Respuesta :
Answer:
α = 8.25°
Explanation:
This is a balancing exercise, let's write Newton's equations
Translational equilibrium
X axis
fr - Fx = 0
Fx = fr
Axis y
N - W = 0
N = W = mg
The force (F) is for the support of the ladder on the wall
Rotational balance
We place our system and reference on the floor, at the point of support, the anti-clockwise turns are positive
fx y -W x = 0
The friction force has as an expression
fr = μ N
fr y = mg x
μ mg y = mg x
Distance is related by trigonometry
sin α = x / L
x = L sin α
cos α = y / L
y = L cos α
The angle α is measured from the wall
μ L cos α = L sin α
sin α / cos α = μ
tan α = μ
α = tan⁻¹ (μ)
α = tan⁻¹ 0.145
α = 8.25°
The maximum angle α, relative to the wall, that the ladder can lean without slipping is 8.25°
Static and dynamic friction
To solve the question, we will use the balancing concept. According to the Newton's equations
Resolving the force along the x-axis
Fr - Fx = 0
Fx = fr
Resolving along the y-axis
N - W = 0
N = W = mg
The force (F) is for the support of the ladder on the wall
For the rotational balance
We place our system and reference on the floor, at the point of support, the anti-clockwise turns are positive
Fx y -W x = 0
The friction force has as an expression
Fr = μ N
Fr y = mg x
μ mg y = mg x
Distance is related to trigonometry
sin α = x / L
x = L sin α
Similarly, cos α = y / L
y = L cos α
The angle α is measured from the wall
μ L cos α = L sin α
sin α/cos α = μ
So that;
tan α = μ
Calculate the required angle
α = tan⁻¹ (μ)
α = tan⁻¹ 0.145
α = 8.25°
Hence the maximum angle α, relative to the wall, that the ladder can lean without slipping is 8.25°
Learn more on friction here: https://brainly.com/question/25534663
