This problem has been solved! See the answer A 21.5 kg person climbs up a uniform 85 N ladder. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface. The angle between the horizontal and the ladder is 59 ◦. The person can climb 78 % up the ladder before the base of the ladder slips on the floor. What is the coefficient of static friction µs between the base of the ladder and the floor? The acceleration of gravity is 9.8m/s 2 .

For this exercise we use the rotational equilibrium condition, where we set a reference system at the top of the ladder where it is in contact with the parent, we will assume the counterclockwise rotations as positive

∑ τ = 0

W_stairs x/2 + W_man x’+ fr y - N x = 0

fr y = -W_stairs x/2 - W_man x’ + N x

let's use trigonometry for distances

* the man up the ladder 78%

l ’= 0.78 L

cos 59 = x ’/ l’

x’= 0.78 L cos 59

* the horizontal distance

cos 59 = x / L

x = L cos 59

* vertical distance

sin 59 = Y / L

y = L sin 59

we substitute

fr L sin 59 = -W_stairs Lcos59 / 2 - W_man 0.78 L cos59 + N L cos59

fr sin 59 = - ½ W_stairs cos 59 - 0.78 W_man cos 59 + N cos 59

fr = ctan 59 (N - ½ W_stairs - 0.78 W_man) (1)

Let's write the translational equilibrium equations

Y axis

N -W_stairs - W_man = 0

N = W_stairs + W_man

N = 85 + 21.5 9.8

N = 295.7 N

we substitute in 1

fr = ctg 59 (295.7 - ½ 85 - 0.78 21.5 9.8)

fr = 0.6 (295.7 - 42.5 - 164.346)

fr = 52.89 N

the expression for the friction force is

fr = μ N

μ = fr / N

μ = 52.89 / 295.7

μ = 0.179