(a) Finding the Number of Books
Let $n$ be the total number of books. The mass of one book is $m = 0.4$ kg. The horizontal clamping force is $F = 120$ N.
Condition 1: Slipping at the Hands
The friction from the hands supports the entire weight of the stack.
Total weight $W = nmg$.
Maximum friction available = $2 \times \mu_{hb} \times F$ (two hands).
$$ nmg \le 2 \mu_{hb} F $$
$$ n(4.0) \le 2(0.40)(120) $$
$$ 4n \le 96 \implies n \le 24 $$
Condition 2: Slipping between Books
The “weakest” interface is between the 1st and 2nd book (from either side). The friction here supports the weight of the inner $(n-2)$ books.
Weight of inner books $W_{in} = (n-2)mg$.
Maximum friction available = $2 \times \mu_{bb} \times F$.
$$ (n-2)mg \le 2 \mu_{bb} F $$
$$ (n-2)(4.0) \le 2(0.25)(120) $$
$$ 4(n-2) \le 60 \implies n-2 \le 15 \implies n \le 17 $$
Conclusion: Since slipping occurs internally at a lower number of books ($n=17$) than at the hands ($n=24$), the maximum number of books is 17.
(b) Friction between 3rd and 4th Book
We consider the system of the central books situated between the 3rd and 4th book on the left, and the corresponding interface on the right (between 14th and 15th book).
These two interfaces support the weight of the inner books.
Total books $n=17$. Removing 3 from each side leaves $17 – 6 = 11$ books in the middle.
Let $f$ be the friction force at one interface (between 3rd and 4th book).
$$ 2f = (\text{mass of 11 books}) \times g $$ $$ 2f = 11 \times m \times g $$ $$ 2f = 11 \times 4.0 = 44 \, \text{N} $$ $$ f = 22 \, \text{N} $$