A person of mass 80 kg is standing on the top of a 18 kg ladder of length 6 m. Upper end of the ladder rests on a smooth vertical wall and thelower end is on the ground 3 m away from the vertical wall. What should be the minimum coefficient of friction between the floor and the ladder so that the system remains in equilibrium ?

A person of mass 80 kg is standing on the top of a 18 kg ladder of len

1.	A person of mass 80 kg is standing on the top of a 18 kg ladder of length 6 m. Upper end of the ladder rests on a smooth vertical wall and thelower end is on the ground 3 m away from the vertical wall. What should be the minimum coefficient of friction between the floor and the ladder so that the system remains in equilibrium ?
Answer» Solution :Given that, for the LADDER weight W acts through the centre of gravity, C (mid -point of AB) of theladder Fig. Weight W. of the MAN acts at B downwards. Normal reaction R. by the WALL acts at B. Normal reaction R. by the ground acts at A. Limiting friction that acts along the floor at A, f = `mu` R where, `mu` is the COEFFICIENT of friction required for equilibrium. At equilibrium, the sum of all horizontal forces and the sum of all vertical forces will be zero separately. `mu R -R. =0 " ""or", muR=R.""cdots (1)` R-W-W.0 `" ""or", R=W+W.""cdots(2)` Again the sum of the moments of all forces taken about A will be zero. `:. R.xxBD-WxxAE-W.xxAD=0` or, `mu(W+W.)xxBD=WxxAE+W.xxAD` `:."" mu = (WxxAE+W.xxAD)/((W+W.))xxBD` Here, AB = 6m, AD = 3 m. `:."" AE=(AD)/(2)=(3)/(2) m, BD = sqrt(AB^(2)-AD^(2))=5.2 m,` W =18 kg `xxg, W. = 80 kgxxg` `:. " " mu =(18xx(3)/(2)+80xx3)/((18+80)xx5.2)=0.52.`