Unformatted text preview: n of the resultant displacement is 0.2837 m = 21° south of east 0.7316 m θ = tan –1 ____________________________________________________________________________________________ ( ) 18. REASONING For both the tug and the asteroid, Equation 2.8 x = v0t + 1 at 2 applies with
2 v0 = 0 m/s, since both are initially at rest. In applying this equation, we must be careful and
use the proper acceleration for each object. Newton’s second law indicates that the
acceleration is given by a = ΣF/m. In this expression, we note that the magnitudes of the net
forces acting on the tug and the asteroid are the same, according to Newton’s actionreaction
law. The masses of the tug and the asteroid are different, however. Thus, the distance
traveled for either object is given by, where we use for ΣF only the magnitude of the pulling
force ΣF 2
x = v0t + 1 at 2 = 1 t
2
2 m SOLUTION Let L be the initial distance between the tug and the asteroid. When the two
objects meet, the distances that each has traveled must add up to equal L. Therefore, L = xT + x A = 1 aT t 2 + 1 a At 2
2
2 ΣF 2 1 ΣF 2 1
1
1 2
L= 1
+
t + 2 t = 2 ΣF t
2m m T mA T mA Solving for the time t gives 172 FORCES AND NEWTON'S LAWS OF MOTION t= 2L
1
1
ΣF + m T mA = 2 ( 450 m ) 1
1
+
( 490 N ) 3500 kg 6200 kg = 64 s ____________________________________________________________________________________________ 19. SSM WWW REASONING We first determine the acceleration of the boat. Then,
using Newton's second law, we can find the net force ∑ F that acts on the boat. Since two
of the three forces are known, we can solve for the unknown force FW once the net force ∑ F is known.
SOLUTION Let the direction due east be the positive x direction and the direction due
north be the positive y direction. The x and y components of the initial velocity of the boat
are then
v0 x = (2.00 m/s) cos 15.0° = 1.93 m/s
v0 y = (2.00 m/s) sin 15.0° = 0.518 m/s Thirty seconds later, the x and y velocity components of the boat are
vx = (4.00 m/s) cos 35.0° = 3.28 m/s
v y = ( 4.00 m/s) sin 35.0° = 2.29 m/s Therefore, according to Equations 3.3a and 3.3b, the x and y components of the acceleration
of the boat are
v −v
3.28 m/s – 1.93 m/s
ax = x 0 x =
= 4.50 × 10−2 m/s 2
t
30.0 s ay = v y − v0 y
t = 2.29 m/s – 0.518 m/s
= 5.91× 10−2 m/s 2
30.0 s Thus, the x and y components of the net force that act on the boat are
∑ Fx = max = (325 kg) (4.50 × 10 –2 m/s 2 ) = 14.6 N ∑ Fy = ma y = (325 kg) (5.91×10 –2 m/s 2 ) = 19.2 N
The following table gives the x and y components of the net force ∑ F and the two known
forces that act on the boat. The fourth row of that table gives the components of the
unknown force FW . Chapter 4 Problems 173 Force xComponent yComponent ∑F 14.6 N 19.2 N F1 (31.0 N) cos 15.0° = 29.9 N (31.0 N) sin 15.0° = 8.02 N F2 –(23.0 N ) cos 15.0° = –22.2 N –(23.0 N) sin 15.0° = –5.95 N FW = ∑ F − F1 − F2 14.6 N – 29.9 N + 22.2 N = 6.9 N 19.2 N – 8.02 N + 5.95 N = 17.1 N The magnitude of FW is given by the Pythagorean theorem as
FW = (6.9 N) 2 + (17.1N ) 2 = 18.4 N The angle θ that FW makes with the x axis is 17.1 N = 68° 6.9 N θ = tan −1 17.1 N θ Therefore, the direction of FW is 68°, north of east . 6.9 N ____________________________________________________________________________________________ 20. REASONING The gravitational force acting on each object is specified by Newton’s law of
universal gravitation. The acceleration of each object when released can be determined with
the aid of Newton’s second law. We recognize that the gravitational force is the only force
acting on either object, so that it is the net force to use when applying the second law.
SOLUTION
a. The magnitude of the gravitational force exerted on the rock by the earth is given by
Equation 4.3 as Frock =
= Gmearth mrock ( 2
rearth
6.67 × 10−11 N ⋅ m 2 / kg 2 5.98 × 1024 kg ( 5.0 kg ) ( )( 6.38 × 106 m ) ) 2 = 49 N The magnitude of the gravitational force exerted on the pebble by the earth is 174 FORCES AND NEWTON'S LAWS OF MOTION Fpebble =
= Gmearth mpebble ( 2
rearth
6.67 × 10−11 N ⋅ m 2 / kg 2 5.98 × 1024 kg 3.0 × 10−4 kg )( ( 6.38 × 10 m )
6 )( 2 )= 2.9 × 10−3 N b. According to the second law, the magnitude of the acceleration of the rock is equal to the
gravitational force exerted on the rock divided by its mass.
arock = Frock
mrock Gmearth =
= ( 2
rearth
6.67 × 10−11 N ⋅ m 2 / kg 2 5.98 × 1024 kg )(
2
( 6.38 × 106 m ) )= 9.80 m /s 2 According to the second law, the magnitude of the acceleration of the pebble is equal to the
gravitational force exerted on the pebble divided by its mass.
apebble = Fpebble
mpebble = Gmearth
2
rearth ( 6.67 × 10−11 N ⋅ m2 / kg 2 )(5.98 × 1024 kg ) =
=
2
( 6.38 × 106 m ) 9.80 m /s 2 ______________________________________________________________________________
21. REASONING AND SOLUTION
a. According to Equation 4.5, the weight of the space traveler of mass m = 115 kg on earth is
W = mg = (11...
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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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