Dynamics: Question 1

Worked solution

Step 1: Recall what a Newton’s third law pair requires

Newton’s third law states that when body $X$ exerts a force on body $Y$, body $Y$ exerts a force back on body $X$ that is:

• equal in magnitude,
• opposite in direction,
• the same type of force (e.g. both contact forces, or both gravitational forces),
• acting on two different bodies (never both forces on the same object), and
• simultaneous (they appear and disappear together).

Step 2: Apply this to the given force

The force described is the table pushing up on the book (a contact/normal force). Its Newton’s third law pair must therefore be:

• the same type of force (a contact/normal force),
• equal in magnitude ($12\text{ N}$),
• opposite in direction (downward),
• acting on the other body in the interaction, the table, not the book.

That is exactly the force in option B: the book pushes down on the table with $12\text{ N}$.

Step 3: Why the other options are wrong

• A (the book’s weight): this force acts on the book itself (not the table), and it is a gravitational force, not a contact force, a different type of force entirely. Its own Newton’s third law pair is actually the gravitational pull the book exerts back on the Earth, not the table’s contact force. The fact that the weight and the normal force are equal here is just a consequence of equilibrium (Newton’s first/second law), not Newton’s third law.
• C (floor on table legs): this is a genuine contact-force pair, and even has a plausible magnitude, but it acts between the wrong pair of bodies (the floor and the table), not between the table and the book as required.
• D (friction on the book): no horizontal force or tendency to slide is mentioned in the scenario, so there is no frictional interaction to pair with anything here; this option is simply the wrong type of force for this situation.

Final answer

• The Newton’s third law reaction pair is $\boxed{12}\text{ N}$ downward, the contact force the book exerts on the table, option B.