4.1: a) For the magnitude of the sum to be the sum of the magnitudes, the forces must
be parallel, and the angle between them is zero. b) The forces form the sides of a right
isosceles triangle, and the angle between them is
90
. Alternatively, the law of cosines
may be used as
 
,cos22
2
2
22

FFFF 
from which
cos 0


, and the forces are perpendicular. c) For the sum to have 0
magnitude, the forces must be antiparallel, and the angle between them is
180
.
4.2: In the new coordinates, the 120-N force acts at an angle of
53
from the
x
-axis,
or
233
from the
x
-axis, and the 50-N force acts at an angle of

4.3: The horizontal component of the force is
N1.745cos)N10( 
to the right and the
vertical component is
N1.745sin)N10( 
down.
4.4: a)
,cos

FF
x

where

is the angle that the rope makes with the ramp (
 30θ
in
this problem), so
.N3.69
30cos
N0.60
cos



x
F
F F

b)

R
,N8.25960sin)N300( 
y
R
so
 
.7.31arctan,N494)N8.259()N420(
420
8.259
22
 θR
4.6: a)
N10.8)9.126(cos)N00.6(120cos)N00.9(
21

xx
FF
N.00.3)9.126(sin)N00.6(120sin)N00.9(
21

yy
FF
b)
N.8.64N)(3.00N)10.8(
2222

yx
RRR
4.7:
2

b) The speed at the end of the first 5.00 seconds is
m/s4.4at
, and the block on the
frictionless surface will continue to move at this speed, so it will move another
m0.22vt
in the next 5.00 s.
4.11: a) During the first 2.00 s, the acceleration of the puck is
2
m/s563.1/ mF
(keeping an extra figure). At
s00.2t
, the speed is
m/s13.3at
and the position is
m13.32/2/
2
 vtat
. b) The acceleration during this period is also
2
m/s563.1
, and the
speed at 7.00 s is
m/s6.26s)00.2)(m/s(1.563m/s13.3
2

. The position at
s00.5t
is
m125s)2.00sm/s)(5.00(3.13m13.3 x
, and at

4.14: a) With
0
0

x
v
,
.m/s1050.2
)m1080.1(2
)m/s1000.3(
2
214
2
262





x
v
a
x
x
b)
s1020.1
8
s/m1050.2
s/m1000.3
214





g
w
F
gw
F
m
F
a
4.17: a)
kg49.4)m/s80.9/()N0.44(/
2
 gwm
b) The mass is the same, 4.49 kg, and
the weight is
.N13.8)m/s81.1)(kg49.4(
2

4.18: a) From Eq. (4.9),
kg.327.0)s/m80.9/()N20.3(/
2
 gwm
b)
N.137)s/m80.9)(kg0.14(
2
 mgw
4.19:

.s/m104.7
)kg100.6(
)s/m80.9)(kg45(
223
24
2
EE
E




m
mg
m
F
a
4.24: (a) Each crate can be considered a single particle:
AB
F
(the force on
A
m
due to
B
m
) and
BA
F
(the force on

N142n
4.28: a)
b)


sinmgT 
N2790.26sin)s/m80.9)(kg0.65(
2

4.29: tricycle and Frank
T is the force exerted by the rope and
g
f
is the force the ground exerts on the tricycle.
spot and the wagon

T

is the force exerted by the rope. T and
T

form a third-law action-reaction pair,

.TT



4.30: a) The stopping time is
.s1043.7
4

-direction to be along
1
F

and the
y
-direction to be along
R

. Then
N1300
2

x
F
and
N1300
2

y
F
, so
N1838
2
F
, at an angle of
135
from
1
F