Ladder Against a Smooth Wall: How Far Can the Painter Climb?
A uniform ladder of length 5 m and mass 20 kg rests with its foot on rough horizontal ground and its top against a smooth vertical wall. The ladder makes an angle of 60° with the ground and the coefficient of friction between the ladder foot and the ground is 0.3. A painter of mass 70 kg climbs the ladder. His employer insists the ladder is safe only if the painter can reach the top rung; a colleague argues that, because the wall is smooth, the ladder will slip almost as soon as he starts climbing. You are asked to settle the argument with a full equilibrium analysis, taking g = 9.8 m/s². Model the ladder as a uniform rod and the painter as a particle on the ladder. The wall reaction is horizontal only; the ground reaction has a normal component and a frictional component, and friction reaches its limiting value exactly when slipping is about to occur. State clearly which point you take moments about and why.
1. Find how far up the ladder the painter can climb before the ladder is on the point of slipping.
Read the worked answer
Let x be the distance climbed, measured along the ladder from the foot. Let N_w be the horizontal wall reaction, and N_g, F the normal and frictional ground reactions.
Force balance. Vertically: N_g = 20g + 70g = 90g. Horizontally: F = N_w.
Moments about the foot. This point is chosen because both N_g and F act through it, so they have zero moment; only N_w and the two weights remain. Taking perpendicular distances:
N_w (5 sin60°) = 20g (2.5 cos60°) + 70g (x cos60°)
N_w × 4.330 = (50g + 70gx)(0.5)
N_w = (25g + 35gx)/4.330.
Limiting friction. Slipping is about to occur when N_w = F = μN_g = 0.3 × 90g = 27g. Hence
(25 + 35x)/4.330 = 27 → 25 + 35x = 116.9 → x = 2.63 m.
The painter can climb about 2.6 m of a 5 m ladder, just past halfway. Reaching the top rung is not safe with μ = 0.3.
2. Evaluate the colleague's claim that a smooth wall means immediate slipping, and explain what would change if the wall were rough.
Read the worked answer
The colleague's claim is wrong. Smoothness of the wall means only that the wall supplies no vertical friction; the entire vertical load is carried by N_g, and the whole horizontal thrust N_w must be balanced by ground friction. Slipping at the first step would require μ ≈ 0 at the ground. With μ = 0.3 there is a genuine margin, and the calculation gives x = 2.63 m.
The employer's claim also fails. Put x = 5: N_w = (25g + 175g)/4.330 = 46.2g, far above the available 27g. The ladder would slip long before the painter reached the top.
If the wall were rough. The wall could then supply an upward friction as well as a horizontal reaction. That would reduce the required N_g (and hence the ground friction) and would add its own moment about the foot, allowing a greater climbing distance. The improvement is not automatic, though: at limiting equilibrium both the wall contact and the ground contact must be checked, and the ladder slips at whichever reaches its limit first.
3. Find the magnitude and direction of the total ground reaction just before slipping, and explain why the angle does not depend on the painter's mass.
Read the worked answer
At x = 2.63 m: N_g = 90g = 882 N and F = 27g = 264.6 N. These are perpendicular components, so
R = √(N_g² + F²) = √(882² + 264.6²) = √(777 924 + 70 013) = √847 937 ≈ 921 N.
The angle to the vertical is arctan(F/N_g) = arctan(264.6/882) = arctan(0.3) = 16.7°.
Why the angle is mass-independent. At limiting friction F = μN_g, so F/N_g = μ exactly. The resultant therefore always makes the friction angle tan⁻¹μ with the normal to the ground, whatever the loads are. The mass affects the magnitude of R (through N_g) but not its direction. This is a useful check: if a candidate's angle is not tan⁻¹μ at limiting equilibrium, an error has been made somewhere in the force or moment equations.