Q.A 2 m long ladder of mass 10 kg is kept against a wall such that its base is 1.2 m away from the wall. The wall is smooth but the ground is rough. The roughness of the ground is such that it offers a maximum horizontal resistive force (for sliding motion) equal to half the normal reaction at the point of contact. A monkey of mass 20 kg starts climbing the ladder. How far can it climb along the ladder? How much is the horizontal reaction at the wall?

[!TLDR] Using torque balance about the base (with weights expressed in kg-weight, kgf) and the maximum-friction condition (half of the normal reaction), the monkey can climb 1.5 m up the ladder before it is about to slip, with the wall's horizontal reaction there equa …

Ladder length 2 m, base 1.2 m from the wall, so by Pythagoras the height reached at the wall is $\sqrt{2^2-1.2^2}=\sqrt{4-1.44}=\sqrt{2.56}=1.6$ m. Working in kg-weight (kgf) units throughout, so that g need not be introduced separately: ground normal reaction $N_g=10+20=30$ kgf (ladder 10 kg + monkey 20 kg), and the maximum available friction at the ground is $f_{max}=\frac{1}{2}N_g=15$ kgf. The wall is smooth, so it supplies only a horizontal reaction $N_w$ (no vertical, no friction there); horizontal equilibrium requires $N_w=f$ (the ground friction must exactly balance the wall's push).

Let s be the distance the monkey has climbed along the 2 m ladder (from the base); its horizontal position from the base is $\frac{s}{2}\times1.2=0.6s$, and the ladder's own weight (10 kgf) acts at its midpoint, horizontal position $\frac{1.2}{2}=0.6$ m. Taking torques about the BASE of the ladder (where $N_g$ and f both act, contributing zero torque about that point) balances the two downward weight-torques against the wall reaction's torque: $N_w(1.6)=10(0.6)+20(0.6s)=6+12s$, giving $N_w=3.75+7.5s$ kgf. The monkey can climb further as long as the required $N_w$ (equivalently, the required friction f, since $f=N_w$) stays below $f_{max}=15$ kgf; setting $N_w=15$: $3.75+7.5s=15\Rightarrow 7.5s=11.25\Rightarrow s=1.5$ m, at which point $N_w=15$ kgf. …

Work in kg-weight (kgf) units so g need not appear separately; use the ladder geometry (Pythagoras) to find the wall-contact height, write the torque balance about the base (which eliminates the ground reactions) including the ladder's and monkey's weight-torques and the wall reaction's …

Assuming the wall reaction equals the FULL weight of the system rather than a torque-balance-determined fraction of it; forgetting that torques about the ladder's BASE conveniently eliminate the …