As shown in the figure,the two sides of a step ladder $BA$ and $CA$ are $1.6 \; m$ long and hinged at $A$. $A$ rope $DE$ of length $0.5 \; m$ is tied halfway up. $A$ weight of $40 \; kg$ is suspended from a point $F$,$1.2 \; m$ from $B$ along the ladder $BA$. Assuming the floor to be frictionless and neglecting the weight of the ladder,find the tension in the rope and the forces exerted by the floor on the ladder. (Take $g = 9.8 \; m/s^2$)

(D) Let $N_B$ and $N_C$ be the normal forces exerted by the floor on the ladder at points $B$ and $C$ respectively. Let $T$ be the tension in the rope $DE$.
Given: $BA = CA = 1.6 \; m$,$DE = 0.5 \; m$,$BF = 1.2 \; m$,$m = 40 \; kg$.
Since $D$ and $E$ are midpoints of $AB$ and $AC$,$DE$ is parallel to $BC$ and $DE = \frac{1}{2} BC$. Thus,$BC = 2 \times DE = 1.0 \; m$.
Let $I$ be the midpoint of $BC$. $AI$ is the altitude of $\triangle ABC$. $BI = IC = 0.5 \; m$.
$AF = BA - BF = 1.6 - 1.2 = 0.4 \; m$.
Since $D$ is the midpoint of $AB$,$AD = 0.8 \; m$. $F$ is at $0.4 \; m$ from $A$,so $F$ is the midpoint of $AD$.
Let $H$ be the intersection of $DE$ and $AI$. $DH = \frac{1}{2} BI = 0.25 \; m$.
Let $G$ be the projection of $F$ on $AI$. Since $F$ is the midpoint of $AD$,$FG = \frac{1}{2} DH = 0.125 \; m$ and $AG = \frac{1}{2} AH$.
In $\triangle ADH$,$AH = \sqrt{AD^2 - DH^2} = \sqrt{0.8^2 - 0.25^2} = \sqrt{0.64 - 0.0625} = \sqrt{0.5775} \approx 0.76 \; m$.
Translational equilibrium: $N_B + N_C = mg = 40 \times 9.8 = 392 \; N$.
Rotational equilibrium about $A$: $N_B \times BI - N_C \times IC + mg \times FG = 0$ (taking clockwise as positive).
$N_B(0.5) - N_C(0.5) + 392(0.125) = 0 \implies 0.5(N_C - N_B) = 49 \implies N_C - N_B = 98$.
Solving $N_B + N_C = 392$ and $N_C - N_B = 98$,we get $N_C = 245 \; N$ and $N_B = 147 \; N$.
For side $AB$,taking moments about $A$: $N_B \times BI - mg \times FG - T \times AG = 0$.
$147 \times 0.5 - 392 \times 0.125 - T \times (0.76/2) = 0$.
$73.5 - 49 = T \times 0.38 \implies 24.5 = 0.38T \implies T \approx 64.47 \; N$.