﻿ Resolução vol 2 Halliday, Resnick & Walker (8ed) - Fenômenos

# Resolução vol 2 Halliday, Resnick & Walker (8ed)

Disciplina:Fenômenos Térmicos e Ópticos11 materiais57 seguidores
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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 “kink” where

G
F is exerted, we

find (since the acceleration is zero) 2T sin θ = F, where θ is the angle (taken positive)
between each segment of the string and its “relaxed” position (when the two segments are
collinear). Setting T = F therefore yields θ = 30º. Since α = 180º – 2θ is the angle
between the two segments, then we find α = 120º.

3. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the
rope, causing a “kink” similar to that shown for problem 10 (see the figure that
accompanies that problem). By analyzing the forces at the “kink” where

G
F is exerted, we

find (since the acceleration is zero) 2T sinθ = F, where θ is the angle (taken positive)
between each segment of the string and its “relaxed” position (when the two segments are
colinear). In this problem, we have

1 0.35mtan 11.5 .
1.72 m

θ − § ·= = °¨ ¸

Therefore, T = F/(2sinθ ) = 7.92 × 103 N.

4. From
G G
τ = ×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
θ be the angle between the rope and the vertical. Then Newton’s
second law gives

vertical component : T cos θ – mg = 0
horizontal component: FN – T sin θ = 0.

(a) We solve the first equation for the tension: T = mg/ cos θ. We
substitute cosθ = +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 θ= .
Using sinθ = +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= − =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

+ − =

+ − =

(a) The second equation gives

1
3.0m (580 N)= 1160 N
1.5m

L dF W
d

§ ·−
= − = − −¨ ¸

which should be rounded off to 31 1.2 10 NF = − × . 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= − =

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= − =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 φ between the force and the horizontal is given by
tan φ = F3/F2 = 830/280 = 2.94,

so φ = 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 θ
be the angle between the ladder and the ground. We use

2 2cos / or sin / d L L d Lθ θ= = − to find θ = 60º. Here L is the length
of the ladder (5.0 m) and d is the distance from the wall to the foot of

(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 Lθ θ θ+ − =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

θ
θ

+ °
= =

°

= ×

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

− =

− − =

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

θ − § ·= = °¨ ¸