I think this is vectors. Sketch the two vectors A and B on the x axis and then find their magnitudes using cosine... i would like to know if i am correct.
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
Answer:
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Answer:
An aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²
Explanation:
Applying Bernoulli's equation, we determine the highest pressure on the aircraft.
where;
P is the highest pressure on the aircraft
is the density of air = 1.204 kg/m³ at sea level temperature.
V is the velocity of the aircraft = 220 m/s
P = 0.5*1.204*(220)² = 29136.8 N/m²
Therefore, an aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²
