I have the following exercise:

A uniform rod of mass $M$ is given a horizontal velocity $v$ on a rough track as shown in the figure. The surface is rough on the right side of the origin $O$ and the surface is smooth and frictionless on the left side of the origin as shown in the figure. Express the velocity of the rod as a function of distance from the origin. Also find the distance before it comes to instantaneous rest. enter image description here

I am not able to deduce what the force of friction on a small length $\mathrm{d}x$ of the rod should be. To get the friction on that part should I consider the normal reaction of that part only or of the whole rod ?

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    $\begingroup$ Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ Apr 25, 2016 at 12:10
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    $\begingroup$ Interesting problem, Hardik. What attempts have you made to solve it? What aspect of the problem is causing difficulty? $\endgroup$ Apr 25, 2016 at 12:43
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    $\begingroup$ A note to prospective close voters: the conceptual question here is in the last two sentences. $\endgroup$
    – David Z
    Apr 25, 2016 at 13:19

3 Answers 3


You are correct : to calculate the friction force, you only need to consider the weight of that part of the rod resting on the rough surface, not the whole of it.

When the block overlaps the rough area by distance $x$, the normal reaction on that portion of the block is $\frac{Wx}L$ and the friction is $F={\frac{\mu Wx}L}$. The work done against friction in moving a short distance $dx$ is $Fdx$. The work done in moving distance $x \le L$ from the start position is ${\frac{\mu Wx^2}{2L}}$. When $x \gt L$ the friction force is $\mu W$ so the work done then is $\mu W(x-L)$.

Work done against friction gradually reduces kinetic energy to zero. The critical point is where $x = L$. If the block stops when $x \le L$ then
\begin{aligned} \frac12Mv^2 &= \frac{\mu Wx^2}{2L} = \frac{\mu Mgx^2}{2L}\\ v^2 &= \frac{\mu gx^2}{L}\\ x &= v\sqrt{\frac L{\mu g}} \end{aligned} If the block stops at $x = L$ then $v_0^2 = \mu gL$. If the block starts with speed $v \gt v_0$ then it will stop where \begin{aligned} \frac 12M(v^2-v_0^2) &= \mu Mg(x-L)\\ v^2 &= v_0^2 + 2\mu g(x-L) = \mu gL + 2\mu g(x-L) = \mu g(2x-L)\\ x &= \frac L2 + \frac{v^2}{2\mu g} \end{aligned}

Summary : If the block starts with speed $v_0 \lt \sqrt{\mu gL}$ then it will stop at after travelling a distance $x ={ v\sqrt{\frac L{\mu g}}= \frac{Lv}{v_0}}$. If it starts with speed $v_0 \gt \sqrt{\mu gL}$ then it will stop at after travelling a distance $x ={ \frac L2+\frac{v^2}{2\mu g} = (1+(\frac v{v_0})^2)\frac L2}$.


Imagine there are two objects instead of one. One object has a certain mass, and experiences friction. The other object has another mass, and no friction. If those two objects were joined together, you would have no difficulty figuring out the equation of motion.

But actually the problem is harder than it looks: the way you have drawn it, the object has finite height; this means that there will be a torque as the object decelerates, and this torque will increase the normal force on the leading edge of the object. But unless the height of the object is given, you cannot solve for that.


Actually there is a normal reaction equal to $\frac{W}{2}$ on the two edges. One causes sliding friction of $\mu \frac{W}{2}$ and the other doesn't.

The amount of material under friction does not matter. You can just lump the mass on the two ends and treat them accordingly.

Edit 1 @Floris is correct, the normal reaction is not even because friction causes an applied torque and the contact forces need to react upon it.

Looking at a free body diagram (below) we have the following balance of forces


$$ \begin{align} F - \mu N_1 & = m \ddot{x} \\ N_1 + N_2 - W & = 0 \\ \frac{\ell}{2} N_1 - \frac{\ell}{2} N_2 - \frac{h}{2} \mu\,N_1 & = 0 \\ \end{align} $$

This is solved for the reaction normal forces $$ \begin{align} N_1 & = \frac{\tfrac{\ell}{2}}{\ell - \mu \tfrac{h}{2} } W\\ N_2 & = \frac{\tfrac{\ell - \mu h}{2} } {\ell - \mu \tfrac{h}{2}} W \end{align}$$

So the frictional force $\mu N_1$ does not depend on the position of the block over the rough surface.

  • $\begingroup$ When the block is not supported at all by the rough area of the track, there is no friction; when it is supported completely by the rough area the friction is μW. The amount of friction depends on the portion of the block on the rough area. $\endgroup$ May 2, 2016 at 2:04
  • $\begingroup$ I think this is wrong. If the entire rod is in contact with the surface, then the normal force should be apportioned in proportion to the contact area - not evenly divided between "has friction" and "does not have friction". Although in reality, the height of the object should matter. As the friction generates torque, the normal force on the leading edge will increase... $\endgroup$
    – Floris
    May 2, 2016 at 3:12
  • $\begingroup$ @Floris I know quite a lot about contact pressure distributions and in this situation the reactions are concentrated on the edges. Similar to physics.stackexchange.com/a/253337/392. Only if the block had curvature the pressure distribution would be "even". $\endgroup$ May 2, 2016 at 12:29
  • $\begingroup$ I looked at your other answer. I agree that in the case of elastic deformation of the substrate / contact interface, there will be a (near) singularity at the corners, but it's not clear to me that the force will be evenly distributed when the rod is sliding with friction on only one corner. Do you agree with me that if there is any friction, the resulting torque will affect the force distribution? Incidentally - it wasn't my downvote... $\endgroup$
    – Floris
    May 2, 2016 at 13:15
  • $\begingroup$ Yes, friction would create an applied torque, which would need to be reacted upon from the load distribution. On the other hand, we don't know where the force to push the block is applied and if it creates a torque also. So there isn't enough information to really solve this. $\endgroup$ May 2, 2016 at 14:01

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