Newton's Third Law for Normal Force

We're learning about normal forces and I came up with the following scenario. Its conclusion doesn't make sense, and I was wondering where I went wrong in my assumptions?

If you put a book on a table, the table exerts a normal force on the book, the book in turn exerts a normal force on the table (Newton's Third Law). If the normal force exerted on the table by the book were NB, the weight of the table were WT, and the weight of the book were WB, would the force that the table exerted on the ground be NB + WT + WB?

This really confused me because the force exerted on the ground by the table would increase by twice the weight of the book, which makes no sense.

If you draw a free body diagram for the book, the forces acting on the book are its weight (downward gravitational force) $W_B$ and the upward force exerted on the book by the table $N_B$. So the force balance on the book is $$W_B-N_B=0$$

If you draw a free body diagram for the table, the forces acting on the table are the downward force of the book $N_B$, the weight of the table (downward gravitational force) $W_T$, and the upward contact force of the ground $W_G$. So, the force balance on the table is $$N_B+W_T-W_G=0$$

When you do a force balance on a body, you include only the forces exerted on the body by other bodies, not the forces exerted by the body on other bodies. A free body diagram on a body helps you immeasurably in identifying the correct forces to include.

• Thanks a lot! The last part "When you do a force balance on a body, you include only the forces exerted on the body by other bodies, not the forces exerted by the body on other bodies." REALLY helped my understanding a lot. Appreciate it! – Andi Gu Sep 30 '16 at 0:31
• I feel this is the correct solution to the OP's issues. – Suzu Hirose Sep 30 '16 at 0:41

No, the force exerted downwards is just the weight of the table and book together. What you called N_B is W_B.