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 fricionless 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 eqilibrium of each side of the ladder separately.)

Explanation

The forces acting on the ladder are shown in Fig. 7.14. Here, `IV=40 kg =40xx9.8 N=392 N, AB=AC=1.6 m, BD=1//2xx1.6 m=0.8 m, BF=1.2 m " and" DE 0.5 m`
In the Fig. `DeltaADE " and " DeltaABC` are similar triangles, hence
`BC=Dexx(AB)/(AD)=(0.5xx1.6)/(0.8)=1.0 m`
Now, considering equilibrium at point B `sum tau=0`
`therefore " " Wxx(MB)=N_(C )xx(CB)`
But `MB=(KBxxBF)/(BA)=(0.5xx1.2)/(1.6)=0.375 m`
Substituting this value in (i), we get
`therefore " " N_(C )=(Wxx(MB))/((CB))=(392xx0.375)/(1)=147 N`
Again considering equilibrium at point C in similar manner, we have
`Wxx(MC)=N_(B)xx(BC)`
`therefore " " N_(B)=(Wxx(MC))/((BC))=(Wxx(BC-BM))/((BC))`,
`(392xx(1-0.375))//1=245 N`
Now, it can be easily shown that tension in the string `T=N _(B)-N_(C )=245-147=98N`.