Problem — Book on a Table (Action–Reaction Pairs)

Problem

A book rests on a table. Draw the book’s interaction diagram, identifying the action–reaction pairs, and explain why the weight of the book and the reaction from the table do not form a third-law pair, even though they have the same magnitude and opposite direction.

Solution

Step 1 — The book’s interactions. The book takes part in only two interactions:

• with the Earth, via the gravitational force;
• with the table, via contact.

Each interaction corresponds to an action–reaction pair, that is, two forces equal in magnitude, opposite in direction and applied to the two different bodies that interact.

Step 2 — The two third-law pairs.

• Gravitational interaction. The Earth attracts the book downward with the weight $\vec P$ (acting on the book). Its twin is the attraction the book exerts on the Earth, $-\vec P$, directed upward and applied to the Earth.
• Contact interaction. The table pushes the book upward with the normal reaction $\vec R$ (acting on the book). Its twin is the force with which the book presses on the table, $-\vec R$, directed downward and applied to the table.

Step 3 — Why $\vec P$ and $\vec R$ are not an action–reaction pair. The weight $\vec P$ and the reaction $\vec R$ both act on the same body, the book. A third-law pair, by contrast, has its two forces on different bodies. So $\vec P$ and $\vec R$ cannot form an action–reaction pair.

The fact that they are equal in magnitude and opposite in direction is the consequence of a different law: Newton’s second law applied to the stationary book. Since the acceleration is zero, the resultant of the forces on the book must vanish, $\vec P + \vec R = \vec 0$, hence $\vec R = -\vec P$. It is equilibrium (the second law), not the third law, that imposes the equality of the magnitudes.

$\ev{\vec P \text{ and } \vec R \text{ act on the same body} \Rightarrow \text{not an action–reaction pair}}$

Links

Topics: Dynamics Concepts: Newton’s third law Skills: Interaction diagram