Answer:
μ = 0.179
Explanation:
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
