Ladder Equilibrium with a Climbing Painter
A uniform ladder of length 5 m and mass 20 kg rests against a smooth vertical wall and on rough horizontal ground, with its foot 3 m from the wall. A painter of mass 60 kg climbs the ladder. The coefficient of friction between the ladder and the ground is 0.4; the wall is smooth. The painter pauses at a distance x metres measured along the ladder from its foot, and the site supervisor must decide whether the ladder will stay in equilibrium at that point. The ladder is light enough to be treated as uniform, and the ground is level. The supervisor is worried because the painter’s mass is three times the ladder’s mass, so the usual ‘ladder alone’ calculation is not enough. Before allowing the climb, the supervisor needs the greatest value of x for equilibrium, an explanation of which contact is likely to fail first, and the minimum angle the ladder would need with the ground if the painter were to reach the top safely while keeping the same friction coefficient.
1. Taking g as the acceleration due to gravity, find the greatest distance x along the ladder at which the painter can pause without the ladder slipping.
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Geometry: the ladder makes angle θ with the ground where cosθ = 3/5, sinθ = 4/5, tanθ = 4/3. The vertical ground reaction is N_G = (20 + 60)g = 80g, so the maximum friction is F_max = μN_G = 0.4 × 80g = 32g. Let R be the normal reaction from the smooth wall. Taking moments about the foot of the ladder: R × (5 sinθ) = 20g × (2.5 cosθ) + 60g × (x cosθ). So R = [20g(2.5)(3/5) + 60g x(3/5)]/(5 × 4/5) = (30g + 36gx)/4 = 7.5g + 9gx. Equilibrium requires R ≤ 32g, since the wall cannot provide friction. Thus 7.5 + 9x ≤ 32, giving x ≤ 24.5/9 ≈ 2.72 m. The painter can pause up to about 2.72 m along the ladder.
2. Explain which contact would fail first if x is increased beyond this value, and calculate the friction force actually needed at x = 3.5 m.
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The wall is smooth, so the only horizontal force that can balance R is friction at the ground. As x increases, R increases. At x = 3.5 m, R = 7.5g + 9g(3.5) = 39g. The ground can supply at most 32g, so the required friction exceeds the limit. The ground contact fails first: the foot tends to slide away from the wall while the top presses into the wall. Equilibrium is impossible at x = 3.5 m with this μ.
3. Find the minimum angle θ with the ground for which the painter can reach the top (x = 5 m) without slipping at the same μ.
Read the worked answer
At the top, x = 5 m. Moments about the foot give R × (5 sinθ) = 20g × (2.5 cosθ) + 60g × (5 cosθ) = (50g + 300g)cosθ = 350g cosθ. So R = 350g cosθ/(5 sinθ) = 70g cotθ. Require R ≤ 32g: 70 cotθ ≤ 32 ⇒ cotθ ≤ 16/35 ⇒ tanθ ≥ 35/16 = 2.1875. Thus θ ≥ arctan(2.1875) ≈ 65.4°. The ladder must be steeper than about 65.4° to the ground for the painter to reach the top without slipping.