Suppose there was a small block on top of a larger block on an incline and there is both kinetic friction between the block and the incline and static friction between the blocks. If both blocks accelerated together down the incline, could you just discount the static friction as they cancel out for the system of the two blocks? I was wondering because when Fg acts on the small block, wouldn't the static friction be pointed up the incline but because the block under it is also moving down, wouldn't there also be a static friction force on that block pointed up the incline as well so there for the entire system there would be 2 static friction forces pointed up the incline? But I don't think that makes sense?
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$\begingroup$ As long as the two blocks are moving together you can treat them as a single block with their masses combined. $\endgroup$– Leo AdbergCommented Sep 21, 2020 at 5:48
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$\begingroup$ Is the friction coefficient between the two blocks the same as between the bottom block and the slide? $\endgroup$– user1079505Commented Sep 21, 2020 at 5:53
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$\begingroup$ See this question: physics.stackexchange.com/questions/51444/…. The scenario is nearly identical to yours except for the fact that there is no friction with the ground. $\endgroup$– KingLogicCommented Sep 21, 2020 at 6:05
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Look first of all , let's call the small box $m_1$ and the larger one to be $m_2$ .
In your first case , you are saying that since $m_1$ is experiencing gravity it must be experiencing $friction force$ (say $f_1$) in the direction up the incline (as shown in figure) . But remember here that $m_1$ also applies the same amount of force on $m_2$ in the opposite direction along the length of the larger box.
In your second case , you are saying that since $m_2$ is slipping down, $m_1$ must apply a frictional force (say $f_2$) on it in the direction up to the incline. But by Newton's third law , $m_2$ also applies the same amount of force on $m_1$ in the opposite direction.
So considering them as a system there is no static friction force in the upward direction because in both the cases the action - reaction pairs cancel each other . So for the system there is only $(m_1+m_2)g$ along the incline and kinetic friction force (say $F$) in the opposite direction to the incline.
Note : I have named the friction force as $f_1$ and $f_2$ just to indicate the friction force in the two scenarios although there is only one friction force acting on $m_1$ due to $m_2$. And also all the friction forces act along the surface in contact . So don't mind with the way they are represented in the figure.
Hope it helps ☺️.
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$\begingroup$ There is only one friction force, $f_1$ (static or kinetic) acting up the incline on the upper block (unless the lower block is being forced downward). The reaction force acts down the incline on the lower block. These are associated with equal and opposite normal forces. $\endgroup$ Commented Sep 21, 2020 at 16:24
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$\begingroup$ @R.W. Bird Yes I know that. I have just used $f_1$ and $f_2$ to show the two scenarios which OP mentioned about. I have written that in the Note section. $\endgroup$– AnkitCommented Sep 21, 2020 at 16:44
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