Question

As shown in the figure, the two sides of a step ladder $BA$ and $CA$ are $1.6m$ long and hinged at $A$. A rope $DE$, $0.5m$ is tied halfway up. A weight $40kg$ is suspended from a point $F$, $1.2m$ 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 forces exerted by the floor on the ladder. (Take $g=9.8m^2$)
(Hint: Consider the equilibrium of each side of the ladder separately)

Answer 1

$\Delta AOD\sim \Delta AGB$

$\Rightarrow \frac{OD}{AD}=\frac{GB}{AB}$

$\Rightarrow \frac{0.25}{0.8}=\frac{x}{1.6}$

$\Rightarrow x=0.5m$

Balancing forces $N_1+N_2=40g=392N$ ...(1)

Balancing torque of AB about A

$\Rightarrow (-N_1\times 0.5)+T\left(OA\right)+w\left(0.4\right)=0$

$\Rightarrow +0.5N_1=0.76T+156.8$

$\Rightarrow 0.5N_1-0.76T=156.8$ ...(2)

Balancing totque of AC about A

$-(N_2\times 0.5)+T\left(0.76\right)=0$

$\Rightarrow -0.5N_2+0.76T=0$ ...(3)

On solving (1), (2) and (3) we get

$N_1=352.8N$

$N_2=39.2N$

$T=25.79N$