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Chapter
5 -- Conservation of Linear Momentum |
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Updated
--- 18 September 2006 |
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Hint |
Answer |
Update Date |
5.1 |
Parts
(a) and (b) are just kinematics. Take a good look at the review exercise on
kinematics starting on page 5-5 and you'll be OK. |
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21Sept05 |
5.2 |
FYI
--- Most rocket sleds that I know of travel in a horizontal plane. When they
travel vertically or leave the ground they are rockets. So assume that
all motion is in a plane perpendicular to the gravity vector. You can
probably solve it for vertical motion, but it won't match my answers. (a) Apply LM and carefully draw an interaction diagram (free-body
diagram) carefully locating all force transports of momentum for the rocket
sled.. (Assume it is a closed system.) |
(a)
k = 16.33 x 10-3 lbf-s2/ft2 (b) 78.8 s; 14.1 x 103 ft |
18Sept06 |
5.3 |
Carefully
sketch your system boundary because the location of your boundary determines
how the forces enter the problem. Is this a steady-state system? |
(a) TCA = 1203 N; (c) TCA = 5628 N |
17Sept05 |
5.4 |
Follow
the hints in the problem statement. Be careful with units. Note that in this
problem you are using two different systems to study the same process. This
is necessary because you are looking for two different pieces of information.
This is a very common occurrence in problem solving. Also remember that if
you want to know the value of a momentum transport, say a force, then you
have to have a boundary at the location where the transport occurs. Be very
careful to draw the forces on your free body diagram. |
(a) 225.1 ft (b) 4398 lbf in tension |
21Sept05 |
5.5 |
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5.6 |
Your
system should cut the flanges; thus you have both flow into and out of the
reducing pipe bend. Carefully apply the conservation of linear momentum to
this system paying particular attention to the net pressure force and the net
transports of linear momentum with mass flow. Note that mass flow rates
across the boundary of a system can carry both positive and negative linear
momentum. The plus/minus sign that appears in the rate-form of the
conservation of linear momentum equation is there to indicate whether the
mass flow rate is into or out of the system. The specific linear momentum V
is a vector and also has a sign associated with it. |
Fbolts = 14.93 kN to the left |
21Sept05 |
5.7 |
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5.8 |
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5.9 |
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5.10 |
Suggest
you take the athlete as a closed system. Write rate form of the LM equation for
the athlete showing a time varying reaction force of the ground on the
athlete. Integrate with respect to time and solve for the average
impulsive force, i.e. the constant force which if it acted on the runner
during the known time interval would have cause the necessary change in
system LM. Giving your answer in terms of the weight of the athlete gives the
result in relative terms. |
(a) If runner is running to the right, the horizontal force is 1.29Wathlete to the left. (b) The vertical component on the athlete is 6.21Wathlete in a direction opposite to gravity. |
21Sept05 |
5.11 |
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5.12 |
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5.13 |
Two
possible systems: Cone alone (a closed system) or enlarged cone-shaped volume
(open system) that includes the conical deflector and the air streams.
Also note that there are really only TWO inlet/outlets for the second system.
The main jet enters from the bottom and then leaves at the top of the
deflector in an annular sheet of moving air. Again answer requires
application of cons. of LM. |
m = 0.109 kg |
21Sept05 |
5.14 |
You
will need two very different systems here. For part (a) this should be a
particle of sand (a cl___ system) as you travel with it. IS this system
steady-state? For part (b) your system should be the region of space icluding
the belt and the sand entering and leaving it (an op__ system). OR you could
just consider the volume of the sand on the belt as an open, stationary,
non-deforming system. Is this a steady-state system? |
(a)
6.09� or 52.9� P = 0.041sin(α)gmdotL + 24mdot With mdot = 100 lbm/s, P is 83.5 lbf with 8.9 lbf from weight of sand and 74.6 lbf from change in LM of sand. |
21Sept05 |
5.15 |
(1)
Recognize that the resistance force can be written as R= kx. |
V25 mm = 565.7 m/s |
21Sept05 |
5.16 |
Apply
conservation of LM to the block. (What does 125 lb mean? A mass
of 125 lbm or a weight of 125 lbf? Would it really make any difference if
process occurs on the earth's surface? IF the process is on the earth and you
are assuming standard gravitational acceleration, then your final answer will
be the same regardless of which interpretation you assume. Try it!) (a) Remember the block will not move until the applied force just
exceeds the maximum static friction force. |
(a)
t = 5 s (b) t = 12 s; Vmax = 49.9 ft/s (c) t = 17.88 s |
24Sept07 |
5.17 |
For
part (a) you must find out if the boxes move together or if they pull
apart. To do this examine each block as if it was alone on the conveyor
belt. IF the acceleration of Block A, all alone, is greater than the
acceleration of Block A, all alone, THEN Block A pushes Block B and they move
together as a single unit. IF the converse is true, then Block B will pull
away from Block A and they will separate. There is only one answer to part
(b), depending upon what you find in part (a). If they move as a single unit,
you must find the actual acceleration of the combined system with the
appropriate friction forces for each box. |
aa = 0.997 ft/s2 |
21Sept05 |
5.18 |
As
is the case with many problems involving friction, you do not know what the
state of motion is a priori. You must therefore assume some state of
motion first, apply CoLM, and then chack that assumption against something
else. I assumed that block A did not slide relative to B,
solved CoLM for the friciton force in between the blocks, and then compared
that value to the critical value of friction when impending motion
would occur (which is...) |
(a)
No sliding. |aA| = |aB| = 0.667
m/s2 |
21Sept05 |
5.19 |
(a)
Application of finite-time form of the cons. of LM to the package and cart
combined. Be sure to show all forces that act on your system when you draw
your system diagram. (If both cart and package are in the system are there
any forces drawn between the package and cart? NO! Why?) (b) Since you want impulse exerted on package by the cart, you want to find the time integral of the forces that the cart exerts on the package during the impact. To do this you must select the system as your package. The x-component is independent of the time interval; however, the vertical component of the impulse depends on the time for the impact. IF Dt << 1 second, the impulse in the positive y-direction is 15.00 N-s. |
(a)
0.742 m/s (b) RxavgDt =
-18.56 N-s |
21Sept05 |
5.20 |
Apply
conservation of LM to combined system of both cars, then solve for the ratio
of Va to Vb. This will tell you who is going faster, then solve B.
Slower car is traveling at the speed limit of 50 km/h. Don't forget
that LM has two components in the x-y plane. Gravity plays no role in this
problem. |
(b)
VA = 192 km/hr |
21Sept05 |
5.21 |
To
find the tension in the rope, where should your system boundary be? What
information have you been given by looking at the speed reduction over time?. |
(a)
8.51 km/h |
21Sept05 |
5.22 |
Clarifications: Pressure at the
inlet to the elbow is 2 psig (or 16.7 psia) The pressure at the exit of the elbow
is atmospheric pressure, 0 psig (or 14.7 psia.) Pressures reported as
psia are measured with respect to a complete vacuum; pressures reported as
psig are measured with respect to the local pressure of the atmosphere. The
abbreviation psig psig stands for "pounds per square inch gauge"
which is a way of reporting pressures relative to atmospheric
pressure, i.e. 2 psig = (2 + 14.7) psia "pounds per square inch
absolute." The KEY issue is that all pressures are reported in the same way for
purposes of the calculation, i.e. all in psig or all in psia. When
atmospheric pressure acts over part of a system, the easiest way is to
account for the net pressure forces is to use all pressures in psig. Then the
only those pressures that differ from atmospheric will produce a net force.
[Note that it is impossible to have negative absolute pressures; however, a
vacuum will have a negative gage pressure, e.g. 12.7 psia = -2.0 psig.] |
---Horizontal
force acting on the elbow by the pipe = 160.9 lbf to the right
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21Sept05 |
5.23 |
Part
(a): the impact is so short, the impulse do to friction can be neglected.
Part (b): assume the blocks immediately start to slip relative to each other
(you will want to use two systems due to the friction between the blocks).
Part (c): when the blocks stop sliding on each other, they will have equal
velocities and accelerations relative to the ground. |
a)
v = 0.86 m/s (right) b)
aA = 0.97 m/s2 (left), aC = 2.94 m/s2
(right) c)
t = 0.22s |
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5.24 |
Determine
the unknown constant A from the given information. Then answer the
other questions as required. Suggest you use a spreadsheet or MAPLE to
graph the various quantities to make life simple for your self. |
(b)
168.5 m |
21Sept05 |
5.25 |
Apply
conservation of linear momentum to the system and recognize that there is no
motion; thus, the system has a constant value of linear momentum. |
TAC= 1376 N; TAB= 1601 N |
18Sept06 |
5.26 |
Apply
conservation of LM to the block. Although any coordinate system will work,
the easiest coordinate system to use is one aligned with the inclined plane,
i.e. x is parallel to the inclined plane and y is perpendicular to the plane. |
(a)
56.97 N to 318.6 N (b) --- |
21Sept05 |
5.27 |
(a)
Carefully consider your system. If you pick both the block A and mass
B as the system, you will have two unknowns: dV/dt and F.
Thus you will need an additional relation between these two variables or know
at least one of them. Is there another system, you can pick. [Hint:
Consider just the mass and note that the link can only transmit a
force that is parallel to the link.] |
(a)
207.4 N |
21Sept05 |
5.28 |
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5.29 |
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5.30 |
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5.31 |
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5.32 |
Do
not be scared by the lack of numbers, just follow the standard COM and COLM
method. Note, the drag force pulls the boat to the right. |
a)
v1 = V_dot/A1, v2 = V_dot/A2 b)
V_dot = vw*sqrt((kA1A2)/(rho(A1-A2cos(theta)))) |
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5.33 |
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5.34 |
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5.35 |
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5.36 |
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5.37 |
Remember
that velocity has a direction; this is important for the m_dot*v terms. Also
consider which inlet or outlet has an associated gauge/vacuum pressure; if
one exists you will need to include it on your FBD. |
R
= 34.9 kN (down) |
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5.38 |
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5.39 |
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5.40 |
Maximum
deceleration can occur only when the maximum friction force is developed. |
dv/dt
= 1.25 m/s2 (left) |
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5.41 |
If
block B immediately moves, you know kinetic friction must be involved. Since
there is friction between the blocks, it will be easiest to use two systems. |
T
= 97.7 N aB
= 2.93 m/s (down the ramp) |
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5.42 |
What
exactly are you trying to find here? Would it make sense to treat the car and
the ball as a single system? Consider the ball as a closed system. Use
LM to relate the projectile trajectory to its initial velocity. Note that
the LM of the ball must be written in terms of its absolute velocity measured
with respect to the ground not its relative velocity with respect to the car
(the given information in the problem). Use relative velocity relations to
relate the initial ball velocity required for the given trajectory to the
the given information and the car velocity. |
(a)
0.792 s |
21Sept05 |
5.43 |
Apply
both conservation of mass and conservation of linear momentum to an open
system that corresponds to the cart and water in the tank. Assume that all of
the water in the tank is moving with the velocity of the cart. Take care when
you differentiate the LM of the open system and use the cons. of mass where
applicable to relate dm/dt to the mass flow rate. Also note that the
mass flow rate must be calculated with a velocity that is measured relative
to the flow area. |
vc,max
= (v0 - F/(rhov0A)(ln(mc+mw)/mc) |
26Sept05 |
5.44 |
Use
a closed system that minimizes the unknown information. Use the relative
velocity relations to relate the absolute velocities of the boat and
swimmer(s) to the absolute velocity of the swimmer(s). |
(a)
1.154 m/s ← |
29Sept05 |
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