Resolução Halliday vol.2 ed.8 (completa), Exercícios de Física

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• Chapter

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 m x 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 m y 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 g x m g m g m g m g m g m g m g

=

=

(d) Similarly, y cog = [0 + (2.00)( m )(7.80) + (4.00)( m )(7.60) + (4.00)( m )(7.40) + (2.00)( m )(7.60) + 0]/(8.00 m + 7.80 m + 7.60 m + 7.40 m + 7.60 m + 7.80 m ) = 1.97 m.

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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) 2 T 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 colinear). In this problem, we have

tan 1 0.35 m 11.. 1.72 m

θ −^

Therefore, T = F /(2sin θ ) = 7.92 × 10^3 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

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; A = (3.05 −1.78) m =1.27 mis the horizontal distance from the rear axle to the center of mass; F 1 is the force exerted on each front wheel; and, F 2 is the force exerted on each back wheel.

(a) Taking torques about the rear axle, we find

2 3 1

(1360 kg) (9.80 m/s ) (1.27 m) 2.77 10 N. 2 2(3.05 m)

Mg F L

= = = ×

A

(b) Equilibrium of forces leads to 2 F 1 (^) + 2 F 2 = Mg ,from which we obtain F 2 = 389. × 103 N.

7. We take the force of the left pedestal to be F 1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F 2 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 x 2 ) equal to zero:

1 2 1

F F W

F d W L d

(a) The second equation gives

1

3.0 m (580 N)= 1160 N 1.5 m

L d F W d

−^ §^ ·

which should be rounded off to F 1 (^) = −1.2 × 10 3 N. Thus, | F 1 | = 1.2 × 103 N.

(b) Since F 1 is negative, indicating that this force is downward.

(c) The first equation gives F 2 (^) = W − F 1 =580 N+1160 N=1740 N

which should be rounded off to F 2 (^) = 1.7 × 103 N. Thus, | F 2 | = 1.7 × 103 N.

(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 F 2 (^) = ( M + m g ) = (75 kg + 10 kg) (9.8 m/s ) = 8.3 ×10 N

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 2^2 + F 32 = (. 2 8 × 102 N) 2 + ( .8 3 × 10 2 N) 2 = 8 8. × 102 N.

(c) The angle φ between the force and the horizontal is given by

tan φ = F 3 / F 2 = 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. F 1 is the force of the window, horizontal because the window is frictionless. F 2 and F 3 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

cos θ= d / L or sin θ= L^2^ − d^2 / L 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 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,

Mg A cos θ+ mg (^) ( L / 2 cos) θ− F L 1 sin θ= 0 ,

and 2 1 2

( / 2) cos [(75 kg) (3.0 m)+(10 kg) (2.5 m)](9.8 m/s ) cos 60 sin (5.0 m) sin 60 2.8 10 N.

M mL g F L

= ×

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

The first of these equations gives F 3 (^) = F 1 = 2 8. × 102 N and the second gives

10. The angle of each half of the rope, measured from the dashed line, is

tan 1 0.30 m 1.. 9.0 m

θ −^

Analyzing forces at the “kink” (where

G

F is exerted) we find

550 N (^) 8.3 10 N. 3 2sin 2sin1.

F

T

= = = ×

(b) Looking at the horizontal forces at that point leads to

T 2 (^) = T 1 sin 35 ° = (49N)sin 35 ° =28 N.

(c) We denote the components of T 3 as Tx (rightward) and Ty (upward). Analyzing horizontal forces where string 2 and string 3 meet, we find Tx = T 2 = 28 N. From the vertical forces there, we conclude Ty = wB =50 N. Therefore,

2 2 T 3 (^) = T (^) x + Ty =57 N.

(d) The angle of string 3 (measured from vertical) is

tan 1 tan 1 28 29. 50

x y

T

T

θ −^ −

© ¹ ©^ ¹

12. (a) Analyzing vertical forces where string 1 and string 2 meet, we find

1

40N

49N.

cos cos

T w^ A

13. (a) Analyzing the horizontal forces (which add to zero) we find Fh = F 3 = 5.0 N.

(b) Equilibrium of vertical forces leads to Fv = F 1 + F 2 = 30 N.

(c) Computing torques about point O , we obtain

( )( ) ( )( ) 2 3

10 N 3.0 m + 5.0 N 2.0 m = + = = 1.3m. v 30 N F d F b F a Ÿ d

15. Setting up equilibrium of torques leads to a simple “level principle” ratio:

2.6 cm (40 N) (40 N) 8.7 N. 12 cm

d F ⊥ (^) L

16. With pivot at the left end, Eq. 12-9 leads to
    • m s g L 2 – Mgx + TR L = 0

where m s is the scaffold’s mass (50 kg) and M is the total mass of the paint cans (75 kg). The variable x indicates the center of mass of the paint can collection (as measured from the left end), and TR is the tension in the right cable (722 N). Thus we obtain x = 0.702 m.

18. Our system consists of the lower arm holding a bowling ball. As shown in the free-body diagram, the forces on the lower arm consist of T

G

from the biceps muscle, F

G

from the bone of the upper arm, and the gravitational forces, mg

G

and Mg

G

. Since the system is in static equilibrium, the net force acting on the system is zero:

0 = ¦ F net, y = T − F − ( m + M g )

In addition, the net torque about O must also vanish:

(^0) net ( )( ) (0) ( )( ) ( ) O

= ¦ τ = d T + F − D mg − L Mg.

(a) From the torque equation, we find the force on the lower arms by the biceps muscle to be 2

2

( ) [(1.8 kg)(0.15 m) (7.2 kg)(0.33 m)](9.8 m/s ) 0.040 m 648 N 6.5 10 N.

mD ML g T d

= ≈ ×

(b) Substituting the above result into the force equation, we find F to be

F = T − ( M + m g ) = 648 N − (7.2 kg +1.8 kg)(9.8 m/s ) 2 = 560 N = 5.6 × 102 N.

19. (a) With the pivot at the hinge, Eq. 12-9 gives TL cos θ – mg L 2 = 0. This leads to

θ = 78 .° Then the geometric relation tan θ = L/D gives D = 0.64 m.

(b) A higher (steeper) slope for the cable results in a smaller tension. Thus, making D greater than the value of part (a) should prevent rupture.