A man of weight 60 kg tries to climb the ladder. How much distance along the ladder he will be able to climb if the coefficient of between ladder and floor as 0.2 and that between ladder and wall as 0.3. Also find the angle the ladder should make with the horizontal such that the man can climb till the top of the ladder

Assume following FBD, where A is a point of ladder in contact with wall and B is in contact with ground

Let $N_1$ and $N_2$ be the normal reaction at B and A respectively

$\therefore F_{s1} =\mu_1N_1=0.2N_1 \rightarrow (1) \\ F_{s2}=\mu_2N_2=0.3N_2 \rightarrow (1) \\ \sum F_x=0 \\ \therefore -F_{s1}+N_2=0 \\ \therefore N_2=F_{s1} \\ \therefore N_2=0.2 N_1 \rightarrow (3) (from 1) \\ Also \sum F_y=0 \\ \therefore F_{s2}+N_1-60g -25g=0 \\ \therefore 0.3N_2 +N_1 =85g (from 3) \\ \therefore 1.06 N_1=85 g \\ \therefore N_1=80.1887 g \\ From (3) , N_2 =0.2\times 80.1887 g=16.0377 g\\ And , \sum M_B=0 \\ \therefore 60 g \times BC + 25 g \times BD -N_2 \times OA -F_{s2} \times OB=0 \\ \therefore 60 g \times x\cos a +25g \times 1.5 \cos a =N_2 \times 3\sin a +0.3 N_2 \times 3 \cos a \\ \therefore 60 g \times x\cos a \times 37.5 g \cos a =16.0377 g \times 3\sin a +16.0377 g \times3\cos a\\ \therefore 20x +125 =16.0377 \tan a +0.3 \times 16.0377 \rightarrow (4) $

Divide by $ 3g\cos d$

Part I : when $a=50^\circ $

From (4) $,20x + 12.5 =16.0377\tan 50 +0.3\times16.0377 \\ \therefore 20x + 12.5 = 23.9244 \\ \therefore 20x =11.4244 \\ \therefore x=0.5712 m$

The man will be able to climb $0.5712m$ along the ladder.

Part I : when $x=3m$

From (4) $,20\times 3 + 12.5=16.0377\tan a +0.3\times 16.0377 \\ \therefore 72.5=16.0377\tan a + 4.8113 \\ \therefore 67.6887 =16.0377\tan a \\ \therefore a =76.6705^\circ $

The ladder should make an angle of $76.6705^\circ$ with the horizontal so that the man can climb till the top.