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Using the results in Eqs. (1) and (3), letting
mD722 kg
denote the mass of the probe in our problem, the
change in kinetic energy of the probe at aphelion is
TjD1
2mv2
aphelion 1
2m.vcirc/2
Jupiter
, which, using (the
full precision values of) the results obtained thus far, can be evaluated to obtain
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1141
Problem 5.191
A water jet is emitted from a nozzle attached to the ground. The jet has a
constant mass flow rate
.Pm
f/nz D15 kg=s
and a speed
vw
relative to the
nozzle. The jet strikes a
12 kg
incline and causes it to slide at a constant
speed
v0D2m=s
. The kinetic coefficient of friction between the incline
and the ground is
kD0:25
. Neglecting the effect of gravity and air
resistance on the water flow, as well as friction between the water jet and
the incline, determine the speed of the water jet at the nozzle if ✓D47ı.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 1143
Problem 5.192
Revisit Example 5.20 and derive the equation of motion of the free end of
the string starting from the force balance for the right branch of the string
modeled as a variable mass system.
Solution
We start with reporting Eq. (9) (on p. 415 of the textbook) from Example 5.20.
XFyR W⇢`RgD⇢`RayR CPmoEvoO|; (1)
1144 Solutions Manual
An intubed fan (a fan rotating within a tube or other conduit) is
mounted on a cart that is connected to a fixed wall by a linear elastic
spring with constant
kD70 N=m
. Assume that in a particular test
the fan draws air that enters the tube at
A
with a speed
vA
. The
outgoing flow at
B
has a speed
vB
. The flow of air through the tube
causes the cart to displace to the left so that the spring is stretched
by
0:25
m from its unstretched position. Assume that the density
of air is constant throughout the tube and equal to
⇢D1:25 kg=m3
.
In addition, let the tube’s cross section be circular, and let the cross-
sectional diameters at
A
and
B
be
dAD3
m and
dBD1:5
m,
respectively. Determine the velocities of the airflow at Aand B.
Solution
Since the airflow is steady, we select as our control volume the interior volume of the tube
delimited by the (vertical) cross sections at
A
and
B
. The FBD of the chosen control volume
is as shown on the right, where
R
is the horizontal force acting on the airflow do to the fan
(using symmetry arguments we can say that no net force acts on the airflow in the vertical
direction). Hence, the force balance for the chosen control volume in horizontal direction is
d2
AvAx Dd2
BvBx;(7)
which is a system of two equations in the two unknowns vAx and vBx whose solution is
A
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