0
$\begingroup$

A particle of mass $m$ is projected up a plane that is inclined at an angle $\alpha$ to the horizontal. At $t=0$, its velocity is $v_0$ and the coefficient of dynamic friction of the slope is $\mu$. Show that the particle comes to rest at time $T = \dfrac{v_0}{g(\mu \cos\alpha + > \sin\alpha)}$

I remember doing these kinds of questions at A level but I'm pretty stuck here. I drew a diagram and considered the forces acting on the particle:

$m\vec{g} = -mg\sin\alpha\hat{i} - mg\sin\alpha\hat{j}$ (Taking the upwards direction to be positive)

$\vec{v_0} = v_{0}\sin\alpha\hat{i} + v_{0}\cos\alpha\hat{j}$

$\vec{R} = R\hat{j} = \mu mg\cos\alpha$

$\vec{F} = \mu\vec{R}$

That's about as far as I can get at this point!

$\endgroup$
6
  • $\begingroup$ Double check your expression for $m\vec g$. The normal force is $\mu mg\cos \alpha$. Also, $v_0$ is the initial velocity in the answer. You have it as the velocity function. $\endgroup$
    – John Douma
    Apr 25, 2013 at 19:47
  • $\begingroup$ I'm not sure how to go about finding $T$ given the forces... $\endgroup$
    – Mathlete
    Apr 25, 2013 at 19:50
  • $\begingroup$ Draw the incline plane. This is a one dimensional problem. The angle $\alpha$ is the angle between the weight vector and the normal vector. The $\sin \alpha$ is the component along the plane and the $\cos \alpha$ is the component perpendicular to the plane. $\endgroup$
    – John Douma
    Apr 25, 2013 at 19:51
  • $\begingroup$ Yep I've done that, I think the forces are all now correct? $\endgroup$
    – Mathlete
    Apr 25, 2013 at 19:52
  • $\begingroup$ You can divide to get rid of the mass and you end up with $\vec g=\ddot{x} = -g\sin \alpha - \mu g\cos \alpha$. Integrate that to get the velocity. Then $v_0$ will be the initial velocity. Set $t=0$ and solve. $\endgroup$
    – John Douma
    Apr 25, 2013 at 19:57

1 Answer 1

0
$\begingroup$

You have correctly set up your forces as $$ma = -mg\sin \alpha -\mu mg \cos \alpha$$.

We use the fact that $a=\ddot{x}$ and divide by $m$ to get $$\ddot{x}=-g\sin \alpha - \mu g\cos \alpha$$.

Integrating with respect to $t$ we get

$$v = \dot{x} = (-g\sin \alpha - \mu g\cos \alpha)t + v_0$$

Finally, setting $v=0$, we get

$$0=(-g\sin \alpha - \mu g\cos \alpha)t + v_0\Rightarrow t= \frac{v_0}{g\sin \alpha + \mu g\cos \alpha}$$

$\endgroup$
8
  • $\begingroup$ How would I incorporate the $\hat{i}$ and $\hat{j}$ components? $\endgroup$
    – Mathlete
    Apr 25, 2013 at 20:15
  • $\begingroup$ You don't. This is a one dimensional problem. Even though the normal force is in the $\bar j$ direction, the frictional force is not. $\endgroup$
    – John Douma
    Apr 25, 2013 at 20:17
  • $\begingroup$ OK...I just feel like you've skipped a few steps in the working out. $\endgroup$
    – Mathlete
    Apr 25, 2013 at 20:20
  • $\begingroup$ Please be more specific. I am happy to clarify anything. $\endgroup$
    – John Douma
    Apr 25, 2013 at 20:21
  • $\begingroup$ The first step, how did you get the $\mu$? $\endgroup$
    – Mathlete
    Apr 25, 2013 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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