Dynamics 2e 697
Problem 4.25
An F/A-18 Hornet takes off from an aircraft carrier, using two separate propul-
sion systems: its two jet engines and a steam-powered catapult. During launch,
a fully loaded Hornet weighing about
50;000 lb
goes from
0mph
(relative to
the aircraft carrier) to
165 mph
(measured relative to surface of the Earth) in a
distance of
300 ft
(measured relative to the aircraft carrier) while each of its two
engines is at full power generating about
22;000 lb
of thrust. Assuming that the
aircraft carrier is traveling in the same direction as the takeoff direction and at a
constant speed of 30 knot, determine:
(a) The total work done on the aircraft during launch.
(b) The work done by the catapult on the aircraft during launch.
(c) The force exerted by the catapult on the aircraft.
In solving this problem, model the aircraft as a particle; assume that its trajectory
is horizontal and that the catapult assists the aircraft the full
300 ft
needed for
takeoff; and finally, let all forces be constant and neglect air resistance and
friction. Use an inertial reference frame attached to the aircraft carrier.
Photo credit: U.S. Navy photo by Mass Communication Specialist 3rd Class Torrey W. Lee
Solution
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698 Solutions Manual
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Dynamics 2e 699
Problem 4.26
Rubber bumpers are commonly used in marine applications to keep boats and ships from getting damaged
by docks. Treating the boat
C
as a particle, neglecting its vertical motion, and neglecting the drag force
between the water and the boat
C
, what is the maximum speed of the boat at impact with the bumper
B
so that the deflection of the bumper is limited to
6in:
? The weight of the boat is
70;000 lb
, and the force
compression profile for the rubber bumper is given by
FBDˇx3
, where
ˇD3:5⇥106lb=ft3
and
x
is the
compression of the bumper.
Solution
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Dynamics 2e 701
Problem 4.27
Packages for transporting delicate items (e.g., a laptop or glass) are designed
to “absorb” some of the energy of the impact in order to protect their contents.
These energy absorbers can get pretty complicated (e.g., the mechanics of
Styrofoam peanuts is not easy), but we can begin to understand how they work
by modeling them as a linear elastic spring of constant
k
that is placed between
the contents (an expensive vase) of mass
m
and the package
P
. Assume that
the vase weighs
6lb
and that the box is dropped from a height of
5ft
. Treat the
vase as a particle, and neglect all forces except for gravity and the spring force.
Determine the maximum displacement of the vase relative to the box and
the maximum force on the vase due to the spring if kD264 lb=ft.
Solution
Referring to the FBD on the right, we model the vase as a particle subject to its
own weight
mg
and to the force
Fv
(
v
stands for ‘vertical’), which is equal to
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Dynamics 2e 703
Problem 4.28
Packages for transporting delicate items (e.g., a laptop or glass) are designed
to “absorb” some of the energy of the impact in order to protect their contents.
These energy absorbers can get pretty complicated (e.g., the mechanics of
Styrofoam peanuts is not easy), but we can begin to understand how they work
by modeling them as a linear elastic spring of constant
k
that is placed between
the contents (an expensive vase) of mass
m
and the package
P
. Assume that
the vase weighs
6lb
and that the box is dropped from a height of
5ft
. Treat the
vase as a particle, and neglect all forces except for gravity and the spring force.
Plot the maximum displacement of the vase relative to the box and the
maximum force on the vase due to the spring as a function of the linear elastic
spring constant
k
. What do these plots tell you about the problem you would
encounter in trying to minimize the force on the vase?
Solution
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Dynamics 2e 705
Problem 4.29
A block
A
moves horizontally under the action of a force
F
whose line of action is parallel to the motion.
If the kinetic energy of
A
as a function of
x
is that shown (
x1
and
x2
are extrema for
TA
), what can you
say about the sign of Ffor x0<x<x
1and x1<x<x
2?
Solution
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Problem 4.30
While the stiffness of an elastic cord can be quite constant (i.e., the force
versus displacement curve is a straight line) over a large range of stretch, as a
bungee cord is stretched, it softens; that is, the cord tends to get less stiff as
it gets longer. Assuming a softening force-displacement relation of the form
kı ˇı3
, where
ı
(measured in ft) is the displacement of the cord from its
unstretched length, considering a bungee cord whose unstretched length is
150 ft
, and letting
kD2:58 lb=ft
, determine the value of the constant
ˇ
such
that a bungee jumper weighing
170 lb
and starting from rest gets to the bottom
of a 400 ft tower with zero speed.
Solution
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