FE/EIT Dynamics Sample Problems
FE/EIT Dynamics Sample Problems
FE/EIT Dynamics Sample Problems
A body in motion can be considered a particle if rotation of the body is absent or insignificant. A particle
does not possess rotational kinetic energy. All parts of
a particle have the same instantaneous displacement,
velocity, and acceleration.
A rigid body does not deform when loaded and can be
considered a combination of two or more particles that
remain at a fixed, finite distance from each other. At
any given instant, the parts (particles) of a rigid body
can have different displacements, velocities, and accelerations if the body has rotational as well as translational
motion.
a
9
r
acceleration
gravitational
acceleration
position
radius
distance
time
velocity
ft/sec2
m/s2
ft/sec2
ft
m/s2
rn
ft
ft
sec
ft/sec
s
m/s
e
w
dr
dt
dv
d2r
dt
dt2
V=-
a=-=-
[position]
14.1
[velocity]
14.2
[acceleration]
14.3
RECTILINEAR MOTION
Symbols
0:
angular acceleration
angular position
angular velocity
rad/sec''
rad
tad/see
rad/s2
rad
rad/s
Subscripts
0
initial
final
f
n
normal
r
radial
t
tangential
transverse
v(t)dt
JJ
a(t)de
14.4
14.5
_ d2s(t)
dt2
14.6
( ) _ dv(t)
a t dt
Rectangular Coordinates
INTROOOCTIONTo"KINEMATICS
Dynamics is the study of moving objects. The subject is
. divided into kinematics and kinetics. Kinematics is the
study of a body's motion independent of the forces on
the body. It is a study of the geometry of motion without consideration of the causes of motion. Kinematics
_______________________________________
(x, y. z)
a(t)
v(t)
ao
s(t) = ao
J
JJ
14.10
dt = va
+ aot
dt
aot2
= So
v (t)
2
~ath
]4.1]
+ vat + -2-
v6 + 2ao(s
- so)
14.12
14.13
of particle
x
k
CURVILINEAR MOTION
z
0\
= xi + yj + zk
14.7
The velocity and acceleration are the first two derivatives of the position vector, as shown in Eqs. 14.8
and 14.9.
v=-
dr
dt
= xi + yj + z k
dv
a=-=dt
14.8
d2r
In polar coordinates, the position of a particle is described by a radius, r, and an angle, B. The position
may also be expressed as a vector of magnitude r and
direction specified by unit vector er. Since the velocity
of a particle is not usually directed radially out from the
center of the coordinate system, it can be divided into
two components, called radial and transverse, which
are parallel and perpendicular, respectively, to the unit
radial vector. Figure 14.2 illustrates the radial and
transverse components of velocity ina polar coordinate
system, and the unit radial and unit transverse vectors,
er and ee, used in the vector forms of the motion equations.
dt2
=xi+jjj+ik
14.9
v = ;e;
[position]
14.14
[velocity)
]4. ]5
[acceleration]
/4./6
+ Veee
= fer + reee
Constant Acceleration
Acceleration is a constant in many cases, such as a free-
.falling body with constant acceleration g. If the acceleration is constant, the acceleration term can be taken
= (..r-r B02) er
+ (rB + 2rB)ee
Kinemati(s
of a particle around a fixed circular path. The behavior of a rotating particle is defined by its angular position, B, angular velocity, w, and angular acceleration,
a. These variables are analogous to the s, v, and a
variables for linear systems .. Angular variables can be
substituted one-for-one in place of linear variables in
most equations.
path of
particle
\8
eQ~ __e~r
__ ~l
w=-
dB
dt
[angular position]
14.19
[angular velocity]
14.20
[angular acceleration]
14.21
dJ.JJ
a=-
dt
d2e
dt2
14~3
A particle moving in a curvilinear path will have instantaneous linear velocity and linear acceleration. These
linear variables will be directed tangentially to the path,
and, therefore, are known as tangential velocity, Vt, and
tangential acceleration, at, respectively. The force that
constrains the particle to the curved path will generally
be directed toward the center of rotation, and the particle will experience an inward acceleration perpendicular to the tangential velocity and .acceleration, known
as the normal acceleration, an. The resultant acceleration, a, is the vector sum of the tangential and normal
accelerations. Normal and tangential components of acceleration are illustrated in Fig. 14.3. The vectors en
and et are normal and tangential to the path, respectively. p is the principal radius of curvature.
_-_
. ...............................
__
= rB
Vt
= rw
14.22
14.23
dVt
at
14.24
= ra = dt
vr
an=-=rw
r
14.25
14.17
14.18
PROJECTILE MOTION
A projectile is placed into motion by an initial impulse.
(Kinematics deals only with dynamics during the flight.
The force acting on the projectile during the launch
phase is covered in kinetics.) Neglecting air drag, once
the projectile is in motion, it is acted upon only by
the downward gravitational acceleration (i.e., its own
weight). Thus, projectile motion is a special case of
motion under constant acceleration.
Consider a general projectile set into motion at an angle of B from the horizontal plane, and initial velocity
Vo, as shown in Fig. 14.4. In the absence of air drag,
the following rules apply to the case of travel over a
horizontal plane.
x
45 .
14-4
FEReview Manual
2. What
= O?
is the acceleration
of the particle
at tiuie
(A) 2 mfs2
(B) 3 mfs2
(0) 5 mfs2
(D) 8 mfs2
Th
#17 394 .
y
v{t)
. is ;
~----
path of projectile
ch<
Solution:
ize
(D
tiv
At t
Fa
= 0,
nO"
a = 8 mffp
Answer is D.
The following solutions to most common projectile
problems are derived from the laws of uniform acceleration and conservation of energy.
ax = 0
14.26
14.27
ay = -g
Vx = Vxo = Vocosf
14.28
14.29
14.30
Vy = VyO- gt = Vosine - gt
x
= vxot = votcose
= Vyot
1
- 29t
= vot sin e -
1
29t
14.31
3. What
ticle?
(A)
(B)
(0)
(D)
21.8
27.9
34.6
48.0
mfs
mjs
mfs
mfs
#78394
>
Solution:
The maximum of the velocity function is found by
equating the derivative of the velocity function to zero
and solving for t.
sAMpLEPROBLEMs
Problems 1-3 refer to a particle whose curvilinear motion is represented by the equation s = 20t + 4t2 - 3t3.
1. What is the particle's
dv
dt
20
+ 8t - 9t2
-18t
= 8
= 0
8
s = 0.444 s
18
t = -
initial velocity?
20 mjs
25 mjs
30 mfs
(A)
(B)
(C)
(D)
is the maximum
Vmax =
32 mfs
20
+ (8)(0.444
s) - (9)(0.444
S)2
21.8 mfs
Answer is A.
#76394
Solution:
v = -d
At t
= 20 + 8t - .9t
0,
v
= 20 + (8) (0)
= 20 mfs
- (9) (0)2
(A) a
(B)
(0) v
(D)
= 9.81
v = Vo
vo
mjs2
a rigid
+ voft
+ aot
+
lot a(t)dt
a = vVr
Answer is A.
Professional Publications, Inc.
---
..
14-6
FE Review Manual
._
Solution:
gt2
e- 2
votsine + y = 0
Y = vot sin
t2
!J...
2
y
= -1500
(9" m)
.~1 2"
2s
t2
m
-
-b
S2
1500 m
(C) 10 m/s
(D) 11 m/s
...jb2 - 4ac
= .
#84394
[quadratic formula]
2a
500
0
(A) 0 m/s
4.905 - 2t - 500 t
V( -500)2
- (4)(4.905)(-1500)
(2) (4.905)
= +104.85 s, -2.9166
= votcos8
(1000 :)
(104.85 s) cos 30
90803 m
(90800 m)
32 m;
45 m;
58 m;
(D) 64 m;
(A)
(B)
(C)
Answer is D.
the
the
the
the
ball
ball
ball
ball
set)
5. Rigid link AB is 12 m long. It rotates counterclockwise about point A at 12 rev/min. A thin disk with radius 1.75 m is pinned at its center to the link at point B.
The disk rotates counterclockwise at 60 rev /min with
respect to point B. What is the maximum tangential
velocity seen by any point on the disk?
(A) 2.61
(B) 2.75
(C)
3.30
(D) 4.12
#82689
(A)
(B)
(C)
(D)
2.00
2.56
3.14
4.00
#83689
IIIIiIII
(A)
(B)
(C)
(D)
6 m/s
26 m/s
33 m/s
45 mjs
.;..
__
.;..
--
-.~
II
I
Kinematics
Hand-
6. A particle has a tangential acceleration of at (represented by the equation given) when it moves around
a point in a curve with instantaneous radius of 1 m.
What is the instantaneous angular velocity (in rad/s)
of the particle?
at
2t - sin t
(A) t2 + cost
+ 3 In
+ 3 cot t
8(t)
= t3 -
2e -
4t
+ 10
[in m/s2]
10. What is the angular velocity at t = 3 s?
(A)
[csc r]
-16 rad/s
(B) -4 rad/s
t2-cost+3lnlcsctl
(B)
(C) t2 - cost + 3 In [sin t]
(D) t2+cost+3Inlsintl
(C)
(D)
11 rad/s
15 rad/s
#2886687
#3555794
7. A
stone
sound
of the
14-7
(A) 2.4 m
(B). 7.2 m
5 s? .
(A) 4 rad/s2
(B) 6 rad/s2
(C) 26 rad/s2
(D) 30 rad/s2
. #2886 687
(C) 28 m
(D) 30 m
#2878687
0.40 rad/s?
2.5 rad/s2
(C) 5.8 rad/s2
(D) 16 rad/s2
#2887687
0.18 m/s2
0.25 m/s2
0.37m/s2
(D) 1.3 m/s2
(A)
(B)
(C)
#2885687
____________________________________
14-8
FEReview Manual
-----------!!IIIIII----------I-------- __
._
(A)
(B)
(C)
(D)
3.8 s
7.7 s
8.9 s
12 s
Vt =
#OOOOMM 298
15. What
projectile?
Soluti
Solution 3:
is the horizontal
and rota-
The
spec'
rw
(21l' ::~)
(60 ~)
mm
(A) 80 m
(B) 400 m
(C) 800 m
(D) 1200 m
=
Vt =
(A)
(B)
72 m
140 m
(C) 350 m
(D) 620 m
(0.3
rad)
m) ( 36.65 -s-
The
the
= 11.0 m/s
#OOOOMM 298
36.65 rad/s
#OOOOMM 298
Th<
Solution 4:
con
per
of
Solution 1:
Y = Vyot -
. ds(t)
= --;It =
v ()t
At t
gt2
= vat sin
"'2
e-
gt2
"'2
2.
.
cos zi -- 4sintJ
When the ball hits the ground, y = 0, and
4 rad,
v(4)
Iv(4)1
-1.3li - (-3.03)j
V(-1.31)2
votsine
gt2
"'2
+ (3.03)2
= 3.30
2va sin ()
t= -'--9
Answer is C.
So
Solution 2:
From Prob. 1,
v( t)
x
2 cos ti - 4 sin tj
dv(t)
at( ) = -= -2Sintldt
a(rr) = -2sin1l'i-4cos1l'j
+ 4.0j
\a(1l') \ = V(O)2 + (4.0)2
=
Answer is D.
Oi
4.0
4costJ
= vatcose
2v2
= _0
sin
2vosine)
= vc (
mr
9.812'
s
Answer is D.
cose
e cos ()
(2) (25
---'----;m~s~
= 63.7 m
sin 45 cos45
Solution 5:
Solution 7:
The elapsed time is the sum of the time for the stone to
fall and the time for the sound to return to the listener.
The distance, x, traveled by the stone under the influence of a constant gravitational acceleration is
= rw = r(27rf)
Vt;BIA
( 12 -.rev)
nun
min
1 2
+ vot + 2,gt
X = Xo
Xo
= 15.08 m/s
1
x = _gt2
2
= rw =r(27r f)
Vt,disklB
(60 re.v)
min
rev
60 _s_
min
11.00 sss]
f!
-- x+ -x = 2.47
c
Vt,disklB
= Vt,BIA + Vt,disklB
m
15.08 -
26.08 m/s
+ 11.00 -
t2 = 2.47 s
9.81 -
342-
S2
Answer is B.
+ ------m
= 2.47 s
Solution 6:
Answer is C.
dVt
at=:"
Vt =
dt
Solution 8:
at dt
=J(2t~sint+3
cott) dt
= t2 + cos t + 3 In [sin tl
t2
Vt
+ cost + 3 In
(70 ~)
[in m/s)
[sin r]
vo
rim
t2
+ cas t + 3 In
Answer is D.
___________________________________
[sin tl
[in rad/s)
3600 ~
= - = ------------~--~
(1000 ~)
19.44
mls
+ vot
2,1 at 2
Professional Publications, Inc.
14-10
FEReview Manual
= Wo
(2)(vot - x)
t2
- at
o=
(15 S)2
ex
0.37 m/s2
.~. .Ans'"
2.51 rad/s
Answer is B.
Answer is C.
Solution 13:
1
B = Bo + wot - 2at
Solution 9:
v
= va + at
krn
= 70 _
+ (1200
x
( -0.5
~)
(158) (3600 ~)
+ .:....:.
__ -'-----=-'-_~'-
",-8
= 43
min
(15 s) - 250 m]
(2) [( 19.44 ~)
Solution 12:
a = ~--'----".,,----'-
1000 k~
e=
r:2d)
(60 rr:n)]
103 rad
km/h
::i:)
[(6 min)
135.4
271
= 21.5
103 rev
Answer is A.
Alternative solution: The average rotational
ing deceleration is
speed dur- .
rev
.
7200 -.
-0
Solution 10:
mm
dB = 3t 2 -
(t) =. -dt
w(3)
..4t - 4
(3)(3)2 - (4)(3) - 4
11 rad/s
2
3600 rev /min
e = wt
rev)
= ( 3600 min
(6 min)
= 21,600 rev
Answer is B.
Answer is C.
Solution 14:
The vertical component
Vy =
Solution 11:
0=
a(t)
dw(t)
dt
= 6t - 4
a(5)
Answer is C.
(6)(5) - 4
26 rad/s2
va sine - gt
(no 7) sin 20 -
(9.81 ~)
t = 3.84 s
Kinematics
solution 15:
Solution 16:
(110
7)
cos200(7.68
=794m
y = vot sin
s)
= (110
Answer is C.
e ~ (~)
7)
- (~)
=72
14-11
gt
(3.84 s) sin 20
(9.81 ~)
(3.848)2
Answer is A.
______________________________________
Kinetics
Subscripts
0
f
k
n
r
R
s
t
initial
friction
dynamic
normal
radial
resultant
static
tangential
transverse
INTRODUCTION TO KINETICS
Nomendature
a
f
F
9
9c
k
m
N
p
r
R
s
t
T
v
acceleration
linear frequency
force
gravitational
acceleration
gravitational
constant (32.2)
spring constant
mass
normal force
linear momentum
position
radius
resultant force
distance
time
period
velocity
weight
ft/sec2
Hz
lbf
m/s2
Hz
N
ft/sec2
m/s2
S
()
j.L
angular
acceleration
deflection
angular position
coefficient of
friction
angle
natural frequency
lbm- ft /lbf-sec2
lbf/ft
lbm
lbf
lbf-sec
N/m
kg
N
N-s
ft
ft
lbf
ft
sec
sec
ft/sec
lbf
rad/sec?
rad/s2
m
rad
ft
rad
.............................................
MOMENTUM
The vector linear momentum
(momentum)
is defined
by Eq. 15.1. It has the same direction as the velocity
vector. Momentum has. units of force x time (e.g., lbfsec or Ns).
p=mv
N
p=-
m
s
s
m/s
N
Symbols
Q
mv
gc
[81]
15.10
[U.S.]
15.1b
NEWrON/SFiRSTANDSECONDiAWSOF
MOTIO~........
.r:
deg
deg
rad/sec
rad/s
15-2
FEReview Manual
external force
F= dp
15.2
dt
For a fixed mass,
15.5
The static coefficient of friction is usually denoted with
the subscript s, while the dynamic coefficient of friction
is denoted with the subscript k. ILk is often assumed
to be 75 percent of the value oflLs. These coefficients
are complex functions of surface properties.
Experimentally determined values for various contacting conditions can be found in handbooks.
For a body resting on a horizontal surface, the normal
force is the weight of the body. If the body rests on an
inclined surface, the normal force is calculated as the
component of weight normal to that surface, as illustrated in Fig. 15.1.
N
mgcos
mgcos
15.6b
dv
dt
=ma
F=
[US.]
---;It
dt
=m-
15.6a
9c
F _ dp _ d(mv)
-
[SI]
ma
gc
[SI]
15.3a
[US.]
15.3b
impending
motion
WEIGHT
The weight, W, of an object is the force the object exerts
due to its position in a gravitational field, g.
W=mg
[SI]
15.4a
W= mg
gc
[US.]
15.4b
Friction is a force that always resists motion or impending motion. It always acts parallel to the contacting
surfaces. If the body is moving, the friction is known
as dynamic friction. If the body is stationary, friction
is known as static friction.
-lJ.Ls
15.1
----
Kinetics
"Figure 15.2 "Frictional Force ~ersus DisturbingForce " " " ... " .. "
.
no rnotron
I(
impending motion
~e~~~b~~~~~~e~
LFn
~sN
Rectangular Coordinates
Equation 15.8 is Newton's second law in rectangular
coordinate form and refers to motion in the x-direction.
Similar equations can be written for the y-direction or
any other coordinate direction.
[81]
15.8
;,
Xo + vxot
"(Fx(t))
---:;:n-
dt
[81]
[81]
15.15
=mae
[81]
15.16
[81]
15.17
.................
FREEVIBRATION
::, .
15.9 '
If Fx is constant (i.e., is independent of time, displacement, or velocity), then the motion equations become
[81]
vx(t)
= v-o + (~)
15.11
t
15.12
x(t) = Xo
+ vxot +
F.
t2
2~
t2
= Xo
(d~t)
=m
15.10
vx(t)dt
e
h
15.14
LFe
KiNETICSOF"A"PARTiCLE""
x(t)
[81]
disturbing force
ri
e
e
:,
m(vI)
4~
= man =
t mat
LF
~motion
vx(t)
15-3
ax
+ vxot + --2
__________________________________
15.13
[81]
15.180
[u.s.]
IS.18b
15-4
FEReview Manual
F=ma
+ Ost) = m/i:
+ Ost) = mii:
mx+kx
=0
mg - k(x
kOst - k(x
[81J
15.19
[81]
15.20
[81J
15.21
15.22
[81]
sAMp(fpROBIEM"S
....................
a~ constant
velocity
.
Solution:
This is a restatement of Newton's first law of motion
which says that if the resultant external force acting
on a particle is zero, then the linear momentum of the
particle is constant.
C1 coswt
+ C2 sinwt
15.23
give a 1 g
accelerate
accelerate
accelerate
Solution:
w={
w= )kf=~=~ T
T= ~=
f
w
gc
m
[81]
15.240
[U.8.]
15.24b
15.25
211"
211"
xocoswt
+ (~)
sinwt
= xocos
wt
m/s2,
m/s2,
m/s2,
m/s2,
15.27
x( t)
Answer is B.
15.26
x(t)
15.28
#88691
5.
Solution:
50e4 N
550 kg
a=-=--F
m
4.96 rn/s2
---.
Kinetics
s(t)
s
50et N
550 kg dt
)0
et 4
1110
= 4.87
m/s
50e N
--k- de
a 550 gas
ti it t
a
= ~87tl~ =
(4.87
7)
= 4.96
it
Solution:
- 11
(4.87
m)
dt
Fx
= mgx
max
max = mgsin45
(4 s) - 0
19.48 rn
15-5
p,=
mgsin45
is
- Ff
- p,mgcos45
-:-max
gsin45
mgcos45
- ax
9 cos 45
Answer is A.
(9.81
Problems
(A) 1.5
:Z) cos 45
0.78
Answer is C.
6. A constant force of 750 N is applied through a pulley
system to lift a mass of 50 kg as shown. Neglecting
the mass and friction of the pulley system, what is the
acceleration of the 50 kg mass?
(B) 3 m
(C) 6 m
(D) 12 m
#89689
F= 750 N
Solution:
v(t)
va
+ aot
m
ao=
v(t) -va
6 -;-0
= _o?...-_
4 s
(C)
(D)
= 1.5 m/s2
sCt)
So + vat
=0+0+
=
+ 2aot
(~)
16.2 m/s2
20.2 m/s2
#91 694
Solution:
(1.5 ~)
(4S)2
12 m
Answer is D.
T,
1.
mass A
#90689
F
T,
pulley B
pulley C
156
a=
mass A:
Tl - mg = ma
pulley B:
2T2 - T, = 0
pulley C:
T2
=F =
2T2 -
Tl - mg
m
750 N
2F -
mg
problE
In sta
- and b
initial
. pulley
F(N)
mg
2
t (5)
0 m/s
0.075 m/s
(C) 0.15 m/s
(D) 0.30 ta]
50 kg
20.2 m/s2
Answer is D.
#94691
(A)
(B)
0.30 s
0.60 s
(C) 0.90 s
(D) 6.3 s
#92394
The 52 kg block shown starts from rest at position A and slides down the inclined plane to position B.
C
C
Solution:
T
1m
= 27rV k = 27r~
C-
F
. are rl
10 ~
(
(
= 6.3 s
(
Answer is D.
, For'
bool
7. \
8.
. #95681
4.
C.
of a
no C
spriJ
(A) 3.23 m
(B) 4.78 rn
(C) 7.78 m
(D) 10.1 m
#96681
.-
--
Kineti(s
15 7
(A) 46 N/m
(B) 56 N/m
(C) 60 N/m
(D) 63 N/m
#39351294
10000 N
o m/s
(B)
3.5 m/s
(B)
0.59
#3942295
(A)
(A) 0.35 m
(C) 0.77 m
(D) 0.92 m
#97994
For the following problems use the NCEES Handbook as your only reference.
7. What is the period of a pendulum
center point 20 times a minute?
(A)
(B)
(C)
(D)
0.2 s
0.3 s
3s
6 s
#3471 193
4600 N-s
22 000 N-s
37000Ns
48 000 Ns
#2142689
e,
#1584 689
(A) -14
(B) 25
(C) 64
(D) 76
#39331294
#4.169 795
15-8
FEReview Manual
(A) 0.19 m
(B) 1.1 m
(C)
1.3 m
(D) 15
m
#2108
689
An:
LFx
= ma
Wx - flN
Solu'
= max
Ref,
Solution 1:
Newton's second law, F = ma, can be applied separately to any direction in which forces are resolved into
components, including the resultant direction.
ax
= 9
sin
fl9 cas
5
[1 3 - (0.15)
(9.81 ~)
2.415 m/s2
= v~ + 2ao(s - so)
va
G~)]
So = 0
= 2aos
=
Ap
Answer is C.
v =
(20 m)
(2) (2.415 ~)
= 96.6
Solution 2:
e-
m2/s2
96.6
~2
Bu
= 9.83 iis]
Answer is D.
Solution 4:
LFx
dp
=ma
dt
FD..t
= m!:::.v
FD..t
D..v
= --
(2) [(~)
(3N)(2S)]
= --=-~'--:,------=-
tri
+ Psine) = ma
40 kg
= 0.15 m/s
= (~)
[mgsine - pcose
- fl(mgcose
+ P sin e)]
Cc
TI:
P
= g sin e - flg cos e - - (cos e + fl sin e)
m
Answer is C.
Solution 3:
(9.81 ~)
52 kg
[1 3_ (383 N)
5
(0.15TG~)]
[12
13
+ (0.15)
(~)]
13
-4.809 m/s2
---
Kinetin
2
'2
9.83 m/s
30
-vo
3
[from Prob. 3]
=0
= 2ao = (2)
= 10.05
(9.83
But F
m)2
S
(-4.809
ma, so
;)
Since the tension in the rope is the same everywhere,
m :
Answer is D.
10000 N - T
10000 N
Solution 5:
Refer to the following free-body diagrams.
T
T - 7500 N
7500 N
T = 8571 N
From block A,
T)
g(WA -
WA
(9.81 ~)
=
But m
= Wig,
(10000:)
9.812'
(10000 N - 8571 N)
10000 N
1.4 m/s2
Answer is B.
so
(aA)
Solution 6:
10000 N - TA
VA = Vo
+ aAt
0 + (1.~ ;)
(2.5 s)
= 3.5 ui]
Answer is B.
Solution 7:
Combine the equations and solve for aA, setting TA
TB, and aA = -aB'
WA - mAaA
aA=
WB - mBaB
WA - WB
mB+mA
m)
= 9.81 s2
= WE + mBaA
g(WA - WB)
WB+WA
(10000 N -7500 N)
7500 N + 10000 N
1.4 m/s2
___________________________________
- __'J1I
I
Alternate Solution:
v =vo+2aO(S-30)
vo
15-9-
60 s
10
6 s
_
Answer is D.
.~h~oo~~~~m~l~
1510
FEReview Manual
Solution 8:
From Hooke's law, the relationship between
and deflection, x, or a linear spring is
force,' F,
(l~OO ~)
3600
F=kx
=
(40 N)cose
(20 cm)
cm
100 m
(100 :)
x2
(50 :)
=
case
= 0.25
Answer is C.
e = cos-1(0.25)
-;; 75.5
:r; = 0.77 m
(76)
Solution 11:
The normal acceleration
the airplane) is
Answer is D.
Solution 9:
an
(perpendicular
vZ
= ..l.
kz =
k1Xl
to the path of
(10
kZX2
~)2
1.5 m
k1Xl
66.7 m/s2
X2
(50 ~)
The tension
force.
(30 cm)
in the string
25 cm
= 60 N/m
(1 kg) (66.7 ~)
= 66.7 N
Answer is C.
Answer is D.
Solution 10:
Use the conservation of momentum equation to determine the velocity of the cannon after the ball is fired.
Initially, the cannon and cannonball are both at rest.
Since the cannon recoils, its velocity direction will be
opposite (i.e., negative) to the direction of the cannonball.
Solution 12:
(100 k:)
v
3600 ~
= 27.78
(10 kg) (0)+(250
=
= 48060
Vc
= -40
7)
N-s
km/h
KE=PE
sii]
P=mv
kg) (0)
(10 kg) (1000 ~)
(1000 :)
= --'-----"---.",.----
Answer is D.
Solution 13:
The law of conservation of momentum states that the
linear momentum is unchanged if no unbalanced forces
act on an object. This does not prohibit the mass and
velocity from changing; only the product .of mass and
m-
"Tfr1
I '
"\1: I." .
Kinetics
15-11
;1
II
Answer is B.
Solution 14:
F
1000 N - 150 N
850 N
a=-
m
850 N
3500 kg
=
0.243 m/s2
1
= vot + zat
+ (~)
1.09 m
(0.243 ~)
(3 s)2
Answer is B.
i ,
I
i
Kinetics of
Rotational Motion
Subscripts
o
c
f
n
o
S
initial
centroidal
friction
normal or natural
origin or center
static
tangential or torsional
Nomenclature
a
A
d
F
9
. ge
G
h
I
J
kt
m
M
r
r
R
t
v
W
acceleration
area
distance
force
gravitational
acceleration
gravitational
constant (32.2)
shear modulus
angular momentum
mass moment
of inertia
area polar moment
of inertia
torsional spring
constant
length
mass
moment
radius
radius of gyration
moment arm
time
velocity
weight
ft/sec2
ft2
ft
lbf
m/s2
m2
ft/sec2
m/s2
lbm-ft.Zlbf-sec''
lbf/ft2
ft-lbf-sec
Pa
Nms
Ibm-ft2
kg.m2
i: =
(y2
+ z2)dm
16.1
ft4
m4
t;
(x
+ z2)dm
16.2
ft-lbf/rad
ft
lbm
ft-lbf
ft
ft
ft
sec
ft/sec
lbf
N-m/rad
rn
kg
N-m
rn
m
m
s
m/s
N
(x2
+ y2)dm
16.3
rad/sec''
rad
deg
rad/s2
rad
deg
Ibm/ft3
rad/sec
rad/sec
kg/m3
radj's
rad/s
m
N
t,
Symbols
Q
o
()
fJ,
p
w
to
angular acceleration
angular position
superelevation angle
coefficient of friction
density
angular velocity
natural frequency
J
=J
=J
.
Iany parallel
axis
Ie
+ md
16.4
-2
16.5
16-2
FEReview Manual
r=ff
]6.6
1= r2m
]6.7
Table 16.1 (at the end of this chapter) lists the mass
moments of inertia and radii of gyration for some standard shapes.
ho =Iw
Iw
ho=gc
[SI]
]6.90
[U.S.]
]6.9b
]6. ]0
M=-
gra
tie~
dho
dt
]6.110
[U.S.]
]6.11b
plane motion
]6.12
16.]3
translation
]6.14
rotation
mv
mv
ho=rx.
9c
[SI]
]6.80
[U.S.]
16.8b
--
ac
in,
se:
N!
th
at
tT'"
be
tr
av
ce
in
_.------------------------------_
crraphic procedure is slightly different if the two velocities are parallel, as Fig. 16.2 shows.) For a rolling wheel,
the instantaneous center is the point of contact with the
supporting surface.
o
.. ,...
Fe
16-3
mvi
= man = -= mrw 2
r
_ man _ mv 2
t
Fe--------ge
ger
_ mrw
ge
[SI]
16.160
[u.s.]
16.16b
BANKING OF CURVES.
IC
IC
v=lw=
lvo
-
16.15
16.17
= J.lsW
e,
tan 8
v2
= -.i.
16.18
gr
CENtRIFUGACFORCf
16.19
)9
16-4
FEReview Manual
16.20
, ;:
.....
B(t)
Bocoswnt
+ (::)
sinwnt
16.21
SAMpLE' PROBLEMs
Solution:
Solut
The
asse
fron
be 3
ity (
VB
= 50
4m
II
------------1,-
m/s
I/IC
1
I
I
13m
#99689
I
I
I
Solution:
As the skater brings her arms in, her radius of gyration and mass moment of inertia decrease. However, in
the absence of friction, her angular momentum, h, is
constant. From Eq. 16.9,
vc-
VB
h
I
= ABwAB
= (5 m) ( 10 rad)
-s- = 50 m/s
w=-
wBC
VB
OB
50 m
= 4 ~ = 12.5 rad/s
[clockwise]
Answer is D.
Answer is D.
2. Link AB of the linkage mechanism shown in the
illustration rotates with an instantaneous counterclockwise angular velocity of 10 rad/s, What is the instantaneous angular velocity of link Be when link AB is
horizontal and link CD is vertical?
4.
ba
fri,
ca:
ill,
So
4 C . -
T
fo
5m
T
D._
~
Professional Publicalions, Inc.
--
16-6
FEReview Manual
f----'--'------j
(A)
10 m/s ~
10.0 m/s
16.2 m/s
18.5 m/s
#106193
14
25
44
54
km/h
km/h
km/h
km/h
#107394
5. Traffic travels at 100 km/h around a banked highway curve with a radius of 1000 m. What banking angle
is necessary such that friction will not be required to resist the centrifugal force?
(A) 1.4
(A) 18.0 N
(B) 2.8
(C) 4.5
(D) 46
(B) 19.6 N
(C) 24.5 N
(D) 29.4 N
#2107689
#104
694
Forthe following problems use the NCEES Handbook as your only reference.
2.
of
in
of
stant?
Professional Publkcfions, Inc.
---
(A)
(B)
(C)
(D)
4.1 kgm2
16 kgm2
41 kgm2
150 kgm2
0.47 kgm2
0.56 kgm2
0.87 kgrrr'
3.7 kgm2
(A)
(B)
(C)
(D)
#3553794
#2095689
(A) 0 kgm2
Cl,W
(B)
0.045 kg-rrr'
lever
#2096689
(A) 32.4 em
(B)
(C)
(D)
3.6.0 em
54.0 em
108 em
(A) 6 m/s2
(B) 12 m/s2
(C)
#3975395
13 m/s2
(D) 18 m/s2
9. A torsional pendulum consists of a 5 kg uniform disk
with a diameter of 50 em attached at its center to a rod
1.5 m in length. The torsional spring constant is 0.625
Nrri/rad. Disregarding the mass of the rod, what is the
natural frequency of the torsional pendulum?
(A) 1.0 rad/s
(B) 1.2 radys
(C) 1.4 rad/s
(D) 2.0 rad/s
#4170
#2893 687
795
10. A 3 kg disk with a diameter of 0.6 m is rigidly attached at point B to a 1 kg rod 1 m in length. The
rod-disk combination rotates around point A. What is
the mass moment of inertia about point A for the combination?
0.6 m
I__
'-
..;.
---------------------Cl~"
L
16-8
FEReview Manual
(A)
(B)
(C)
(D)
3g
16
(A) 8
(B) 21
(C) 36
(D) 78
3g
#1582689
3g
4
#2911
687
1.
4
Solution 1:
(B)
(D)
(A)
(C)
W
3
4W
7
7W
12
#2912687
15. A 1530 kg car is towing a 300 kg trailer. The coefficient of friction between all tires and the road is 0.80.
How fast can the car and trailer travel around an unbanked curve of radius 200 m without either the car or
trailer skidding?
(A) 40.0 km/h
(B) 75.2 km/h
(C) 108.1 km/h
(D) 143 km/h
an
= man = T
LPn
=
=
man
- Wsin60
+ mg sin 60
= 18.0 N
#1579689
2 m
= 0.5 m/s2
16. A 1530 kg car is towing a 300 kg trailer. The coefficient of friction between all tires and the road is 0.80.
The car and trailer are traveling at 100 km/h around
a banked curve of radius 200 m. What is the necessary banking angle such that tire friction will not be
necessary to prevent skidding?
2 (1 7)2
v
= -...1. = -'---"'-.:....-
Answer is A.
Solution 2:
tension
= centripetal
force
---
------------------------------
mv2t
T= __
r
mv2
--
= rmv = r mw
ho
m/s
(12.13 ~) (3600 ~)
v=
5.77 rad/s
2
= vf1gr
=
100 N
(2 kg)(1.5 m)
=
=f1mg
16-9
1000
= (1.5 m)2 (2 kg) (5.77 r:d)
r:
= 43.67 km/h
26.0 Nms
Answer is C.
Answer is B.
Solution 5:
Since there is no friction force, the superelevation
B, can be determined directly.
Solution 3:
Use the instantaneous center of rotation to solve this
problem. Assume the wheel is pinned at point A.
angle,
v2
tanB = ~
gr
e = tan "
(;!)
(100 ~)
(
(1000 ~))
3600 ~
=tan-l~--~--~~------~~
(9.81 ~) (1000 m)
Z2
= (2)(0.25
4.50
[lawof cosines]
Answer
= 0.2134 m2
I
= VO.2134 m2
Zvo
vp
= ---;;:=
=
= 0.462 m
Solution 6:
(0.462 m) (10 ~)
0.25 m
The wheel will lose contact with the top of the rail when
the reaction on the rail is zero. Refer to the following
illustration. Wheel A is the inner wheel.
18.5 m/s
is C.
Answer is D.
YCG
= 0.5
Solution 4:
The car uses friction to resist the centrifugal force.
______________________________
16-10
FEReview Manual
The forces acting on the car are the centrifugal force, Fe,
its weight, and the reaction at the outer wheel. (The
reaction at the inner wheel is zero.) Take moments
about rail B.
L MB = 0 = W XCG -
Solution 9:
The radius of the disk is
R= .~D
2
FeYCG
50 em
mv;YcG
r
=mgxCG-
_...::..::.....:....:c.
v = JgXCG'r
YCG
(~)
of inertia
of the disk is
= ~MR2
(9.81 ~)
TJ
0.25 m
(15 m)
= (~) (5 kg)(0.25
m)2
0.5 m
= 0.15625 kg-rrr'
12.1 m/s
Answer is B.
kt
Solution 7:
Find the formula for Ix in the Mass Moments of Inertia
table.
m(3R2 + 4h2)
Ix = ---''------"12
(50 kg)[(3)(0.5
m)2
+ (4)(3
153.1 kg-rrr'
= w2I
w=/
~----,--,-0.625 Nm
rad
0.15625 kgm2
m)2]
12
=
2 rad/s
(150 kgrn")
Answer is D.
Answer is D.
Solution 10:
Solution 8:
The centroidal
moment
of inertia is
ML2
= -3-
Irod,A
I = ~mR2
2
= 0.09
The acceleration
a
at
t =
= 3t =
=
(1 kg)(1 m)2
kgm2
12 s is
(3
r:d)
(12 s)
of inertia
MR2
(3 kg)
Ia
r=F
(Tr
2
2
(0.09 kgm ) (36 ~)
The distance
10 N
0.324 m
0.135 kgm2
AC is
AC
tl
J AB2 + BC
(32.4 em)
(1 m)2
Answer is A.
Idisk,c = -2-
36 rad/s2
Mo=Fr=Ia
0.33 kg.m2
..
1.04
+ (O.~
ill
rn
The resultant
acceleration is
a Va; + a~
=
+ mdiskAC2
= 0.135 kgm2 + (3 kg) (1.04 m)2
= Idisk,e
(6 ~)
= 3.38 kgrrr'
+ (12
~)
13.4 m/s2
Irod,A
Answer
Idisk,A
+ 3.38
0.33kgm2
kgm"
= 3.71 kg.m2
Answer
is C.
Solution 13:
Point C is L/4 from the center of gravity of the rod.
The mass moment of inertia about point C is
is D.
Ie
Solution 11:
=
=
Answer is A.
=
leG
+ md2
(112) mL
+m (~
Solution 12:
The angular acceleration
at t
2 s is
a(t)dt
WL mgL
J(6t - 4)dt
2
= 3t
4t +wo
The angular acceleration is
=2
LMe
Ie
mgL
4
(:8) mL2
= 4
The tangential
a=---
rad/s
= (0.75 m)
=
rad)
(8 ~
The tangential
6 m/s2
12g
7L
acceleration of point B is
= (~)
= (0.75 m) ( 4
=
12 m/s2
(~1)
3g
rad)2
Answer
is C.
,\
II
16-12
Solution 14:
Solution 16:
The velocity is
(100 k:)
(1000 :)
LMCG
3600 ~
27.78 m/s
angle is
=Rc (~)
MCG =
ICGacG
Rc= 4W
7
Answer is C.
Answer is B.
Solution 15:
Solution 17:
To keep the vehicle and trailer from skidding, the centripetal force must be less than or equal to the frictional
force. At the limit,
Fe =Ff
The unbalanced
= --
= mblockR(g
=p,g
The acceleration
is given by
(0.8) (9.81 ~)
M=Ia
mblockR(g
=
=
m/s
(39.6 ~)
==
mwheelr2a
(7.848 ~) (200 m)
= 39.6
- Ra)
mblockRg
Janr
- a)
- Ra)
= 7.848 m/s2
Vt
= mR(g
M = F R = (mg - ma)R
p,N
an=-
m
p,mg
+ (100 kg)(0.75
m)2
10.7 rad/s
Answer is C.
(3600 ~)
1000 :
= 143 km/h
Answer is D.
----
Nomenclature
a
e
E
F
g
\
\gc
h
I
Imp
k
m
p
t
v
W
X
acceleration
coefficient of
restitution
energy
force
gravitational
acceleration
gravitational
constant (32.2)
height above
datum
mass moment
of inertia
impulse
spring constant
mass
linear momentum
distance
time
velocity
work
displacement
w=
ft-lbf
lbf
J
N
Ibm-ftj'lbf-sec-'
angular velocity
17.1
Kinetic Energy
Kinetic energy is a form of mechanical energy associated with a moving or rotating body. The linear kinetic
energy of a body moving with instantaneous linear velocity v is
ft
KE = Zmv
lbm-ft2
Ibf-sec
lbf/ft
lbm
lbm-ft/sec
ft
sec
ft/sec
ft-lbf
ft
kgm2
Ns
KE= mv
2gc
rad/sec
rad/s
[SI]
17.20
[U.S.]
17.2b
N/m
kg
kg-m.'s
m
s
m/s
J
m
Symbols
w
l=
ENERGYANDWORIC
The energy of a mass represents the capacity of the mass
to do work. Such energy can be stored and released.
There are many forms that the stored energy can take,
including mechanical, thermal, electrical, and magnetic
energies. Energy is a positive, scalar quantity, although
[S1]
17.30
KE = Iw2
2gc
[U.S.]
/7.3b
For general plane motion in which there are translational and rotational components, the kinetic energy is
the sum of the translational and rotational forms.
The change in kinetic energy is calculated from thedifference of squares of velocity, not from the square of the
velocity difference.
.I
17.40
17.4b
17-2
FEReview Manual
Potential Energy
Potential energy (also known as gravitational potential
energy) is a form of mechanical energy possessed by
a mass due to its relative position in a gravitational
field. Potential energy is lost when the elevation of a
mass decreases. The lost potential energy usually is
converted to kinetic energy or heat.
PE
mgh
PE= mgh
[81]
[u.s.]
17.5a
LINEAR IMPULSE
Impulse is a vector quantity equal to the change in momentum. Units of linear impulse are the same as those
for linear momentum: Ibf-sec or N-s. Figure 17.1 illustrates that impulse is represented by the area under the
force-time curve.
11.5b
t2
Imp =
17.9
Fdt
17.6
11.10
'E,E = constant
17.7
F(t2 - h)
Fdt
= 6.(mv)
= mdv
F=
17.8
Generally, the principle of conservation of energy is applied to mechanical energy problems (i.e., conversion of
work into kinetic or potential energy).
17.11
= 6.p
00
VI
Tl:
to
gc
A spring is an energy storage device because a compressed spring has the ability to perform work. In a
perfect spring, the amount of energy stored is equal to
the work required to compress the spring initially. The
stored spring energy does not depend on the mass of the
spring. Given a spring with spring constant (stiffness)
k, the spring's elastic potential energy is
mdv
-=ma
dt
[81]
11.12
[81]
11.13
[81]
11.14
IMPACTS
According to Newton's second law, momentum is conserved unless a body is acted upon by an external force
such as gravity or friction. In an impact or collision,
contact is very brief, and the effect of external forces is
insignificant. Therefore, momentum is conserved, even
though energy may be lost through heat generation and
deforming the bodies.
Figl
_---------------
1_
17-3
Solution:
VI
1
2mv2
KE=
d(KE)
dt
(2) (~)
(mv) (dV)
dt
=mva
This combination of variables (kgm2/s3) corresponds
to a watt (i.e., power).
.~
Answer is D.
m1
V1
A simple way of determining whether the impact is elastic or inelastic is by calculating the coefficient of restitution, e. The coefficient of restitution is the ratio of
relative velocity differences along a mutual straight line.
The collision is inelastic if e < La, perfectly inelastic if
e = a, and elastic if e = 1.0. (When both impact velocities are not directed along the same straight line, the
coefficient of restitution should be calculated separately
for each velocity component.)
5m
What
point A?
2.
is the tangential
(A) 6.52
m/s
mls
#109
689
Solution:
11.17
(5 m) sin 60
4.33 m
This drop in elevation corresponds to a decrease in potential energy and an increase in kinetic energy.
t.PE = t.KE
mgt.h
#108
691
V2gf:lh
9.22
Answer is B.
mls
V (2)
mv2
-2-
(9.81 ~)
(4.33 m)
FEReview ~anual --
17-4
What is the instantaneous acceleration in the direction of travel of the block at point B?
3.
(A) 0 m/s2
(B)
2.45
m/s2
#110689
Solution:
#112691
. Solution:
From the definition of coefficient of restitution,
e
vi - v~
Vi -Vi
=1
2
V2 -
Answer is A.
VI
e(v2 - VI)
m-2S
m).
= (0.5) ( -2 S
= -2
mls
[I]
#111
193
Since
Solution:
VI
= 2 mls
VI
At the top of the ramp, all of the energy is gravitational potential energy; at the bottom, the energy is
spring potential energy. Neglecting the small toss of
gravitational potential energy in deflecting the spring,
the energy balance is
mgh
and
V2
-2
mis,
+ V2 = 2 ~ + (-2
~)=
So,
1
"2kX2
vi
by adding them.
-1 mls
v; = 1 mls
X=
J2~gh'
Answer is A.
(2) (1 kg) (9.81 ~) (1 m)
2000 N
m
=
0.099 m
(9.9 cin)
is A.
6.
7.
5
fi
t<
(A) 9.9 cm
(B) 11.4 em
(C) 11.7 cm
(D) 14.1 em
Answer
Sol
---
Solution:
17-5
(KE)i - (KE) f
(2)
-A
2 mv2)
( m;i --t
~)21
=6N-m
compressed
position
Answer is C.
__
(A) 0.96 m
(B) 1.3 m
(C)
30.0 m/s
(D) 42.9 m/s
(C) lAm
(D) 1.8 m
#114 689
#115 689
Solution:
vI
(2 kg) (40
7)
+ (5 kg) (-10
7) = (2
v'
(A) 19.8 J
(B) 219 J
(C) 392 J
(D) 2350 J
#116689
kg+5
kg)v'
4.29 m/s
Answer is A.
3.
(A)
(B)
(C)
(D)
3.13 m/s
4.43 m/s
9.80 m/s
19.6 m/s
#117 689
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