Unformatted text preview: ts of F31 and F32: Chapter 4 Problems F3 = 2 F cos θ = 2 GMm3
r2 181 cos θ The acceleration of sphere 3 is given by Newton's second law:
a3 = F3
m3 =2 (6.67 ×10
GM
cos θ = 2
2
r −11 N ⋅ m 2 /kg 2 ) ( 2.80 kg ) (1.20 m )2 cos 30.0° = 2.25 × 10 –10 m/s 2
____________________________________________________________________________________________ 33. SSM WWW REASONING AND SOLUTION There are two forces that act on the
balloon; they are, the combined weight of the balloon and its load, Mg, and the upward
buoyant force FB . If we take upward as the positive direction, then, initially when the
balloon is motionless, Newton's second law gives FB − Mg = 0 . If an amount of mass m is
dropped overboard so that the balloon has an upward acceleration, Newton's second law for
this situation is
FB − ( M − m) g = ( M − m)a But FB = mg , so that Mg – ( M − m ) g = mg = ( M – m ) a Solving for the mass m that should be dropped overboard, we obtain Ma
(310 kg )(0.15 m/s 2 )
=
= 4.7 kg
g + a 9.80 m/s 2 + 0.15 m/s 2
______________________________________________________________________________
m= 34. REASONING AND SOLUTION
acceleration, it follows that Since both motions are characterized by constant
yJ
yE = 1 a t2
2 JJ
1 a t2
2 EE where the subscripts designate those quantities that pertain to Jupiter and Earth. Since both
objects fall the same distance (yJ = yE), the above ratio is equal to unity. Solving for the
ratio of the times yields
tJ
tE = 2
aE
GM E / RE RJ
=
=
2
aJ
RE
GM J / RJ ME
1
= (11.2 )
= 0.628
MJ
318 ____________________________________________________________________________________________ 182 FORCES AND NEWTON'S LAWS OF MOTION 35. REASONING The gravitational force that the sun exerts on a person standing on the earth
2
is given by Equation 4.3 as Fsun = GM sun m / rsunearth , where Msun is the mass of the sun, m
is the mass of the person, and rsunearth is the distance from the sun to the earth. Likewise,
the gravitational force that the moon exerts on a person standing on the earth is given by
2
Fmoon = GM moon m / rmoonearth , where Mmoon is the mass of the moon and rmoonearth is the
distance from the moon to the earth. These relations will allow us to determine whether the
sun or the moon exerts the greater gravitational force on the person.
SOLUTION Taking the ratio of Fsun to Fmoon, and using the mass and distance data from
the inside of the text’s front cover, we find GM sun m
Fsun
Fmoon 2
M
rsunearth
=
= sun
M
GM moon m moon
2
rmoonearth rmoonearth rsunearth 1.99 × 1030 kg 3.85 × 108 m = 7.35 × 1022 kg 1.50 × 1011 m 2 2 = 178 Therefore, the sun exerts the greater gravitational force.
____________________________________________________________________________________________ 36. REASONING The gravitational attraction between the planet and the moon is governed by
Newton’s law of gravitation F = GMm/r2 (Equation 4.3), where M is the planet’s mass and
m is the moon’s mass. Because the magnitude of this attractive force varies inversely with
the square of the distance r between the center of the moon and the center of the planet, the
maximum force Fmax occurs at the minimum distance rmin, and the minimum force Fmin at
the maximum distance rmax. The problem text states that Fmax exceeds Fmin by 11%, or
Fmax = 1.11 Fmin. This expression can be rearranged to give the ratio of the
forces: Fmax/ Fmin = 1.11. We will use Equation 4.3 to compute the desired distance ratio in
terms of this force ratio.
SOLUTION From Equation 4.3, the ratio of the maximum gravitational force to the
minimum gravitational force is Fmax
Fmin = GMm
2
rmin
GMm
2
rmax 1
= 2
rmin 1
2
rmax = 2
rmax
2
rmin Chapter 4 Problems 183 Taking the square root of both sides of this expression and substituting the ratio of forces
Fmax/ Fmin = 1.11 yields the ratio of distances: rmax Fmax = rmin Fmin = 1.11 = 1.05 The moon’s maximum distance from the center of the planet is therefore about 5% larger
than its minimum distance. 37. REASONING We place the third particle (mass = m3) as shown in the following drawing:
L
D
m3
m
2m
The magnitude of the gravitational force that one particle exerts on another is given by
Newton’s law of gravitation as F = Gm1m2/r2. Before the third particle is in place, this law indicates that the force on each particle has a magnitude Fbefore = Gm2m/L2. After the third
particle is in place, each of the first two particles experiences a greater net force, because the
third particle also exerts a gravitational force on them.
SOLUTION For the particle of mass m, we have
Gmm3
Fafter
Fbefore = Gm 2m
2
D
L2 = L m3 + 1
Gm 2m
2mD 2
2
L
2 + For the particle of mass 2m, we have G 2mm3 Fafter
Fbefore = ( L – D) 2 + Gm2m
L2 Gm2m
L2 = L2 m3 m ( L – D) 2 +1 Since Fafter/Fbefore = 2 for both particles, we have L2 m3
2mD 2 +1 = L2 m3 m ( L – D) 2 +1 or 2D2 = ( L – D ) 2 184 FORCES AND NEWTON'S LAWS OF MOTION Expanding and rearranging this result gives D 2 + 2 LD − L2 = 0 , which can be solved for D
using the quadratic formula: D= –2 L ± ( 2 L )2 – 4 (1) ( – L2 )
=
2 (1)...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details