Respuesta :
Answer:
54.2°
Explanation:
forces are shown on LADDER in equilibrium . applying both conditions of equilibrium we get the result.
see attachment
Newton's second law for linear and rotational motion allows us to find the minimum angle so that the ladder does not slide is:
θ= 54.2º
Newton's second law for stable rotational motion that torque is equal to the product of the moment of inertia times the angular acceleration, when the angular acceleration is zero we have an equilibrium condition.
Σ τ = 0
τ = F r sin θ
Where θ is the torque, F the force, (r sin θ) the perpendicular distance.
In the attached we have a free-body diagram of the forces.
x- axis
R - fr = 0
y -axis
N - W = 0
N = W
The friction force has the expression.
fr = μ N
we substitute
R = μ W (1)
We use Newton's second rotational law where the pivot point is the base of the ladder.
R y - W x = 0
Let's use trigonometry.
sin θ = y / L
cos θ = x / L
y = L sin θ
x = L cos θ
Let's substitute.
R L sin θ = [tex]W \frac{L}{2} cos \theta[/tex]
R sin θ = [tex]\frac{1}{2} W cos \theta[/tex]
We use equation 1.
μ W sin θ =[tex]\frac{1}{2}[/tex] W cos θ
tan θ = [tex]\frac{1}{2 \mu }[/tex]
θ = tan⁻¹ [tex]\frac{1}{2 \mu}[/tex]
We calculate.
θ = tan⁻¹ [tex]\frac{1 }{2 \ 0.36}[/tex]
θ = 54.2º
In conclusion, using Newton's second law we can find the minimum angle for the ladder not to slide is:
tae 54.2º
Learn more here: brainly.com/question/2026476
