Post on 19-Mar-2021
107
The Laws of Motion
SOLUTIONS TO PROBLEMS Section 4.3 Mass P4.2 Since the car is moving with constant speed and in a straight line, the resultant force on it must be
zero regardless of whether it is moving
(a) toward the right or (b) the left.
Section 4.4 Newton’s Second Law⎯The Particle Under a Net Force P4.3
m = 3.00 kgra = 2.00i + 5.00 j( ) m s2
rF = mra = 6.00i + 15.0 j( ) N!rF! = 6.00( )2 + 15.0( )2 N = 16.2 N
108 The Laws of Motion
P4.4 (a) !
rF =
rF1 +
rF2 = "9.00i + 3.00 j( ) N
Acceleration
ra = ax i + ay j = !rF
m=
"9.00i + 3.00 j( ) N2.00 kg
= "4.50i + 1.50 j( ) m s2
Velocity
rv f = vx i + vy j = rv i +
rat =rat
rv f = !4.50i + 1.50 j( )m s2( ) 10 s( ) = !45.0i + 15.0 j( ) m s
(b) The direction of motion makes angle θ with the x-direction.
! = tan"1 vy
vx
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15.0 m s45.0 m s
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! = "18.4° + 180° = 162° from + x-axis
(c) Displacement: x-displacement = x f ! xi = vxit +
12
axt2 =12!4.50 m s2( ) 10.0 s( )2 = !225 m
y-displacement = y f ! yi = vyit +
12
ayt2 =12+1.50 m s2( ) 10.0 s( )2 = +75.0 m
!
rr = "225i + 75.0 j( ) m
(d) Position:
rrf =
rri + !rr
rrf = !2.00i + 4.00 j( ) + !225i + 75.0 j( ) = !227i + 79.0 j( ) m
P4.7 (a)
rF! =
rF1 +
rF2 = 20.0i + 15.0 j( ) N
rF! = mra : 20.0i + 15.0 j = 5.00ra
ra = 4.00i + 3.00 j( ) m s2
or
a = 5.00 m s2 at ! = 36.9° (b)
F2x = 15.0cos 60.0° = 7.50 NF2y = 15.0sin 60.0° = 13.0 NrF2 = 7.50i + 13.0 j( ) NrF! =
rF1 +
rF2 = 27.5i + 13.0 j( ) N = m
ra = 5.00ra
ra = 5.50i + 2.60 j( ) m s2 = 6.08 m s2 at 25.3°
FIG. P4.7
Chapter 4 109
P4.8
rF! = mra reads
!2.00i + 2.00 j + 5.00i ! 3.00 j ! 45.0i( ) N = m 3.75 m s2( ) a
where a represents the direction of
ra
!42.0i ! 1.00 j( ) N = m 3.75 m s2( ) a
rF! = 42.0( )2 + 1.00( )2 N at
tan!1 1.00
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below the –x-axis
rF! = 42.0 N at 181° = m 3.75 m s2( ) a .
For the vectors to be equal, their magnitudes and their directions must be equal.
(a) ! a is at 181° counterclockwise from the x-axis
(b) m =
42.0 N3.75 m s2 = 11.2 kg
(d)
rv f =
rv i +rat = 0 + 3.75 m s2 at 181°( )10.0 s so
rv f = 37.5 m s at 181°
rv f = 37.5 m s cos181°i + 37.5 m s sin181° j so
rv f = !37.5i ! 0.893 j( ) m s
(c) rv f = 37.52 + 0.8932 m s = 37.5 m s
Section 4.5 The Gravitational Force and Weight P4.9 (a) Fg = mg = 120 lb = 4.448 N lb( ) 120 lb( ) = 534 N
(b) m =
Fg
g=
534 N9.80 m s2 = 54.5 kg
P4.11 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just
canceled out by a glass of tomato juice. By subtraction,
Fg( )p = mgp and
Fg( )C = mgC give
!Fg = m gp " gC( ) .
For a person whose mass is 88.7 kg, the change in weight is
!Fg = 88.7 kg 9.809 5 " 9.780 8( ) = 2.55 N . A precise balance scale, as in a doctor’s office, reads the same in different locations because it
compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference.
110 The Laws of Motion
P4.13 (a) F! = ma and v f2 = vi
2 + 2ax f or a =
v f2 ! vi
2
2x f.
Therefore,
F! = mv f
2 " vi2( )
2x f
F! = 9.11 # 10"31 kg7.00 # 105 m s2( )2 " 3.00 # 105 m s2( )2$
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2 0.050 0 m( )= 3.64 # 10"18 N .
(b) The weight of the electron is
Fg = mg = 9.11 ! 10"31 kg( ) 9.80 m s2( ) = 8.93 ! 10"30 N
The accelerating force is
4.08 ! 1011 times the weight of the electron. P4.14 We find acceleration:
rrf !rri =
rv it +12
rat2
4.20 mi ! 3.30 mj=0+ 12
ra 1.20 s( )2 = 0.720 s2rara = 5.83i ! 4.58 j( ) m s2 .
Now
rF! = mra becomes
rFg +
rF2 = mrarF2 = 2.80 kg 5.83i ! 4.58j( ) m s2 + 2.80 kg( ) 9.80 m s2( ) jrF2 = 16.3i + 14.6 j( ) N .
Section 4.6 Newton’s Third Law P4.17 (a) 15.0 lb up
(b) 5.00 lb up
(c) 0
Chapter 4 111
Section 4.7 Applications of Newton’s Laws P4.22 (a) Isolate either mass
T ! mg = ma = 0T = mg .
The scale reads the tension T, so
T = mg = 5.00 kg 9.80 m s2( ) = 49.0 N . (b) Isolate the pulley
rT2 + 2
rT1 = 0T2 = 2 T1 = 2mg = 98.0 N .
(c)
rF! =
rn +rT + mrg = 0
Take the component along the incline
rnx +
rTx + m
rgx = 0 or
0 +T ! mg sin 30.0° = 0
T = mg sin 30.0° = mg2
=5.00 9.80( )
2= 24.5 N .
FIG. P4.22(a)
FIG. P4.22(b)
FIG. P4.22(c)
112 The Laws of Motion
P4.24 (a) First construct a free body diagram for the
5 kg mass as shown in the Figure 4.24a. Since the mass is in equilibrium, we can require T3 ! 49 N = 0 or T3 = 49 N . Next, construct a free body diagram for the knot as shown in Figure 4.24a. Again, since the system is moving at constant velocity, a = 0 and applying Newton’s second law in component form gives
Fx! = T2 cos 50° "T1 cos 40° = 0
Fy! = T2 sin 50° +T1 sin 40° " 49 N = 0
Solving the above equations
simultaneously for T1 and T2 gives
T1 = 31.5 N and T2 = 37.5 N and above
we found T3 = 49.0 N . (b) Proceed as in part (a) and construct a free
body diagram for the mass and for the knot as shown in Figure 4.24b. Applying Newton’s second law in each case (for a constant-velocity system) we find:
T3 ! 98 N = 0T2 !T1 cos 60° = 0T1 sin 60° !T3 = 0
Solving this set of equations we find:
T1 = 113 N T2 = 56.6 N and T3 = 98.0 N
FIG. P4.24(a)
FIG. P4.24(b)
Chapter 4 113 P4.26 The two forces acting on the block are the normal force, n, and the
weight, mg. If the block is considered to be a point mass and the x-axis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle θ is the angle of inclination of the plane. Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have
Fy! = n " mg cos# = 0 : n = mg cos!
Fx! = "mg sin# = ma : a = !g sin"
FIG. P4.26 (a) When ! = 15.0°
a = !2.54 m s2 (b) Starting from rest
v f2 = vi
2 + 2a x f ! xi( ) = 2ax f
v f = 2ax f = 2 !2.54 m s2( ) !2.00 m( ) = 3.18 m s
P4.27 Choose a coordinate system with i East and j North.
rF! = mra = 1.00 kg 10.0 m s2( ) at 30.0°
5.00 N( ) j +rF1 = 10.0 N( )!30.0° = 5.00 N( ) j + 8.66 N( ) i
!F1 = 8.66 N East( )
FIG. P4.27 P4.28 First, consider the block moving along the horizontal. The only
force in the direction of movement is T. Thus, Fx! = ma
T = 5 kg( ) a (1) Next consider the block that moves vertically. The forces on it
are the tension T and its weight, 88.2 N. We have Fy! = ma
88.2 N !T = 9 kg( ) a (2)
FIG. P4.28
Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be
added to give 88.2 N = 14 kg( ) a . Then
a = 6.30 m s2 and T = 31.5 N .
114 The Laws of Motion
P4.30 m1 = 2.00 kg , m2 = 6.00 kg , ! = 55.0° (a) Fx! = m2 g sin" #T = m2a and
T ! m1g = m1a
a = m2 g sin" ! m1gm1 + m2
= 3.57 m s2
(b) T = m1 a + g( ) = 26.7 N
FIG. P4.30
(c) Since vi = 0 , v f = at = 3.57 m s2( ) 2.00 s( ) = 7.14 m s .
Additional Problems P4.44 (a) Following the in-chapter Example about a block on a frictionless incline, we have
a = g sin! = 9.80 m s2( )sin 30.0°
a = 4.90 m s2
(b) The block slides distance x on the incline, with sin 30.0° = 0.500 m
x
x = 1.00 m : v f2 = vi
2 + 2a x f ! xi( ) = 0 + 2 4.90 m s2( ) 1.00 m( )
v f = 3.13 m s after time ts =
2x f
v f=
2 1.00 m( )3.13 m s
= 0.639 s .
(c) Now in free fall y f ! yi = vyit +
12
ayt2 :
!2.00 = !3.13 m s( )sin 30.0°t ! 12
9.80 m s2( )t2
4.90 m s2( )t2 + 1.56 m s( )t ! 2.00 m = 0
t =!1.56 m s ± 1.56 m s( )2 ! 4 4.90 m s2( ) !2.00 m( )
9.80 m s2
Only one root is physical
t = 0.499 s
x f = vxt = 3.13 m s( )cos 30.0°[ ] 0.499 s( ) = 1.35 m
(d) total time = ts + t = 0.639 s + 0.499 s = 1.14 s (e) The mass of the block makes no difference.
Chapter 4 115 P4.47 F! = ma For m1 : T = m1a For m2 : T ! m2 g = 0 Eliminating T,
a = m2 g
m1
For all 3 blocks:
FIG. P4.47
F = M + m1 + m2( ) a = M + m1 + m2( ) m2 g
m1
!
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