A 67 kg person climbs up a uniform 12 kg ladder. The ladder is 5 m long; its lower end rests on a rough horizontal floor (static
friction coefficient 0.39) while its upper end rests against a frictionless vertical wall. The angle between the ladder and the horizontal is 43◦.
Let d denote the climbing person’s distance from the bottom of the ladder (see the above diagram). When the person climbs too far (d > dmax), the ladder slips and falls down (kaboom!). Calculate the maximal distance dmax the person will reach before the ladder slips. The
acceleration of gravity is 9.8 m/s*s. Answer in two decimal places max.

1.7 m

Explanation:

Draw a free body diagram of the ladder.  There are 5 forces:

Normal force N pushing up at the base of the ladder.

Friction force Nμ pushing right at the base of the ladder.

Weight force mg pushing down a distance d up the ladder.

Weight force Mg pushing down a distance L/2 up the ladder.

Reaction force R pushing left at the top of the ladder.

Sum of forces in the x direction:

∑F = ma

Nμ − R = 0

Sum of forces in the y direction:

∑F = ma

N − mg − Mg = 0

Sum of moments about the base of the ladder:

∑τ = Iα

mg (d cos θ) + Mg (L/2 cos θ) − R (L sin θ) = 0

Use the first equation to substitute for R:

mg (d cos θ) + Mg (L/2 cos θ) − Nμ (L sin θ) = 0

Use the second equation to substitute for N:

mg (d cos θ) + Mg (L/2 cos θ) − (mg + Mg) μ (L sin θ) = 0

Simplify and solve for d:

m (d cos θ) + M (L/2 cos θ) − (m + M) μ (L sin θ) = 0

m (d cos θ) = (m + M) μ (L sin θ) − M (L/2 cos θ)

d = [ (m + M) μ (L sin θ) − M (L/2 cos θ) ] / (m cos θ)

Plug in values and solve:

d = [ (67 kg + 12 kg) (0.39) (5 m sin 43°) − (12 kg) (2.5 m cos 43°) ] / (67 kg cos 43°)

d = 1.70 m

Rounded to two significant figures, the maximum distance is 1.7 m.

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