In many engineering uses, the value of "g," the acceleration due to gravity, is taken as a constant. However, g is actually dependent upon the distance from the center of the Earth. A more accurate expression for g is: g = g 0 ( R e R e + A ) 2 Here, g 0 is the acceleration of gravity at the surface of the Earth, A is the altitude above the Earth's surface, and R, is the radius of the Earth, approximately 6,380 kilometers [km]. Assume g 0 = 9.8 meters per second squared [m/s 2 ]. If the value of g is 9 meters per second squared [m/s 2 ], what is the altitude in units of miles [mi]?
In many engineering uses, the value of "g," the acceleration due to gravity, is taken as a constant. However, g is actually dependent upon the distance from the center of the Earth. A more accurate expression for g is: g = g 0 ( R e R e + A ) 2 Here, g 0 is the acceleration of gravity at the surface of the Earth, A is the altitude above the Earth's surface, and R, is the radius of the Earth, approximately 6,380 kilometers [km]. Assume g 0 = 9.8 meters per second squared [m/s 2 ]. If the value of g is 9 meters per second squared [m/s 2 ], what is the altitude in units of miles [mi]?
Solution Summary: The author calculates the altitude (A) above the earth's surface in units of miles, 173.699. The acceleration due to gravity is 9 meters per second squared.
In many engineering uses, the value of "g," the acceleration due to gravity, is taken as a constant. However, g is actually dependent upon the distance from the center of the Earth. A more accurate expression for g is:
g
=
g
0
(
R
e
R
e
+
A
)
2
Here, g0 is the acceleration of gravity at the surface of the Earth, A is the altitude above the Earth's surface, and R, is the radius of the Earth, approximately 6,380 kilometers [km]. Assume g0 = 9.8 meters per second squared [m/s2]. If the value of g is 9 meters per second squared [m/s2], what is the altitude in units of miles [mi]?
1. Plot the moment (M), axial (N), and shear (S) diagrams as functions of z.
a)
b)
F₁ = 1250 N
F₁ = 600 N
M₁ = 350 000 N mm
F2 = 500 N
200 N
a = 600 mm
b=1000 mm
a=750 mm
b = 1000 mm
d)
M₁ = 350 000 N mm
F₁ = 600 N
F₂ =200 N
a = 600 mm
b = 1000 mm
M₁ 175 000 Nmm
F = 900 N
a-250 mm
b-1000 mm
-250 mm.
Figure 1: Schematics problem 1.
Given the following cross-sections (with units in mm):
b)
t=2
b=25
h=25
t = 1.5
b=20
b=25
t=2
I
t = 1.5
a=10
b=15
h-25
b=15
t=3
T
h=25
Figure 3: Cross-sections for problem 2.
1. For each of them, calculate the position of the centroid of area with respect to the given coordinate system
and report them in the table below.
2. For each of them, calculate the second moments of inertia I...
and I, around their respective centroid
of area and report them in the table below. Note: use the parallel axes theorem as much as possible to
minimize the need to solve integrals.
Centroid position
x
y
box
Moment of inertia
lyy
by
a)
b)
c)
d)
e)
Problem 1: Analyze the canard-wing combination shown in Fig. 1. The canard and wing are made
of the same airfoil section and have
AR AR, S = 0.25, and = 0.45%
1. Develop an expression for the moment coefficient about the center of gravity in terms of the
shown parameters (, and zg) and the three-dimensional aerodynamic characteristics of the
used wing/canard (CL C and CM).
2. What is the range of the cg location for this configuration to be statically stable?
You may simplify the problem by neglecting the upwash (downwash) effects between the lifting
surfaces and the drag contribution to the moment. You may also assume small angle approximation.
Figure 1: Canard-Wing Configuration.
Chapter 7 Solutions
Thinking Like an Engineer: An Active Learning Approach (4th Edition)
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