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As shown in Fig. the two sides of a step...

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.)

Text Solution

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`M=40 kg, BACA=1.6m, BF=1.2 m`
`N_(1) and N_(2)` are reactions at points B and C respectively.
`DE=0.5mm`
As `DG-0.25m`
`:.FM=(1)/(2)DG=0.125 m`
For translational equilibrium of ladder
`N_(1)+N_(2)-Mg=0`
`N_(1)+N_(2)=Mg=40xx9.8=-392`
`N_(1)+N_(2)=392 " "...(i)`
For rotational equilibrium of ladder. Taking momentum about point A, we get,
`-N_(1)xxBNxxMgxxFM+N_(2)xxCN+T xx AG-TxxAG=0`
`(N_(1)-N_(2))0.5=40xx9.8xx0.125`
`N_(1)-N_(2)=9.8 " "...(ii)`
Adding (i) and (ii) we get
`2N_(1)=490`
`N_(1)=245N`
Putting this value in (i)
`N_(1)=N_(2)=98`
`245-N_(2)=98`
`N_(2)=147N`
For rotational equilibrium of side AB of step ladder, let us take moments about A.
`MgxxFM-N_(1)xxBN+TxxTxxAG=0`
`40xx9.8xx0.125-245xx0.5+Txx0.76=0`
Solving for T we get `T=97N`
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