Painter on a Ladder in Limiting Equilibrium
A uniform ladder AB of length 5 m and mass 15 kg leans against a smooth vertical wall at B and rests on rough horizontal ground at A. The ladder makes an angle of 60° with the ground. The coefficient of friction between the ladder and the ground is 0.4. A painter of mass 70 kg climbs the ladder. The wall is smooth, so the reaction at the wall is horizontal; the ground reaction has both normal and frictional components. The painter's position is measured as distance x metres from A along the ladder. The ladder is in limiting equilibrium when the painter is at a certain point. A safety officer wants to know how far the painter can climb before the foot begins to slip. Use g = 9.8 m/s² and treat the painter as a particle. The ladder is uniform, so its weight acts at its midpoint.
1. By taking moments about A and resolving forces, find the greatest distance x along the ladder that the painter can reach before slipping. State the limiting-friction condition clearly.
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Let N_A be the normal reaction at the ground, F the friction at the ground and N_B the normal reaction at the wall. Resolving vertically: N_A = 15g + 70g = 85g. At limiting equilibrium, F = μN_A = 0.4(85g) = 34g. Resolving horizontally: N_B = F, so N_B = 34g at the limiting point. Take moments about A. The wall reaction has moment N_B(5 sin 60°). The ladder weight acts at 2.5 m from A and has moment 15g(2.5 cos 60°). The painter's weight has moment 70g(x cos 60°). Hence 34g(5 sin 60°) = 15g(2.5 cos 60°) + 70g(x cos 60°). Cancel g: 34(4.330) = 18.75 + 35x. The left side is about 147.22, so 35x = 147.22 − 18.75 = 128.47, giving x ≈ 3.67 m. Thus the painter can climb about 3.67 m along the ladder before the foot is about to slip.
2. Interpret the result: what happens if the painter climbs to the top of the ladder? Would increasing the coefficient of friction to 0.5 allow the painter to reach the top? Justify with a calculation.
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At the top, x = 5 m. The moment equation gives N_B(5 sin 60°) = 15g(2.5 cos 60°) + 70g(5 cos 60°), so N_B(4.330) = 18.75g + 175g = 193.75g, giving N_B ≈ 44.74g. The friction required to balance this is 44.74g, but the limiting friction at μ = 0.4 is 34g, so the ladder would slip before the painter reaches the top. If μ is increased to 0.5, the limiting friction becomes 0.5(85g) = 42.5g. This is still less than the required 44.74g, so the painter still cannot reach the top. To reach the top, the coefficient would need to satisfy μ(85g) ≥ 44.74g, so μ ≥ 44.74/85 ≈ 0.526. Thus μ = 0.5 is insufficient, and the safety officer should not permit the painter to climb to the top under the stated conditions.