$N_{ B }=$ Force exerted on the ladder by the floor point $B$

$N_{ C }=$ Force exerted on the ladder by the floor point $C$

$T=$ Tension in the rope

$BA = CA =1.6 m$

$DE =0.5 m$

$BF =1.2 m$

Mass of the weight, $m=40 kg$

Draw a perpendicular from A on the floor $BC$. This intersects $DE$ at mid-point $H$.

$\triangle ABI$ and $\triangle AIC$ are similar

$BI = IC$

Hence, $I$ is the mid-point of $BC$.

$DE \| BC$

$BC =2 \times DE =1 m AF =$

$BA – BF =0.4 m \ldots(i)$

$D$ is the mid-point of $AB$.

Hence, we can write:

$A D=\frac{1}{2} \times B A=0.8 m \ldots(ii)$

Using equations $(i)$ and $(i i),$ we get:

$FE =0.4 m$

Hence, $F$ is the mid-point of $AD$.

$FG$ $\|$ $DH$ and $F$ is the mid-point of $AD$. Hence, $G$ will also be the mid-point of $AH$.

$\triangle AFG$ and $\triangle ADH$ are similar

$\therefore \frac{ FG }{ DH }=\frac{ AF }{ AD }$

$\frac{ FG }{ DH }=\frac{0.4}{0.8}=\frac{1}{2}$

$FG =\frac{1}{2} DH$

$=\frac{1}{2} \times 0.25=0.125 m$

In $\triangle ADH$ :

$AH =\sqrt{ AD ^{2}- DH ^{2}}$

$=\sqrt{(0.8)^{2}-(0.25)^{2}}=0.76 m$

For translational equilibrium of the ladder, the upward force should be equal to the downward force.

$N_{ c }+N_{ B }=m g=392 \ldots(\text {iii})$

For rotational equilibrium of the ladder, the net moment about $A$ is $=$ $-N_{ B } \times Bl +m g \times FG +N_{ C } \times Cl +T \times AG -T \times AG =0$

$-N_{ n } \times 0.5+40 \times 9.8 \times 0.125+N_{ c } \times(0.5)=0$

$\left(N_{ C }-N_{ B }\right) \times 0.5=49$

$N_{c}-N_{B}=98\dots(iv)$

Adding equations $(iii)$ and $(iv)$, we get

$N_{ c }=245 N$

$N_{ n }=147 N$

For rotational equilibrium of the side $AB$, consider the moment about $A$. $-N_{ n } \times BI +m g \times FG +T \times AG =0$

$-245 \times 0.5+40+9.8 \times 0.125+T \times 0.76=0$

$0.76 T=122.5-49$

$\therefore T=96.7 N$