Answer:
we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
Coefficient of static friction = 0.26
Explanation:
Given that:
length of the ladder = 16.0 m
weight of the ladder = 520 N
angle θ = 65.0°
(a) We are to find the horizontal and vertical forces the ground exerts on the base of the ladder when an :
Force = 850 N
distance of the climber from the base of the ladder = 4.20 m
The diagrammatic illustration representing what the given information entails can be seen from the attached file below.
Let consider the Ladder being at point A with the horizontal layer of the ground.
From the whole system; the condition for the equilibrium at the point A can be computed as :
[tex]N_2 (16 \ Sin\ 65) = 850(4.2 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(1.7749 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) = 1508.665+1758.016[/tex]
[tex]N_2 (14.50) = 3266.681[/tex]
[tex]N_2 =\dfrac{ 3266.681}{14.50}[/tex]
[tex]N_2 =225.28 \ N[/tex]
[tex]N_1 = mg+F\\[/tex]
where ;
w =mg
[tex]N_1 = 520+850[/tex]
[tex]N_1 = 1370 \ N[/tex]
Therefore; we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?
the coefficient of static friction between ladder and ground when the firefighter is 9.40 m from the bottom can be calculated as:
[tex]N_2 (16 \ Sin\ 65) = 850(9.4 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(3.9726 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) =3376.71+1758.016[/tex]
[tex]N_2 (14.50) =5134.726[/tex]
[tex]N_2 =\dfrac{5134.726}{14.50}[/tex]
[tex]N_2 =354.12 \ N[/tex]
Therefore; the coefficient of the static friction is;
[tex]\mu = \dfrac{f_s}{N_1}[/tex]
[tex]\mu = \dfrac{354.12}{1370}[/tex]
[tex]\mu[/tex] = 0.26
Coefficient of static friction = 0.26
