A19.2 kg person climbs up a uniform ladder
with negligible mass. the upper end of the
ladder rests on a frictionless wall. the bottom
of the ladder rests on a floor with a rough
surface where the coefficient of static friction
is 0.1 . the angle between the horizontal and
the ladder is θ . the person wants to climb
up the ladder a distance of 0.49 m along the
ladder from the ladder’s foot.
what is the minimum angle θmin (between
the horizontal and the ladder) so that the
person can reach a distance of 0.49 m without
having the ladder slip? the acceleration of
gravity is 9.8 m/s^2
answer in units of ◦

63°

Explanation:

Draw a free body diagram of the ladder.  There are 4 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 x up the ladder.

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

Sum of forces in the x direction:

∑Fₓ = ma

Nμ − R = 0

R = Nμ

Sum of forces in the y direction:

∑Fᵧ = ma

N − mg = 0

N = mg

Sum of moments about the base of the ladder:

∑τ = Iα

R (L sin θ) − mg (x cos θ) = 0

R (L sin θ) = mg (x cos θ)

Substituting:

Nμ (L sin θ) = mg (x cos θ)

mgμ (L sin θ) = mg (x cos θ)

μ (L sin θ) = x cos θ

tan θ = x / (μL)

θ = atan(x / (μL))

Given x = 0.49 m, μ = 0.1, and L = 2.5 m:

θ = atan(0.49 m / (0.1 × 2.5 m))

θ ≈ 63°

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