Question

A 0.79-kg ball is placed on top of a 2.5-kg book which is at the top of a table. What is the magnitude of the force exerted by the table to the two objects?

Answer 1

Let's apply the Newton's Second Law of Motion:

$$\sum F_y=ma_y=0,$$ $$N_{table}-m_{book}g-m_{ball}g=0,$$ $$N_{table}=(m_{book}+m_{ball})g,$$ $$N_{table}=(2.5\ kg+0.79\ kg)\cdot 9.8\ \dfrac{m}{s^2}=32.24\ N.$$

Answer:

$N_{table}=32.24\ N.$