Normal force is the component of a contact force that is perpendicular to the surface that an object contacts.
Above is the Wikipedia definition of Normal Force. The confusion is regarding how the following scenario satisfies Newton's Third Law.
The scenario is:
A book is placed on a table and the following forces are involved :
- Gravitational force by earth on the book.
- Gravitational force by the book on earth.
- Normal force by table on the book.
- The push on the table by the book.
My confusion is that the force the book experiences due to gravity is the same force experienced by the table due to push from the book, same as in its the same force, not just the magnitude. So there are only three forces acting. There isn't action reaction pair for the normal force.
So, how does this work?
Best Answer
They are not the same force. See the free body diagrams (FBD) for the book and table below.
The force the book experiences due to gravity is $F_1$. It is not a contact force. In Newtonian mechanics it's a force experienced due to a field, in this case the local gravitational field. The force experienced by the table due to push from the book is $F_3$. It is a contact force (the result of electrical forces between atoms). The two forces happen to have the same magnitude only because the book is in equilibrium.
The forces shown in the FBD's are as follows:
$F_1$ = the force of gravity on the book = $M_{B}g$
$F_2$ = the normal force of the table on the book which, since the book is in equilibrium, = $M_{B}g$
$F_3$ = the normal force of the book on the table = $M_{B}g$
$F_4$ = the force of gravity on the table = $M_{T}g$
$F_5$ = the total normal force of the Earth on the table which, since the table is in equilibrium = $F_{3}+F_{4}=M_{B}g+M_{T}g$
$F_6$ = the total normal force of the table on the Earth = $F_{3}+F_{4}=M_{B}g+M_{T}g$
Not shown is the gravitational force that exists between the book and table. This force is infinitesimal compared to all the other forces.
The two Newton 3rd law equal and opposite forces pairs are $F_{2},F_{3}$ and $F_{5},F_{6}$, circled in red.
The forces acting on the book and the table for the application of Newton's 2nd law forces are circled in blue. The equilibrium requirements for the book and table are, respectively
$F_{1}+F_{2}=-M_{B}g+M_{B}g=0$
$F_{3}+F_{4}+F_{5}=-M_{B}g-M_{T}g+(M_{B}g+M_{T}g)=0$
Hope this helps.