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I'm interested in equations of motion when friction is present for a little graphical side project I am working on. I'm really rusty with physics, so I apologize if this is a basic question or doesn't make sense...

Consider an object on a plane in three dimensions. It has initial position $ \vec{x_0} $ on the plane and initial velocity $ \vec{v_0} $ tangential to the plane. Assume it has acceleration (i.e. downhill gravity), $ \vec{a_G} $, which is constant and tangential to the plane. Now assume it has an additional acceleration (from friction, which I'm modeling as an acceleration), $\vec{a_F}(t) $, with constant magnitude $ k $ but variable direction, opposite that of velocity.

Is there a function $ \vec{x}(t) $ that gives position with respect to time? Something in the vein of the $$ x(t) = x_0 + v_0*t + 0.5*a*t^2 $$ that I learned in high school?

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In one dimension (it is easier that way) we have $$x''=a-kx'\\kx'+x''=a\\x'=\frac ak+ce^{-kt}\\c=v_0-\frac ak\\x=\frac akt-\frac {1}k\left(v_0-\frac ak\right)e^{-kt}+x_0$$

Added: the above misreads the problem to involve a frictional force proportional to the velocity. If the friction force is constant in magnitude, then as long as the velocity is in the same direction as the external acceleration we will just have $$x=x_0+v_0t+\frac 12(a-f)t^2$$

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  • $\begingroup$ Thanks! It makes sense when viewed as a differential equation! Can I generalize to three dimensions by just taking everything componentwise? $\endgroup$
    – Physics Newbie
    Jan 24, 2014 at 1:22
  • $\begingroup$ ..Wait, doesn't $ x'' = a - kx' $ "scale" k in one dimension by $x'$? Shouldn't it be something closer to $x'' = a - k *sign(x') $ ? $\endgroup$
    – Physics Newbie
    Jan 24, 2014 at 1:53
  • $\begingroup$ I took it as friction, which is linear in velocity. You are correct that if the resistant force is constant in magnitude that is what it should be. This makes it discontinuous at zero, but as long as the sign of $x'$ doesn't change, you just take your high school solution, replace $a$ by $(a-k)$ and quit. $\endgroup$
    – Ross Millikan
    Jan 24, 2014 at 2:26
  • $\begingroup$ Well I had derived my scenario given that I can decompose the applied forces on the object and use $ F_k = \mu N $ (which is independent of velocity if I'm not mistaken) on the part normal to the plane and divide by mass to get acceleration. Substituting with $ (a-k) $ won't work with vectors, though, since the direction to apply $ k $ is no longer constant with each component. Thanks for your thoughts, I think I can play around with it some more now. $\endgroup$
    – Physics Newbie
    Jan 24, 2014 at 3:12
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Constant force in the direction opposite to velocity complicates the problem.

Let's define $x$ as the direction directly down the place and $y$ as the horizontally within the plane.

Then:

$$a_{x} = a_{G} -k \hat{v}_{x} $$ $$a_{y} = -k \hat{v}_{y} $$

Now:

$$ \hat{v}_{x} = \frac{v_{x}}{|v_{x}^2+v_{y}^2|} $$

with likewise for $\hat{v}_{y}$.

As you can see, this couples the $x$ and $y$ accelerations. I'd suggest you go with the plan of working with the accelerations directly and just doing step-wise calculations.

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  • $\begingroup$ I see what you're getting at. I was hoping for an explicit formula so I could do some interesting ideas I had with motion and time. Thanks. $\endgroup$
    – Physics Newbie
    Jan 24, 2014 at 4:47

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