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Unformatted text preview: tude is 98 N, because
it supports the 98N weight hanging from the pulley system. Other forces also act on the
arm at the shoulder joint, but we can ignore them. The reason is that their lines of action
pass directly through the axis at the shoulder joint, which is the axis that we will use to
determine torques. Thus, these forces have zero lever arms and contribute no torque. SOLUTION The magnitude of each individual torque is the magnitude of the force times
the corresponding lever arm. The forces and their lever arms are as follows:
Force Lever Arm 98 N 0.61 m 47 N 0.28 M (0.069 m) sin 29° Each torque is positive if it causes a counterclockwise rotation and negative if it causes a
clockwise rotation about the axis. Thus, since the net torque must be zero, we see that ( 98 N ) ( 0.61 m ) − ( 47 N ) ( 0.28 m ) − M ( 0.069 m ) sin 29° = 0
Solving for M gives
M= ( 98 N ) ( 0.61 m ) − ( 47 N ) ( 0.28 m ) = 1400 N
( 0.069 m ) sin 29° Chapter 9 Problems 489 75. SSM WWW REASONING AND SOLUTION The figure below shows the massless
board and the forces that act on the board due to the person and the scales. 2.00 m
person's feet person's head center of gravity
of person x
425 N 315 N W a. Applying Newton's second law to the vertical direction gives
315 N + 425 N – W = 0 or W = 7.40 ×102 N, downward b. Let x be the position of the center of gravity relative to the scale at the person's head.
Taking torques about an axis through the contact point between the scale at the person's
head and the board gives
(315 N)(2.00 m) – (7.40 × 102 N)x = 0 or x = 0.851 m 76. REASONING AND SOLUTION
a. The net torque about an axis through the point of contact between the floor and her shoes is
2 Στ = − (5.00 × 10 N)(1.10 m)sin 30.0° + FN(cos 30.0°)(1.50 m) = 0 FN = 212 N
b. Newton's second law applied in the horizontal direction gives Fh − FN = 0, so
Fh = 212 N .
c. Newton's second law in the vertical direction gives Fv − W = 0, so Fv = 5.00 × 102 N . 77. SSM REASONING When the modules pull together, they do so by means of forces that
are internal. These pulling forces, therefore, do not create a net external torque, and the
angular momentum of the system is conserved. In other words, it remains constant. We will
use the conservation of angular momentum to obtain a relationship between the initial and 490 ROTATIONAL DYNAMICS final angular speeds. Then, we will use Equation 8.9 (v = rω) to relate the angular speeds
ω0 and ωf to the tangential speeds v0 and vf.
SOLUTION Let L be the initial length of the cable between the modules and ω0 be the
initial angular speed. Relative to the centerofmass axis, the initial momentum of inertia of
the twomodule system is I0 = 2M(L/2)2, according to Equation 9.6. After the modules pull
together, the length of the cable is L/2, the final angular speed is ωf, and the momentum of
inertia is If = 2M(L/4)2. The conservation of angular momentum indicates that I f ω f = I 0ω 0
123
123
Final angular
momentum Initial angular
momentum LM F I O = LM F I O
L
L
2
ω
2
ω
M GJ P M GJ P
M4PM2P
N HKQ N HKQ
2 2 f 0 ω f = 4ω 0
According to Equation 8.9, ωf = vf/(L/4) and ω0 = v0/(L/2). With these substitutions, the
result that ωf = 4ω0 becomes
vf
L/4 =4 Fv I or
G/ 2 J
L
HK
0 b g v f = 2 v 0 = 2 17 m / s = 34 m / s 78. REASONING The drawing shows the rod in its initial
(dashed lines) and final (solid lines) orientations. Since
both friction and air resistance are absent, the total
mechanical energy is conserved. In this case, the total
mechanical energy is
E= 1
Iω 2
224
13
4
Rotational
kinetic energy + Center of mass mgh
1 24
43 Pivot Gravitational
potential energy In this expression, m is the rod’s mass, I is its moment of
inertia, ω is its angular speed, and h is the height of its
center of mass above a reference level that we take to be
the ground. Since the rod is uniform, its center of mass is
located at the center of the rod. The energyconservation
principle will guide our solution. hf Center of mass
h0 v0 Chapter 9 Problems 491 SOLUTION Conservation of the total mechanical energy dictates that
1
I ωf2 + mghf
2
144
244
3 = 1
2
I ω0 + mgh0
24
1442444
3 Final total
mechanical energy (1) Initial total
mechanical energy Since the rod is momentarily at rest in its final orientation, we know that ωf = 0 rad/s, so that
Equation (1) becomes
2
mghf = Iω0 + mgh0
1
2 1
2 or 2
I ω0 = mg ( hf − h0 ) (2) We can relate the initial angular speed ω0 to the initial linear speed v0 by using
v
Equation 8.9: ω0 = 0 . With this substitution, the fact that hf − h0 = L (see the drawing),
L
1
3 and the fact that I = mL2 (see Table 9.1 in the text), Equation (2) indicates that
1
2 2
I ω0 = mg ( hf − h0 ) Solving for v0 gives ( 11
mL2
23 or ( ) 2 v0 = mgL
L or 12
v
60 = gL ) v0 = 6 gL = 6 9.80 m/s2 ( 0.80 m ) = 6.9 m/s 79. REASONING The drawing shows the forces acting
on the board, which has a length L. Wall 2 exerts a
normal force P2 on the lower end of the board. The
maximum force of static friction that wall 2 can apply
to the lower end of the board is µsP2 and is directed
upward in the drawing. The weight W acts downw...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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