin(t),t>

    We can immediately say V(t) = dR/dt = <-sin(t),cos(t),1>
    Then a = dv/dt = <-cos(t),-sin(t),0>

    and F=ma, so F=<-mcos(t),-msin(t),0>

    It gets more complicated if you aren't given R as R(t). But for your case that's it.
     
  5. Nov 8, 2015 #4
    Absolutely! There is not much more to say about the acceleration. It gets a little more complicated if you want to relate the acceleration to all the forces. For example, if the z-axis is vertically up, the fact that there is no acceleration in the z-direction means that there is some component of the force from the car or harness or whatever, which is exactly equal to gravity but upwards directed. The rest of that force provides the centripetal force.
     
  6. Nov 9, 2015 #5
    thanks for the help. now this may just be restating my question but its what has me going in circles

    in 2D Fg is split into y= Fg sin(theta) x= Fg cos(theta). for an inclined plane. and theta is just the arctan(dy/dx)

    however for a curve in 3D space Fg is split in to 3 component forces. all of who's combined magnitude would equal Fg.

    the total acceleration in Z is zero be cause of the opposing Fn. however from any point on the curve the only way Fg(z) could be 0 is if the curve were vertical in z. then Fg would only exist in the xy-plane from point reference of the object on the curve. you need Fgz to even know Fn

    this is why i keep thinking i need the partial derivatives of R(x,y,z) to find the force and velocity components.

    and then integrate across the parameter to find the vectors for velocity and acceleration at any given point. because velocity and acceleration are constantly changing.

    thanks again everyone
     
  7. Nov 9, 2015 #6

    sophiecentaur

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    There are two causes of acceleration on the roller coaster. There is g, which is always in the same direction and the centripetal acceleration, which will be towards the centre of the curve (varying all the time) and depending on speed and curvature.
    I was just wondering whether you could get away with using the Energy of the system, which would give you the speed at all points of the curve, based on loss of mgh. That could make life easier. But perhaps you want to solve it parametrically.
     
  8. Nov 9, 2015 #7
    chriss85: I need some clarification. I am quoting a part of your message: "the total acceleration in Z is zero be cause of the opposing Fn. however from any point on the curve the only way Fg(z) could be 0 is if the curve were vertical in z. then Fg would only exist in the xy-plane from point reference of the object on the curve. you need Fgz to even know Fn"
    What is Fg? If it is the force of gravity, and if z is the vertical direction. then Fg(z) is never zero. It is aleways constant, and equal to mg in the downward direction. Fg would then never exist in the xy plane.
     
  9. Nov 9, 2015 #8

    HallsofIvy

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    No, it is not the force of gravity because there is no mention of "gravity" in the problem, it is just asking for the force necessary to move an object along this curve.
    You said in your first post, "I know i need to find the partial derivatives of the curve to find the gradient vector and the directional derivative". No, you do not need to find, nor can you find, any partial derivatives because everything here is a function of a single variable, t. Given that "x= cos(t)", "y= sin(t)", "z= t" then the "position vector is cos(t)i+ sin(t)j+ tk. Differentiating with respect to t, the velocity vector is -sin(t)i+cos(t)j+ k and the acceleration vector is -cos(t)i-sin(t)j. Since "F= ma" the force necessary to produce that path is the mass of the object times -cos(t)i- sin(t)j.
     
  10. Nov 9, 2015 #9
    chandra: i'm talking about the view from a person on the curve. to them the z-axis is always perpendicular to the curve. which is not always in the same direction as the force of gravity. that means that each component is directed away from the force of gravity by some angle. for example in 2D if Fg=10n for an object and it is on a 45 degree slope then Fgy and Fgx both have a magnitude of approx 7n. so if the angle were 90 degrees the Fgy =0 and Fgx=10n

    I'm using the conservation of energy to check my results. but that only helps if i don't want to and friction in to the problem which is something i want to do eventually.
     
  11. Nov 9, 2015 #10

    mfb

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    Not with the curve given in post 1. The person would have to move in the x/y-plane for that.
    Acceleration is always perpendicular to the z-axis, yes, and you get it by simply differentiating the position vector twice with respect to velocity. Add gravity to the vector to get the apparent acceleration.

    I don't see the problem.
     
  12. Nov 9, 2015 #11
    It was my understanding from the original wording of the problem that the curve given in post 1 was a curve described in terms of a spatially fixed coordinate system, and evidently there is gravity in the problem. I quote one of the sentences from the original problem "and because of that i can visualize vector components of gravity". If the parametric equations given by chriss85 are not for a spatially fixed coordinate system, then the problem is much more complicated. Differentiating the coordinates twice wrt t does not give the acceleration proportional to the external force, since it is not an inertial system. So I think these things must be clarified by chriss85, and then my original question remains: what is Fg? If there is gravity, is the z-axis vertical?
     
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