Answer:
a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36
Explanation:
a) This is a balancing act.
Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive
-w x - W x₂ + R y = 0 (1)
usemso trigonometry to find distances
cos 60.08 = x / 7.5
x = 7.5 cos 60.08
x = 3.74 m
fireman
cos 60.08 = x₂ / 4
x2 = 4 cos 60
x2 = 2 m
wall support
sin 60.08 = y / 15
y = 15 are 60.08
y = 13 m
we substitute in equation 1
R y = w x + W x2
R = (w x + W x2) / y
R = (500 3.74 +800 2) / 13
R = 266.92 N
now let's write the expressions for the translational equilibrium
X axis
R -fr = 0
R = fr
fr = 266.92 N
Y Axis
Fy - w-W = 0
fy = 500 + 800
fy = 1300 N
b) ask the friction coefficient
the firefighter's distance is
cos 60.08 = x₃ / 9.00
x₃ = 9 cos 60
x₃ = 5.28 m
from equation 1
R = (w x + W x₃) / y
R = 500 3.74 + 800 5.28) / 13
R = 468.769 N
we saw that
fr = R = 468.769
The expression for the friction force is
fr = μ N
in this case the normal is the ratio to pesos
N = Fy
N = 1300 N
μ N = fr
μ = fr / N
μ = 468,769 / 1300
μ = 0.36
For Anya:v 1 = 3.3 mph, v 2 = 8.3 mph5 : 2 = 2.5 mi ( half of the distance )
t1= d/v1 = 2.5 mi/ 3.3 mph = 0.7576 ht2 = d/v2 = 2.5 mi/8.3 mph = 0.3 hIn total: 0.7576 + 0.3 = 1.0576 hFor Ilya: 8.3 t / 2 + 3.3 t / 2 = 5 / * 28.3 t + 3.3 t = 1011.6 t = 10t = 10 : 11.6t = 0.86 hHe covers the distance for 0.86 h.His average speed:v = 5/0.86 = 5.81 mph.
Friction, engine thrust, normal reaction and weight, these are the forces acting on motor when it moves in a straight line
I believe it's Gamma Rays. Hope my answer helped :)