Solving for velocity for a body with distance varying force

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This discussion addresses the calculation of velocity and displacement for a body subjected to three forces: a constant force, a time-varying force, and a distance-varying force. The method involves treating each force separately using Newton's second law, F=ma, and applying quadrature for the second-order differential equation m(d²x/dt²) = F(x). The solution requires integrating the force functions, leading to the equation v = √(2/m ∫ F(x)dx). The discussion highlights challenges in solving integrals for non-elementary functions, such as those arising from inverse square laws.

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Zexter
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Dear all,
I have three forces acting on a body
1. a constant force
2. a time varying force
3. a distance varying force

So If I need to calculate the velocity and displacement of the body at any particular time what method should I follow?Is there any general procedure to solve such an equation. Since time varying force changes the displacement of the body, the body is made to move in the distance varying force field...


Another bad thing is that I have only discrete values of my distance varying force and not an equation describing it. So it would be helpful if someone gives me a clue to how to solve this problem


thank you
 
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Well, you are not giving us much to go on! Fortunately, F= ma is "linear" so we can treat each force separately then add the accelerations. For constant F, a is just F/m. For F varying with time, we have
[tex]a= \frac{dv}{dt}= \frac{F(t)}{m}[/tex]
so that
[tex]v= \frac{1}{m}\int F(t)dt[/tex]

If F is a function of x, we need to say that [itex]a= d^2x/dt^2[/itex] so we have a second order differential equation:
[tex]m\frac{d^2x}{dt^2}= F(x)[/tex]
Since t does not appear expicitely in that equation, a standard method of solving is "quadrature". Let v= dx/dt so that
[tex]\frac{d^2x}{dt^2}= \frac{dv}{dt}= \frac{dx}{dt}\frac{dv}{dx}= v\frac{dv}{dt}[/tex]
by the chain rule.
That is, the equation is
[tex]mv\frac{dv}{dx}= F(x)[/tex]
so that
[tex]\int mv dv= \frac{1}{2}mv^2= \int F(x)dx[/tex]
and
[tex]v= \sqrt{\frac{2}{m}\int F(x)dx[/tex].

For general F, even if you can integrate F(x), the integral for x can be non-elementary. For example, with an inverse square law, you get what are called "elliptic integrals" (since the integrals are involved in finding the orbits of planets) which cannot be integrated in terms of elementary functions.
 
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