Midterm Exam #2
Feb 23, 2015 (Mon 50 min)
Chapters: 1-7,10.1-3
Closed book
1 pg notes (2-sided)
Review session on Thur (Feb 19)
AA210 University of Washington - Winter 2015
Chapter 1
Equilibrium: F = 0 & Mp = 0
then: v v(t) or v = 0
F = FR
Action = Reacti
Name_
1. A truss supports two loads W=160 N. It is supported with a pin at S and roller at T. h=w=4 m.
a. Determine the force in members HA and AG. (10 pts) AG=226.3 N (T), HA=160 N (C)
b. Determine the forces in members CD and CK. (10 pts) CD=480 N (T),
70
Chapter 3. Kinetics of Particles
Applying Newtons 2nd Law, we obtain
L
x 2 + L2 L
Ey = mxEx
2 + L2
x
(3.68)
Using the Ex -component of the last equation, we obtain
K
x
x 2 + L2 L
E + N +K
2 + L2 x
x
mx = K
x 2 + L2 L
x
x 2 + L2
(3.69)
Rearranging th
51
Now we can obtain an expression for F B in terms of the basis cfw_ur , u , u by
expressing ez in terms of ur and u as
ez = sin ur + cos u
(2.286)
Consequently,
F
B = (sin ur + cos u ) u = sin ur u + cos u
(2.287)
Then, the velocity of point P in refer
60
Chapter 2. Kinematics
Dierentiating et in Eq. (2.345), we obtain
F
det
= x sin xEx + x cos xEy
dt
Consequently,
F
det
dt
= x = Fv
(2.347)
(2.348)
which implies that
en =
F
F
det /dt
det /dt
=
x sin xEx + x cos xEy
= sin xEx + cos xEy
x
(2.349)
Then, us
54
Chapter 2. Kinematics
Now the two terms required to obtain F v are given as
F
dr
= r er
(2.305)
dt
F B
r = ( cos er sin e + ez ) r er = r e + r sin (2.306)
ez
Therefore, the velocity of the particle in reference frame F is
F
v = r er + r e + r sin ez
62
Chapter 3. Kinetics of Particles
Solution to Question 31
Kinematics
Let F be a reference frame xed to the track. Then, choose the following coordinate system xed in reference frame F :
Ex
Ez
Ey
Origin at point O
=
=
=
Along OP when = 0
Out of page
Ez E
56
Chapter 2. Kinematics
Next, the angular velocity of reference frame A in reference frame F is given as
F
A = ez
(2.314)
The velocity of point P in reference frame F is then given as
F
vP =
F
F
d
d
rP /O = F vO + F vP /O
(rO ) +
dt
dt
(2.315)
Now since
67
Rearranging Eq. (3.48) and separating variables, we obtain
g
d = cos 2d
r0
(3.49)
Integrating this last equation, we obtain
1 2 2
g
0 =
[sin 2]0
2
2r0
(3.50)
Noting that (t = 0) = 0, this last equation simplies to
g
2
2 = 0
sin 2
r0
(3.51)
Solvin
75
From the geometry we have that
Nn
= Nn en
(3.100)
Nb
= Nb en
(3.101)
mg = mgez
(3.102)
F = Nn en + Nb eb mgez
(3.103)
Consequently,
Now from Eqs. (3.86) and (3.93) we have
et
eb
e + ez
1 + 2
e ez
=
1 + 2
=
(3.104)
Using Eq. (3.104) we can obtain an ex
64
Chapter 3. Kinetics of Particles
which implies that
2 cos er 2 sin e
2 cos er 2 sin e
en =
= cos er sin e
(3.17)
The principal unit bi-normal vector to the track is then obtained as
eb = et en = ( sin er + cos e ) ( cos er sin e ) = ez
(3.18)
The acce
57
Now, since F vO is expressed in the basis cfw_Ex , Ey , Ez , we have that
F
aO =
F
d
dt
F
vO = xEx
(2.327)
Furthermore, since F vP /O is expressed in the basis cfw_er , e , ez and cfw_er , e , ez
rotates with angular velocity F A , we can obtain F aP
55
Question 221
A slender rod of length l is hinged to a collar as shown in Fig. P2-21. The collar
slides freely along a xed horizontal track. Knowing that x is the horizontal
displacement of the collar and that describes the orientation of the rod relati
65
Now we know that the reaction force is orthogonal to the track while gravity
acts vertically downward. Consequently, we have that
N = Nn en + Nb
mg = mgEy
(3.23)
(3.24)
Then, using the expression for Ey from Eq. (3.2), we obtain the force of gravity
as
73
We then obtain
F
s F s0 = R 1 + 2 ( 0 )
(3.82)
Solving Eq. (3.82) for s, the arclength is given as
F
s = F s0 + R 1 + 2 ( 0 )
(3.83)
Now the tangent vector in reference frame F is given as
et =
Fv
(3.84)
Fv
Using the speed from Eq. (3.79) and the veloc
68
Chapter 3. Kinetics of Particles
Question 32
A collar of mass m slides without friction along a rigid massless rod as shown
in Fig. P3-2. The collar is attached to a linear spring with spring constant K and
unstretched length L. Assuming no gravity, de
72
Chapter 3. Kinetics of Particles
following coordinate system to describe the motion of the particle:
er
ez
e
Origin at O
=
=
=
Along Radial Direction of Circle
Ez from Reference Frame F
ez er
Now, since is the angle formed by the helix with the horizon
79
Now we know that since the maximum distance is obtained when the velocity of
the bead is zero, we must have that 1 = 0. Furthermore, since the initial value
of is zero, we have that 0 = 0. Consequently, Eq. (3.141) reduces to
1
2
mR 2 0 (1 + 2 ) = mgR1
78
Chapter 3. Kinetics of Particles
Furthermore, since en and eb both lie in the direction orthogonal to et , we have
that
en et = 0
(3.132)
eb et = 0
Furthermore, since F v = F vet , we know that
Fnc = (Nn en + Nb eb ) F vet = 0
(3.133)
Consequently,
d
d
66
Chapter 3. Kinetics of Particles
Then, using the expression for en from Eq. (3.17), we have that
er en
= er ( cos er sin e ) = cos
(3.36)
e en
= e ( cos er sin e ) = sin
(3.37)
Substituting the results of Eq. (3.36) and Eq. (3.37) into Eq. (3.35) giv
71
Solution to Question 33
A bead of mass m slides along a xed circular helix of radius R and constant
helical inclination angle as shown in Fig. P3-3. The equation for the helix is
given in cylindrical coordinates as
z = R tan
(3.71)
Knowing that gravit
Chapter 3
Kinetics of Particles
Question 31
A particle of mass m moves in the vertical plane along a track in the form of a
circle as shown in Fig. P3-1. The equation for the track is
r = r0 cos
Knowing that gravity acts downward and assuming the initial
59
Arc-length Parameter as a Function of x
Now we recall the arc-length equation as
d
dt
F
s = F v = x 1 + tan2 x = x sec x
(2.337)
Separating variables in Eq. (2.337), we obtain
F
ds = sec xdx
(2.338)
Integrating both sides of Eq. (2.338) gives
F
s F s0
58
Chapter 2. Kinematics
Question 223
A particle slides along a xed track y = ln cos x as shown in Fig. P2-23 (where
/2 < x < /2). Using the horizontal component of position, x, as the variable to describe the motion and the initial condition x(t = 0) =
74
Chapter 3. Kinetics of Particles
We then obtain the principal unit bi-normal vector as
e + ez
e ez
eb = et en =
(er ) =
1 + 2
1 + 2
(3.93)
Furthermore, the curvature is given as
=
F
1
det /dt
=
Fv
R(1 + 2 )
(3.94)
The acceleration in reference frame
47
Curvature of Trajectory in Reference Frame F
First, we know that
F
det
= F ven
dt
(2.270)
Taking the magnitude of both sides, we have that
F
det
dt
Solving for , we have that
=
= Fv
F
(2.271)
det /dt
Fv
(2.272)
Substituting the expression for F det /dt
42
Chapter 2. Kinematics
Now we know that the position of point P is given as
rP = rQ + rP /Q
(2.235)
= Lux
(2.236)
= Rir
(2.237)
where
rQ
rP /Q
Because the basis cfw_ux , uy , uz is xed in reference frame B, we can apply the
transport theorem to rQ betw
81
The relationship between the bases cfw_Ex , Ey , Ez and cfw_er , e , ez is given as
er
= cos Ex + sin Ey
(3.146)
e
= sin Ex + cos Ey
(3.147)
The position of the particle is then given as
r = r er = r0 ea er
(3.148)
Furthermore, the angular velocity o
80
Chapter 3. Kinetics of Particles
Question 35
A collar of mass m is constrained to move along a frictionless track in the form
of a logarithmic spiral as shown in Fig. P3-5. The equation for the spiral is given
as
r = r0 ea
where r0 and a are constants
44
Chapter 2. Kinematics
Question 217
A particle slides along a track in the form of a spiral as shown in Fig. P2-17. The
equation for the spiral is
r = a
where a is a constant and is the angle measured from the horizontal. Determine (a) expressions for t