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My question is that when we apply a force equal to $F$ to an object and $F$ is greater than $F_{s,\text{max}}$, then it is clear that the object moves. But with what acceleration? It moves with an acceleration of $(F-F_{s,\text{max}})/m$? So what happens to kinetic friction? Should we subtract the result of static friction with $F$ from kinetic friction? Well, we cannot say that F must only counteract $F_{s,\text{max}}$ to move because when $F_{s,\text{max}}$ decreases from $F$, now we have $F_k$ that prevents movement. So in order for an object to move, it must counteract $F_{s,\text{max}}+F_k$ to be able to move, but this does not seem logical. So how will the calculation be?

naturallyInconsistent
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Isumi
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3 Answers3

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You have so many different $F$ in your question that it is confusing to read for such a simple question.

Your $F_{s,\text{max}}$ should be replaced by $\mu_s N$ and $F_k$ should be replaced by $\mu_kN$, as are standard notation. The relationship then is given by $0<\mu_k<\mu_s$ and that succinctly covers everything.

Now because the maximum value of static friction is $\mu_sN$, this means that whenever $F\leqslant\mu_sN$ the object does not move at all.

However, once the object moves, the acceleration is clearly $a=\dfrac{F-\mu_kN}m=\dfrac{F-F_k}m$ and static friction completely disappears from any moving scenario, as the name ``static" requires it to! Your thing about $F_{s,\text{max}}+F_k$ is just silly.

naturallyInconsistent
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Since the kinetic friction force is generally less than the limiting static friction force, because $\mu_{k}\lt \mu_s$, if the applied force is maintained at the level that caused motion to start (equal to the max static friction force), the net force will be $\mu_{s}N-\mu_{k}N$, where the term $\mu_{s}N$ is the applied force, not the static friction force which ceases once motion starts. The acceleration $a$ will be

$$a=\frac{(\mu_{s}N-\mu_{k}N)}{m}$$

Hope this helps.

Bob D
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2nd Law of Newton still applies. Get the net loading (sum of the loads with proper signs) and use it to calculate the acceleration.

So you have an applied force $F_a$ and some frictional force $F_f$ on the contact surface.

Newtons 2nd law states that $$F_a - F_f = m a \tag{1}$$

but the above is not sufficient to solve the problem as both $a$ and $F_f$ are not known.

You need an assumption

  • Assume sticking, if the body are assumed to be stuck to the ground then $a=0$ is to find that $$F_a - F_f = 0 \tag{2}$$ and thus $F_f = F_a$

  • If $F_f > F_{\rm s,max}$ then the above assumption cannot be true and now you have to assume slipping. Here $F_f = F_{\rm s,max}$ and thus $$ F_a - F_{\rm s,max} = m a \tag{3}$$ and thus $a = \frac{F_a - F_{\rm s,max}}{m}$

jalex
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