The forces acting on the ladder are shown in Fig. Here, $m=40 \mathrm{~kg} \Rightarrow$ weight $=40 \times 9.8 \mathrm{~N}=392 \mathrm{~N}, A B=$
$$
A C=1.6 \mathrm{~m}, B D=\frac{1}{2} \times 1.6 \mathrm{~m}=0.8 \mathrm{~m} \text {, }
$$
$B F=1.2 \mathrm{~m}$ and $D E=0.5 \mathrm{~m}$,
In the Fig. $\triangle A D E$ and $\triangle A B C$ are similar triangles, hence,
$$
\begin{aligned}
&B C=D E \times \frac{A B}{A D}=\frac{0.5 \times 1.6}{0.8}=1.0 \mathrm{~m} \\
&\therefore \quad W \times(M B)=N_C \times(C B)
\end{aligned}
$$
But $M B=\frac{K B \times B F}{B A}=\frac{0.5 \times 1.2}{1.6}=0.375 \mathrm{~m}$
Substituting this value in (i), we get
$$
N_C=\frac{W \times(M B)}{(C B)}=\frac{392 \times 0.375}{1}=147 \mathrm{~N}
$$


Again considering equilibrium at point $C$ in similar manner, we have
$W \times(M C)=N_B \times(B C)$
$\therefore \quad N_B=\frac{W \times(M C)}{(B C)}=\frac{W \times(B C-B M)}{(B C)}$
$=\frac{393 \times(1-0.375)}{1}=245 \mathrm{~N}$
Now, it can be easily shown that tension in the string $T=N_B-N_C=245-147=98 \mathrm{~N}$.