0
$\begingroup$

Let’s say there is a block of mass $1$ kg resting on a surface. I studied in my textbook that coefficient of static friction is usually greater than the coefficient of kinetic friction for a given pair of surfaces. So let’s say $\mu_s$ = 0.6 and $\mu_k$ = 0.4 being the friction coefficients between the contact surfaces. Here magnitude of $f_{s,max}$ = $6N$, and magnitude of kinetic friction is $4N$.

We start applying a horizontal force of $4N$ on the block, of course it stays at rest because the force is insufficient to overcome $f_{s,max}$ . We increase the applied force to $6N$ and the block is on the verge of sliding. The moment $F_{applied}$ exceeds $6N$, the block starts sliding, i.e it accelerates. But what is the magnitude of its acceleration? My textbook says that we take into account kinetic friction (not static friction) for a moving body. So what is the magnitude of acceleration of the block? Let’s say $F_{applied}$ is $6.00001N$, for example.

Does the block accelerate at $\frac{F_{applied}-6}{m}$ ? Or does it accelerate at $\frac{F_{applied}-4}{m}$ ?

Or will it initially accelerate at $\frac{F_{applied}-6}{m}$ momentarily, and then its acceleration will increase to $\frac{F_{applied}-4}{m}$ ?

Thanks!

$\endgroup$

1 Answer 1

1
$\begingroup$

If the applied force of 6N equals the max static friction force and motion starts, and the applied force of 6N continues to be applied, then the net force on the block will be 6N minus the kinetic friction force of 4N, or a net force of 2N. That will give an acceleration of 2N/m. That's because once motion starts, the friction force immediately drops to the kinetic level of 4N.

What about the exact moment at which motion starts? The instant at which the block begins to slide.

Actually the value of friction at the instant motion starts is undefined. It's the nature of the transition. See the diagram below (based on an article on the Hyperphysics website on friction). Note that for the transition from static friction to kinetic friction the actual friction force is undefined (in the diagram this is shown as $f_{f}$ = ??).

Hope this helps.

enter image description here

$\endgroup$
4
  • $\begingroup$ Thanks a lot. What about the exact moment at which motion starts? The instant at which the block begins to slide. Is $F_{net}$ still 6.00001N - 4N ? Or is it 6.00001N - 6N at that moment? Although I know that kinetic friction comes into the picture as soon as motion starts, it still doesn't make sense that $F_{net}$ would be equal to 6.00001N - 4N at the very instant at which motion starts. It's so counterintuitive because the friction force of 6N was the one that kept the block from sliding $\endgroup$
    – 4d_
    Apr 22, 2020 at 19:14
  • 1
    $\begingroup$ @πtimese Actually the value of friction at the instant motion starts is undefined. It's the nature of the transition. I will include a diagram in my answer to show you that. $\endgroup$
    – Bob D
    Apr 22, 2020 at 19:21
  • 1
    $\begingroup$ @πtimese I have added the diagram. I don't know if you are familiar with statics. But when you look at the shear force on a beam there is a sudden transition in shear when one encounters a concentrated load where the actual value of shear is undefined at the transition. Hope this helps. $\endgroup$
    – Bob D
    Apr 22, 2020 at 19:31
  • $\begingroup$ So value of friction at the moment motion starts is undefined. That’s what I wanted to understand. Looking at the $f$ vs $F_{applied}$ graph, it makes sense that the value of friction at that instant should be undefined because the graph at that instant is a straight line parallel to $y-axis$. I know a little bit about shear forces, but I don’t know enough. Although I can understand (kinda) the way u explained it in terms of shear, in your last comment. Thanks a lot. $\endgroup$
    – 4d_
    Apr 23, 2020 at 7:06

Your Answer

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

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