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How are the forces on a resting book different from the action-reaction pair between the book and the table?
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mechanics
Figure 1: (A) The vector sum C = A + B = B + A. (B) The vector difference A + (−B) = A − B = D. (C, left) A cos θ is the component of A along B and (right) B cos θ is the component of B along A. (D, left) The right-hand rule used to find the direction of E = A × B and (right) the right-hand rule used to find the direction of −E = B × A.
In the study of mechanics, it is important to distinguish between the forces that maintain an object's equilibrium and the forces defined by Newton’s third law.
For a book resting on a table, two distinct sets of forces are at play:
- Equilibrium Forces: The upward force exerted by the table on the book is balanced by the downward force of gravity acting on the book. Because these two forces act on the same body (the book) and are equal and opposite, they result in a net force of zero, keeping the book in an immobile state.12
- Action-Reaction Pair: The action and reaction pair consists of the downward force the book exerts on the table and the equal, opposite upward force the table exerts on the book. These forces act on different bodies (one on the table, one on the book) and are governed by Newton’s third law. They do not balance each other out because they are not applied to the same object.12