Answer:
Down
F1ₓ = 219.6N
[tex]F1_{y}[/tex] = 1038.8 N
Top
F2ₓ = 219.6 N
[tex]F2_{y}[/tex] = 0
Explanation:
For this exercise we must make a free body diagram of the ladder, see attached, then use the balance equations on each axis
Transnational Balance
X axis
F1ₓ -F2ₓ = 0
F1ₓ = F2ₓ
Y Axis
[tex]F1_{y}[/tex] - [tex]F2_{y}[/tex] - W - W_man = 0 (1)
Rotational balance
The reference system is placed at the bottom of the stairs and we can turn the anti-clockwise direction of rotation as positive
F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0
Let us write the data they give, the masses of the ladder (m = 14.0 kg), the mass of man (m_man = 92 kg), the center of mass of the ladder that is 2m from the bottom (the height) and the position of the man which is 3 m high
Let's look with trigonometry for distances
The angle of the stairs is
cos θ = x / L
θ = cos⁻¹ x / L
θ = cos⁻¹ 2 / 5.6
θ = 69⁰
Height y
tan 69 = y / x
y = x tan 69
y = 2 tan 69
y = 5.21 m
Distance x
tan 69 = 2 / x
x = 2 / tan 69
x = 0.7677 m
The distance x_man
x_man = 3 / tan 69
x_man = 1,152 m
They indicate that between the scalars and the support there is no friction so the vertical force at the top is zero
[tex]F2_{y}[/tex] = 0
Let's replace in the translational equilibrium equation
F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0
F2ₓ 5.21 -0 - 14.0 9.8 0.7677 - 92.0 9.8 1,152 = 0
F2ₓ = 1143.97 / 5.21
F2ₓ = 219.6 N
We use equation 1
[tex]F1_{y}[/tex] + 0 - W - W_man = 0
[tex]F1_{y}[/tex] = W + W_man
[tex]F1_{y}[/tex] = (m + m_man) g
[tex]F1_{y}[/tex] = (14 +92) 9.8
[tex]F1_{y}[/tex] = 1038.8 N
We can write the force on each part of the ladder
Down
F1ₓ = 219.6N
[tex]F1_{y}[/tex] = 1038.8 N
Top
F2ₓ = 219.6 N
[tex]F2_{y}[/tex] = 0
