Resolução física vol 2. 8ª edição, Halliday, Resnick & Walker

Resolução física vol 2. 8ª edição, Halliday, Resnick & Walker


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1. (a) The center of mass is given by 
com
0 0 0 ( )(2.00 m) ( )(2.00 m) ( )(2.00 m) 1.00 m.
6
m m mx
m
+ + + + +
= =
(b) Similarly, we have 
com
0 ( )(2.00 m) ( )(4.00 m) ( )(4.00 m) ( )(2.00 m) 0 2.00 m.
6
m m m my
m
+ + + + +
= =
(c) Using Eq. 12-14 and noting that the gravitational effects are different at the different 
locations in this problem, we have 
6
1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6
cog 6
1 1 2 2 3 3 4 4 5 5 6 6
1
0.987 m.
i i i
i
i i
i
x m g
x m g x m g x m g x m g x m g x m gx
m g m g m g m g m g m gm g
=
=
+ + + + +
= = =
+ + + + +
¦
¦
(d) Similarly, ycog = [0 + (2.00)(m)(7.80) + (4.00)(m)(7.60) + (4.00)(m)(7.40) + 
(2.00)(m)(7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m. 
2. The situation is somewhat similar to that depicted for problem 10 (see the figure that 
accompanies that problem). By analyzing the forces at the \u201ckink\u201d where 
G
F is exerted, we 
find (since the acceleration is zero) 2T sin \u3b8 = F, where \u3b8 is the angle (taken positive) 
between each segment of the string and its \u201crelaxed\u201d position (when the two segments are 
collinear). Setting T = F therefore yields \u3b8 = 30º. Since \u3b1 = 180º \u2013 2\u3b8 is the angle 
between the two segments, then we find \u3b1 = 120º. 
3. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the 
rope, causing a \u201ckink\u201d similar to that shown for problem 10 (see the figure that 
accompanies that problem). By analyzing the forces at the \u201ckink\u201d where 
G
F is exerted, we 
find (since the acceleration is zero) 2T sin\u3b8 = F, where \u3b8 is the angle (taken positive) 
between each segment of the string and its \u201crelaxed\u201d position (when the two segments are 
colinear). In this problem, we have 
1 0.35mtan 11.5 .
1.72 m
\u3b8 \u2212 § ·= = °¨ ¸
© ¹
Therefore, T = F/(2sin\u3b8 ) = 7.92 × 103 N. 
4. From 
G G
\u3c4 = ×r F , we note that persons 1 through 4 exert torques pointing out of the 
page (relative to the fulcrum), and persons 5 through 8 exert torques pointing into the 
page.
(a) Among persons 1 through 4, the largest magnitude of torque is (330 N)(3 m) = 990 
N·m, due to the weight of person 2. 
(b) Among persons 5 through 8, the largest magnitude of torque is (330 N)(3 m) = 990 
N·m, due to the weight of person 7. 
G
5. Three forces act on the sphere: the tension force 
G
T of the rope 
(acting along the rope), the force of the wall NF
G
 (acting horizontally 
away from the wall), and the force of gravity mgG (acting 
downward). Since the sphere is in equilibrium they sum to zero. Let 
\u3b8 be the angle between the rope and the vertical. Then Newton\u2019s 
second law gives
 vertical component : T cos \u3b8 \u2013 mg = 0
 horizontal component: FN \u2013 T sin \u3b8 = 0.
(a) We solve the first equation for the tension: T = mg/ cos \u3b8. We 
substitute cos\u3b8 = +L L r/ 2 2 to obtain 
2 2 22 2 (0.85 kg)(9.8 m/s ) (0.080 m) (0.042 m)
9.4 N
0.080 m
mg L rT
L
++
= = = .
(b) We solve the second equation for the normal force: sinNF T \u3b8= .
Using sin\u3b8 = +r L r/ 2 2 , we obtain 
2 2 2
2 2 2 2
(0.85 kg)(9.8 m/s )(0.042 m) 4.4 N.
(0.080 m)N
Tr mg L r r mgrF
L LL r L r
+
= = = = =
+ +
6. Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.05 m is 
the horizontal distance between the axles; (3.05 1.78) m 1.27 m= \u2212 =A is the horizontal 
distance from the rear axle to the center of mass; F1 is the force exerted on each front 
wheel; and, F2 is the force exerted on each back wheel. 
(a) Taking torques about the rear axle, we find 
2
3
1
(1360kg) (9.80m/s ) (1.27 m) 2.77 10 N.
2 2(3.05m)
MgF
L
= = = ×
A
(b) Equilibrium of forces leads to 1 22 2 ,F F Mg+ = from which we obtain F2
3389 10= ×. N . 
7. We take the force of the left pedestal to be F1 at x = 0, where the x axis is along the 
diving board. We take the force of the right pedestal to be F2 and denote its position as x
= d. W is the weight of the diver, located at x = L. The following two equations result 
from setting the sum of forces equal to zero (with upwards positive), and the sum of 
torques (about x2) equal to zero: 
1 2
1
0
( ) 0
F F W
F d W L d
+ \u2212 =
+ \u2212 =
(a) The second equation gives 
1
3.0m (580 N)= 1160 N
1.5m
L dF W
d
§ ·\u2212
= \u2212 = \u2212 \u2212¨ ¸
© ¹
which should be rounded off to 31 1.2 10 NF = \u2212 × . Thus, 
3
1| | 1.2 10 N.F = ×
(b) Since F1 is negative, indicating that this force is downward. 
(c) The first equation gives 2 1 580 N+1160 N=1740 NF W F= \u2212 =
which should be rounded off to 32 1.7 10 NF = × . Thus, 
3
2| | 1.7 10 N.F = ×
(d) The result is positive, indicating that this force is upward. 
(e) The force of the diving board on the left pedestal is upward (opposite to the force of 
the pedestal on the diving board), so this pedestal is being stretched.
(f) The force of the diving board on the right pedestal is downward, so this pedestal is 
being compressed. 
2 2
2 3
1
1 2
2
(80kg) (9.8m/s ) (3.5m)+(60kg) (9.8m/s ) (2.5m)
5.0m
8.4 10 N.
w sm g m gF += =
+
= ×
A A
A A
(b) Equilibrium of forces leads to 
2 3
1 2 (60kg+80kg) (9.8m/s ) 1.4 10 Ns wF F m g m g+ = + = = ×
which (using our result from part (a)) yields F2
253 10= ×. N . 
8. Let A1 15= . m and 2 (5.0 1.5) m 3.5 m= \u2212 =A . We denote tension in the cable closer to 
the window as F1 and that in the other cable as F2. The force of gravity on the scaffold 
itself (of magnitude msg) is at its midpoint, A3 2 5= . m from either end. 
(a) Taking torques about the end of the plank farthest from the window washer, we find 
2 2
2 ( ) (75kg 10kg) (9.8m/s ) 8.3 10 NF M m g= + = + = ×
The magnitude of the force of the ground on the ladder is given by the square root of the 
sum of the squares of its components: 
F F F= + = × + × = ×2
2
3
2 2 2 22 8 10 8 3 10 8 8 10( . ( . .N) N) N.2 2
(c) The angle \u3c6 between the force and the horizontal is given by 
tan \u3c6 = F3/F2 = 830/280 = 2.94, 
so \u3c6 = 71º. The force points to the left and upward, 71º above the horizontal. We note that 
this force is not directed along the ladder. 
9. The forces on the ladder are shown in the diagram on the right. F1 is 
the force of the window, horizontal because the window is frictionless. 
F2 and F3 are components of the force of the ground on the ladder. M is 
the mass of the window cleaner and m is the mass of the ladder. 
The force of gravity on the man acts at a point 3.0 m up the ladder and 
the force of gravity on the ladder acts at the center of the ladder. Let \u3b8
be the angle between the ladder and the ground. We use 
2 2cos / or sin / d L L d L\u3b8 \u3b8= = \u2212 to find \u3b8 = 60º. Here L is the length 
of the ladder (5.0 m) and d is the distance from the wall to the foot of 
the ladder (2.5 m). 
(a) Since the ladder is in equilibrium the sum of the torques about its 
foot (or any other point) vanishes. Let A be the distance from the foot of the ladder to the 
position of the window cleaner. Then,
( ) 1cos / 2 cos sin 0Mg mg L F L\u3b8 \u3b8 \u3b8+ \u2212 =A ,
and
2
1
2
( / 2) cos [(75kg) (3.0m)+(10kg) (2.5m)](9.8m/s )cos 60
sin (5.0m)sin 60
2.8 10 N.
M mL gF
L
\u3b8
\u3b8
+ °
= =
°
= ×
A
This force is outward, away from the wall. The force of the ladder on the window has the 
same magnitude but is in the opposite direction: it is approximately 280 N, inward. 
(b) The sum of the horizontal forces and the sum of the vertical forces also vanish: 
F F
F Mg mg
1 3
2
0
0
\u2212 =
\u2212 \u2212 =
The first of these equations gives F F3 1
22 8 10= = ×. N and the second gives 
10. The angle of each half of the rope, measured from the dashed line, is 
1 0.30 mtan 1.9 .
9.0 m
\u3b8 \u2212 § ·= = °¨ ¸
© ¹
Analyzing forces at the \u201ckink\u201d (where 
G
F is exerted) we