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I'm confused about the motion on two blocks sliding one over the other when more that one force is acting on them.

Consider two blocks $1$ and $2$, with different masses, with $2$ placed over $1$. There is friction between the blocks.

Suppose that two different forces (of a generic magnitude) $F_1$ and $F_2$ are applied to the two blocks in opposite directions, in the picture that is situation $(1)$.

What will the forces of friction on the masses be?

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I included two diagrams $(2)$ and $(3)$ in the picture. In $(2)$ there is only one action reaction pair, called $F_A$, due to friction between the two blocks, but is it the only one?

I mean the blocks are trying to slide in opposite directions and each one tries to bring the other behind, so shouldn't there be a second pair of forces, due to friction too, as in diagram $(3)$, where it is named $F_B$?

If the correct diagram is $(2)$ how can one determine the directions of the only one pair? If I got that right, friction (the pair of forces in question) is always opposing relative motion between two masses. So must I determine the direction of relative motion first and then draw the pair in the opposite direction?

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3 Answers 3

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If the top block is sliding to the left relative to the bottom block, or if it would slide to the left if there was zero friction, then the friction on the top block due to its interaction with the bottom block would be to the right, as you have shown it. Newton's third law says that there must also be a force on the bottom block to the left. This also agrees with the frictional interaction which actually produces that force. diagram 2 is correct, conditional on the beginning "if"s in my first sentence.

The reason 3 is wrong is because forces always come in pairs, and you don't double the pairs. Force on top because of bottom is the partner of force on bottom because of top. They are a simultaneous pair. Calling them action-reaction is misleading at best, because it sounds like one is causing the other. That is a bad conclusion.

Yes, the direction of friction is determined by going opposite the relative sliding of two objects across each other.

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First isolate block 1 and 2 as a system; ignore the floor the bottom block is on. If the force applied to the block on top is to the left, friction simply acts toward the right because it opposes motion. Simple. Also, the action-reaction pair of the applied force $$F_1$$ (the one acting on the top block)is not friction. It is the block pulling back on whatever supplies the force.

Now isolate the bottom block (block 1) and the floor. There is a force applied on it toward the right. Using the same logic as above, friction acts toward the left, assuming friction exists between block 1 and the floor.

Now, as we can see with the isolation of the systems, block 2 (top) will simply fall off if $$F_1$$ exceeds the force of friction.So don't overthink that. In diagram (3) $$F_b$$ does not exist, which you should now know following our analysis.

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The direction of the frictional force is such as to oppose relative motion.

In diagram 1 the top block wants to go to the left relative to the bottom block.
So the frictional force on block 1 due to block 2 is to the right.
With a similar argument you also conclude that the frictional force on block 2 due to block 1 is to the left.
These two frictional forces are a Newton third law pair and so are equal and opposite.

The fact that each block has an external force acting on it does not matter it is the possibility of relative motion or the actual relative motion which is important. The frictional force will act to oppose that relative motion.

If F1 did not exist the frictional force directions would still be the same.

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