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\begin{center}
    \begin{tikzpicture}[scale=2]
        \foreach \i in {1, 2, ..., 3} {
         \draw (\i, 0) -- (\i, 3) -- (\i+1,3) -- (\i+1, 0) -- cycle;
        }

\draw[-latex, blue, ultra thick] (1, 1.5) -- ++(0.4, 0) node[above] {$Q$};
        \draw[-latex, blue, ultra thick] (1, 1.5) -- ++(0, 1) node[above left] {$\frac{nW}{2}$};
        \draw[-latex, blue, ultra thick] (2, 1.5) -- ++(-0.4, 0) node[above] {$Q$};
        \draw[-latex, orange, ultra thick] (2, 1.5) -- ++(0.4, 0) node[above] {$Q$};
        \draw[-latex, blue, ultra thick] (2, 1.5) -- ++(0, -0.8) node[below left] {$\frac{(n-2)W}{2}$};
        \draw[-latex, orange, ultra thick] (2, 1.5) -- ++(0, 0.8) node[above right] {$\frac{(n-2)W}{2}$};
        
        \draw[-latex, blue, ultra thick] (1.5, 1.5) -- ++(0, -0.5) node[below] {$W$};

\end{tikzpicture}
\end{center}

The force acting vertically on each of the outer books must be (by symmetry) $\frac{nW}{2}$.

The force acting horizontally on the outer books (and between each book in the horizontal direction) will be the same (we might as well say $Q$ since increasing this force doesn't make any task less achievable.

Looking at an end book, it will have force $\frac{nW}{2}$ acting on one side, but it this force needs to not slip, ie $\frac{nW}{2} \leq \mu Q$

\begin{align*}
&& \frac{nW}{2} &\leq \mu Q \\
\Rightarrow && n &\leq \frac{2\mu Q}{W}  \\
&& \frac{nW}{2} & \leq P \\
&& n & \leq \frac{2P}{W} \\
\Rightarrow && n &\leq  \frac2{W}\min \left (P, \mu Q  \right)
\end{align*}

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