Question

As shown in Fig., the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 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 forces exerted by the floor on the ladder. (Take g = 9.8 m/s${}^2$) (Hint: Consider the equilibrium of each side of the ladder separately.)

Answer 1

The normal forces from the floor on the ladder and the tension in the rope is as shown in the figure.

Since D is the mid point of AB, from geometry, $BI=2DH$

Thus $BC=2DE=1m$

and $AD=\frac{1}{2}BA=0.8m$

$FG=\frac{1}{2}DH=0.125m$

In $△ADH$

$AH=\sqrt{A{D}^2-D{H}^2}=0.76m$

From translational equilibrium of the ladder,

$N_B+N_C=mg$ ......(i)

From rotational equilibrium of the ladder, balancing moment about A,

$N_B\times BI+T\times AH=N_C\times CI+T\times AH+mg\times FG$

$⟹N_C-N_B=98$ .......(ii)

Solving (i) and (ii) gives

$N_B=147N$

$N_C=245N$

Hence $T=96.7N$