PROBLEM 12.66
An advanced spatial disorientation trainer allows the cab to rotate around
multiple axes as well as extend inwards and outwards. It can be used to
simulate driving, fixed wing aircraft flying, and helicopter maneuvering. In
one training scenario, the trainer rotates and translates in the horizontal plane
where the location of the pilot is defined by the relationships
r 10 2 cos
3
0.1 2t 2 t , where r,
t and
and t are expressed
in feet, radians, and seconds, respectively. Knowing that the pilot has a mass
of 175 lbs, (a) find the magnitude of the resulting force acting on the pilot at
t= 5 s (b) plot the magnitudes of the radial and transverse components of the
force exerted on the pilot from 0 to 10 seconds.
SOLUTION
r 10 2 cos
Given:
t ft
3
0.1 2t 2 t rad
175 lbs
32.2 ft/s 2
m
5.435 slugs
Using Radial and Transverse Coordinates:
r
r
2
sin t ft/s
3
3
2
2
9
cos
3
0.1 4t 1 rad/s
0.4 rad/s 2
t ft/s 2
Free Body Diagram of pilot (top and side view) showing net forces in the radial and transverse directions:
Equations of Motion:
Fr
Fr
mar
F
F
2
2
m
9
ma
cos
3
0
Fz
10 2 cos
t
3
t
0.12 4t 1
2r
m r
m 10 2 cos
Fz
2
m r r
3
t
0.4
0.4
sin
t
3
3
mg
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4t 1
2
PROBLEM 12.66 (Continued)
(a) Evaluate forces at t=5 s:
Fr
221.78 lbs
F
61.37 lbs
Fz
175 lbs
F
Fr2
F 2 Fz2
F
289.1 lb
(b) Plot of Fr and Fθ from t=0 to t=10 s:
Radial and Transverse Components of Force for t=0 to 10 seconds
100
0
-100
Force, (lbs)
-200
-300
-400
-500
-600
-700
-800
Radial Component
Transverse Component
0
1
2
3
4
5
6
Time, (sec)
7
8
9
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10
PROBLEM 12.70
Pin B weighs 4 oz and is free to slide in a horizontal plane along the
rotating arm OC and along the circular slot DE of radius b 20 in.
Neglecting friction and assuming that
15 rad/s and
20 , determine for that position
250 rad/s 2 for the position
(a) the radial and transverse components of the resultant force
exerted on pin B, (b) the forces P and Q exerted on pin B,
respectively, by rod OC and the wall of slot DE.
SOLUTION
Kinematics.
From the geometry of the system, we have
Then
r
and
ar
r
a
r
2
2
r
2
Now
and
r
2b cos
(2b sin )
2
2b( sin
(2b cos )
2
cos )
2b( sin
2(15 rad/s)2 cos 20 ]
2( 2b sin )
20
ft [(250 rad/s 2 )cos 20
12
2
2b( sin
cos ) (2b cos )
20
ft [(250 rad/s2 )sin 20
12
2r
r
2b( cos
2
2
cos )
1694.56 ft/s2
2
2
sin )
2(15 rad/s)2 sin 20 ] 270.05 ft/s2
Kinetics.
(a)
We have
Fr
mar
and
F
ma
Fr :
(b)
or
1
lb
4
32.2 ft/s2
1
4
32.2 ft/s 2
Fr
Q
F: F
or
lb
P
( 1694.56 ft/s2 )
(270.05 ft/s 2 )
Fr
13.1565 lb
13.16 lb
F
2.0967 lb
2.10 lb
Q cos 20
1
(13.1565 lb) 14.0009 lb
cos 20
P Q sin 20
(2.0967 14.0009sin 20 ) lb
6.89 lb
P
6.89 lb
70
Q 14.00 lb
40
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PROBLEM 12.71
The two blocks are released from rest when r 0.8 m and
30 .
Neglecting the mass of the pulley and the effect of friction in the
pulley and between block A and the horizontal surface, determine
(a) the initial tension in the cable, (b) the initial acceleration of block
A, (c) the initial acceleration of block B.
SOLUTION
Let r and be polar coordinates of block A as shown, and let yB be the
position coordinate (positive downward, origin at the pulley) for the
rectilinear motion of block B.
Constraint of cable:
r
yB
constant,
r
vB
0,
Fx
For block A,
r
aB
mAaA : T cos
+
For block B,
Fy
0
mAaA or T
mB aB : mB g
Adding Eq. (1) to Eq. (2) to eliminate T, mB g
r
or
aB
(1)
mAaA sec
(2)
mB aB
(3)
T
mAaA sec
mBaB
(4)
Radial and transverse components of a A.
Use either the scalar product of vectors or the triangle construction shown,
being careful to note the positive directions of the components.
r
r
2
ar
a A er
(5)
a A cos
0, using Eq. (1) to eliminate r, and changing
Noting that initially
signs gives
aB
a A cos
(6)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
aA
mB g
m A sec
mB cos
From Eq. (6), aB
5.48cos30
(a)
From Eq. (2), T
(b)
Acceleration of block A.
(c)
Acceleration of block B.
(25) (9.81)
20sec30
25cos 30
5.48 m/s 2
4.75 m/s 2
(20)(5.48)sec30
126.6
T
5.48 m/s 2
aA
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126.6 N
aB
4.75 m/s 2