I have the following problem (NO, this is not to ask for a solution to this problem).

A block is placed on a plane inclined at an angle $\theta$. The coefficient of friction between the block and the plane is $\mu=\tan{\theta}$. The block is given a kick so that it initially moves with speed $V$ horizontally along the plane (that is, in the direction perpendicular to the direction pointing straight down the plane). What is the speed of the block after a very long time?

The solution to this problem does not require explicit equations for the components of velocity. But I was wondering if it is possible to derive such equations.

I attempted to find the components of the force vector and managed to write those components in terms of the horizontal and vertical components of velocity at any given time $t$. Upon integration of those equations, I obtained two complicated explicit equations relating the components of velocity and time, though I am not sure they are correct.

More generally, my question is is there a nice form for $\mathbf{v}(t)$ when $\mathbf{F}$'s direction changes with $\mathbf{v}$?


1 Answer 1


I do not know how to solve this in a common case, but the 1-dimensional case and in the special case , when the force along x-axe is depended only on the speed along this axe, i.e $$\textbf{F}(\textbf{v}) = f_x(v_x)\textbf{i} + f_y(v_y)\textbf{j} +f_z(v_z)\textbf{k}$$ In that special case it can be solved as: $$m\frac{dv_x}{dt} = f_x(v_x) \Rightarrow \frac{dv_x}{f_x(v_x)} = \frac{dt}{m}$$ $$\frac{t}{m} = \int \frac{v_x}{f_x(v_x)}$$ And the same for $v_y$ and $v_z $ $$\frac{t}{m} = \int \frac{dv_y}{f_y(v_y)}$$ $$\frac{t}{m} = \int \frac{dv_z}{f_z(v_z)}$$ from this you can find $v_x$, $v_y$ and $v_z$ as functions of t.

Remark: Another special case when it can be solved is when F can be represented as $$\textbf{F} = A\textbf{v}$$ where A is 3x3 constant matrix

  • 2
    $\begingroup$ In the general case when $f_x=f_x(v_x,v_y,v_z)$ and same for $f_y$ and $f_z$ your answer is still valid, the only difference being that the integral in that case would be multivariate. The integral need to be integrated by considering the other variables as constant. $\endgroup$
    – Richard
    Jul 10, 2019 at 6:25
  • $\begingroup$ @Richard Yes, you are right....It somehow didn't occur to me $\endgroup$ Jul 10, 2019 at 8:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.