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The force diagrams show the weights $W_A$ and $W_B$ of the blocks, directed down. $F_1$ is the force of block $B$ on block $A$, while $F_2$ is the force of $A$ on $B$. Reference:- An introduction to mechanics by Kleppner D And Kolenkow R

My doubt:- I am confused with $F_2$ and $W_A$. From where do this $F_2$ coming? Already $W_A$ is the force acted by $A$ on $B$ and $F_1$ is the reaction force.

Math geek
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1 Answers1

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You are jumping several steps ahead. The force diagrams keep $F_1, F_2$ and $W_A$ separate because they make no assumptions about the blocks being in equilibrium or about Newton's Third Law.

By applying Newton's Third Law, we can see that $F_1$ and $F_2$ are equal and opposite i.e. $|F_1|=|F_2|$. And if block $A$ is in equilibrium then we know that the net force on block $A$ must be zero, so $|F_1| = |W_A|$. If block $B$ is also in equilibrium then the net force on block $B$ must also be zero, so

$|N| = |F_2| + |W_B| = |F_1| + |W_B| = |W_A| + |W_B|$

as we expect.

Newton's Third Law will always hold, so we can always be sure that $|F_1|=|F_2|$. But the blocks are not necessarily in equilibrium. Suppose they are in a lift that is accelerating upwards at constant acceleration $a$. Then

$|F_1| = |W_A| + M_Aa\\|N| = |F_2| + |W_B| + M_Ba = |F_1| + |W_B| + M_Ba = |W_A| + |W_B| + (M_A + M_B)a$

gandalf61
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  • Sir, My question was what is the role of $F_2$ here? – Math geek Apr 13 '20 at 11:42
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    @Mathgeek $F_2$ is the force exerted by block $A$ on block $B$. We know $F_2$ is equal in magnitude to $F_1$ by applying Newton's Third Law, and we know that $F_1$ is in turn equal in magnitude to $W_A$ if we assume block $A$ is in equilibrium. But without these assumptions all we can say is that there is some force exerted by block $A$ on block $B$, and we just call this force $F_2$ to give it a name. – gandalf61 Apr 13 '20 at 12:25
  • Suppose a single block lying on the floor, Why don't we assume some force is acting on the floor. we take only normal force and the weight of the block. – Math geek Apr 13 '20 at 12:29
  • @Mathgeek In that scenario there is a force exerted by the block on the floor, equal and opposite to the normal force exerted by the floor on the block. But we don't often think about that reaction force because we are usually only interested in the forces acting on the block, and not the forces acting on the floor. – gandalf61 Apr 13 '20 at 14:09