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

Here, `BA = CA = 1.6 m, DE = 0.5 m, M = 40 kg , BF = 1.2 m`
Let `T =` tension in the rope,
`N_(1),N_(2) =` normal reaction at `B and C` respectively, i.e. forces extered by the floor on the ladder.
In Fig. we find,
As `DE = 0.5m , BC = 10 m , FH = (1)/(2)DG = (1)/(2) xx 0.25 = 0.125 m`
`AG = sqrt(AD^(2) - DG^(2)) = sqrt(0.8^(2) - 0.25^(2)) = 0.76 m`
For translational equilibrium of the step ladder,
`N_(1) + N_(2) - Mg = 0`
or `N_(1) + N_(2) = Mg = 40 xx 9.8 = 392` ..(i)
For rotational equilibrium of the step ladder, taking moments about `A`, we find
`-N_(1) xx BK + Mg xx FH + N_(2) xx CK + T xx AG - T xx AG = 0`
`-N_(1) xx 0.5 + 40 xx 9.8 xx 0.125 + N_(2) xx 0.5 = 0 or (N_(1) - N_(2)) 0.5 = 40 xx 9.8 xx 0.125`
Add (i) and (ii), `2N_(1) = 490` ...(ii)
`N_(1) = (490)/(2) = 245N`
From (i), `N_(2) = 392 - N_(1) = 392 - 245 = 147 N`
For rotational equilibrium of side `AB` of the step ladder, taking moments about `A`, we get
`Mg xx FH - N_(1) xx BK + T xx AG = 0`
`40 xx 9.8 xx 0.125 xx 0.5 + T xx 0.76 = 0`
`:. T xx 0.76 = 245 xx 0.5 - 40 xx 9.8 xx 0.125 = 122.5 - 49 = 73.5`
`T = (73.5)/(0.76) = 96.7N`