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Suppose A and B experience friction with each other but they move on a smooth floor. A is given a force F which is not enough to overcome friction; a frictional force equal to F, thus, acts on B. Now, one would expect both the objects to move together with a common acceleration, due to this friction. In which case, it can be said that

$\frac{F-f}{m_A}=\frac{f}{m_B}={a_A}={a_B}$

But this would imply that the accelerations of A and B are zero as $F=f$ for $F<{f_L}$. So does that mean that a force exerted on A that is less than the limiting value of static friction, will not result in collective acceleration of A and B?

This seems to contradict empirical behavior. For example, in case of a sufficiently heavy object on top of another, the application of a force that is not sufficient for the top object to overcome friction against the bottom object, still results in a collective acceleration.

I have attached a photo containing diagrams for reference.

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Your mistake is assuming that the frictional force that acts on B, which is $f$, is equal to the force $F$ exerted on A. Since A exerts a force $f$ on B, then by Newton's 3rd Law, B exerts an equal and opposite force on A. If $f=F$ then the net force on A is zero and A will not accelerate. But we know that A does accelerate, so we must have $f < F$.

The net force on A is $F-f$, so the acceleration of A is $a_A = \frac {F-f} {m_a}$. Similarly, the net force on B is $f$ and the acceleration of B is $a_B = \frac f {m_B}$. If the blocks move together then they must have equal acceleration so

$a_A=a_B \\ \Rightarrow \displaystyle \frac {F-f} {m_A} = \frac f {m_B} \\ \Rightarrow (F-f)m_B = f m_A \\ \Rightarrow \displaystyle F m_B = f(m_A + m_B) \\ \Rightarrow \displaystyle a_A = a_B = \frac f {m_B} = \frac F {m_A + m_B}$

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  • $\begingroup$ If the force applied is less than the limiting friction, then won't the friction basically act to stop the motion of A? I assumed that A "accelerates" only because it is affixed to B due to resistance from friction, while the reaction to that frictional force pulls B forward. $\endgroup$ Oct 4, 2022 at 5:00
  • $\begingroup$ @SashankSriram A accelerates because the force $F$ is applied to it. B accelerates because the friction force $f$ acts so as to reduce relative motion between A and B. The reaction $-f$ to the friction force acts on A to reduce its acceleration. But if B moves faster than A then friction would act in the opposite direction, to accelerate A and decelerate B. $\endgroup$
    – gandalf61
    Oct 4, 2022 at 8:10
  • $\begingroup$ My question is, won't f basically stop the relative motion between A and B due to F, as F is lesser than the limiting value of static friction (refer diagram where I have specificed F < $f_L$)? Your explanation makes sense if the friction acting is kinetic, but in this case it is static. So it is going to increase with F until the limit, and any relative motion between A and B is only going to happen beyond this point. $\endgroup$ Oct 4, 2022 at 10:54
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    $\begingroup$ @SashankSriram Yes, $f$ will stop any relative motion between A and B. And if there is no relative motion between A and B then they must have the same acceleration, which is $\frac F {m_A+m_B}$. $\endgroup$
    – gandalf61
    Oct 4, 2022 at 11:01
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    $\begingroup$ @SashankSriram No, no, no. I think you are misunderstanding the term "relative motion". No relative motion between A and B just means that each is stationary relative to the other. They can - and they do - both still accelerate, as long as they both accelerate at the same rate. Since A is accelerating then $f$ is not equal to $F$. $\endgroup$
    – gandalf61
    Oct 4, 2022 at 13:39

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