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Force as a function of position, given x(t)

  1. May 15, 2015 #1
    Let me make this clear, this is not a homework question, I'm just curious. So I know a common question is finding velocity as a function of time given force as a function of position. I was curious, however, as to go backwards, starting with position as a function of time, and then finding force as a function of position. Is it possible or is it something that needs to be measured, I was interested in calculating the work done by a force that was not explicitly given as a function of position, I guess you could say I'm trying to derive a way to calculate work given a position as a function of time, or even given force as a function of time. I'm assuming initial conditions are initial velocity is at rest and the acceleration increases at a linear rate, so the force is not constant. This is my first post here, so please try not to misconstrue this.
     
  2. jcsd
  3. May 15, 2015 #2
    Hi Helixkirby and welcome to PF,
    Yes it's allways possible to derive force as function of position, but let's try to make it easier, re-arrange everything up and solve for position to get it as a function of force, but what is force, according to newton second's law : F = ma.
    Knowing that acceleration is the second order derivative of position, you have got a 2nd ODE which is fairly hard to solve and as it is a 2nd ODE, the initial condition is position and velocity, being more clear, let's name X position, m mass and A acceleration
    X = f(mA), if we assume f doesn't depend on the mass you have X = m*f(A) giving X = m*f(X'') good luck solve that, this does not directly answer your question but it's like that, if and object moves at a force field, that position is linked to a force which itself is a function of mass, you'll still have a 2nd ODE but this time not in the form i've written above, So i hope I've made it a little bit clearer :) .
     
    Last edited: May 15, 2015
  4. May 19, 2015 #3
    Sorry for the really late reply, but I was looking at a similar problem, where someone was changing v(t) into v(x), I haven't tried it yet, but I think that since there is a varying force that means da/dt =/= 0, then I can say da/dx=(da/dt)/(dx/dt), so then I just separate variables and integrate to find a(x) right? Assuming da/dt=k, a=kt, v=.5kt^2, x=(1/6)kt^3, assuming all initial conditions are zero, then I'd just say da=k/.5kt^2 dx, dx/dt=v dx=v dt, so da=2v/t^2 dt= 2(.5kt^2)/t^2 so da=k dt, now I'm stuck at the beginning, how do I solve da=k/.5kt^2 dx to give me a(x), or is it right to say that a=kx/.5kt^2?
     
  5. May 19, 2015 #4
    Hello, It sound like a good question to me, It's in fact very challenging, So
    da/dx = k/(.5kt2)
    d2v/(dx)2 = k/v, and you get one hell of a solution involving exp and error functions, I've managed to get it for you when initial conditions set to zero, v(t) = e^(-erf-1(-2x2/π))
    Hope i've made slightly bit clearer and good luck
     
  6. May 21, 2015 #5
    Ok so I don't really know how to use latex so this is going to look nasty, but here's my attempt at doing this

    So, I've been thinking about this some more, you know how there's a kinematic equation that relates velocity to position, what would that look like for a situation with a constant jerk, then I could differentiate that using the chain rule to find acceleration as a function of position, so I was thinking, I'll let all initial conditions be a(0)=0, v(0)=0, and x(0)=0, if j(t)=k, then a(t)=kt, and v(t)=.5kt^2, lastly x(t)=(1/6)kt^3, how would I find v(x) where x is position, would I just solve x(t) for t and plug it into v(t), so t=cbrt(6x(t)/k), then v(x)=.5k(6(x(t))/k)^(2/3), so then a=(dv/dx)*(dx/dt)=v*dv/dx. dv/dx=.5cbrt(k)*cbrt(36)*(2/3)/cbrt(x)=(1/3)cbrt(36k/x). So a=(1/6)cbrt(36k)cbrt(36k/x)=cbrt(1296k^2/216x)
    =cbrt(6k^2)/cbrt(x)=a(x), so then F(x)=mcbrt(6k^2)/cbrt(x) so then work would just be mcbrt(6k^2)S from 0 to x, of x^-(1/3)= mcbrt(6k^2)*(3/2)x^(2/3). Let me know if you can follow this and if I did it right, cbrt means cube root if it's not obvious.
     
  7. May 22, 2015 #6
    My handwriting is atrocious, but I tried to write it out, maybe this is easier to follow? a(x).jpg
     
  8. May 23, 2015 #7
    $$ a(0) = 0 ,\, v(0) = 0 ,\, x(0) = 0 $$$$a(t)=kt ,\, v(t)=\frac{1}{2}kt^2 ,\, x(t) = \frac{1}{6}kt^3 $$$$t=\left(\frac{6x}{k}\right)^{1/3} \Rightarrow v(x) = \frac{1}{2}k\left(\frac{6x}{k}\right)^{2/3} $$
    $$ a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v \Rightarrow \frac{dv}{dx} = v\frac{dv}{dt} = 2\frac{d(v^2)}{dt} $$
     
  9. May 23, 2015 #8
    The procedure is quite simple. If you have the function ##x(t)##, then ##F(t)=m## ##\frac{d^2(x(t))}{dt^2}## (Newton's second law). If a closed form inverse of the original function exists, i.e. ##t(x)## is expressible in closed form, then substituting this value of t into ##F(t)## will give you ##F(x)## . In most cases though, an inverse does not exist. In such scenarios, try to see if the second order derivative of ##x(t)## is related to the original with a constant. You should then be able to make a substitute and find ##F(x)##, although this is unlikely to be the case if ##x(t)## does not entirely consist of trigonometric/exponential functions.

    If none of this works, then ##F(x)## does not exist.
     
  10. May 23, 2015 #9
    Man, that latex, it just looks beautiful, so I could have just said t(x)=cbrt(6x/k) and then substituted it into F(t)=mkt

    which makes mk(6x/k)^1/3=(mk^(2/3))/(x^1/3)

    Also, in Theodore's post, I'm confused as to what you said at the last part, I understand why dv/dt=(dv/dx)(dx/dt), but what did you do after that?
     
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