Chapter 5 – Student Solutions Manual
5. We denote the two forces
G
G
F
1
and
F
2
1
−
. According to Newton’s second law,
GG
G
G
G
G
FFm
a
Fm
aF
12
2
+=
=s
o
,.
(a) In unit vector notation
G
F
1
20 0
=
.
±
Ni
bg
and
()
(
)
(
)(
)
22
2
ˆˆ
ˆ
12.0 sin 30.0 m/s
i
12.0 cos 30.0 m/s
6.00 m/s
i
10.4m/s
j.
j
a
=−
°
−
°
= −
−
G
2
ˆ
Therefore,
( )
(
)
2
2.00kg
6.00 m/s
i
2.00 kg
10.4 m/s
j
20.0 N i
32.0 N i
20.8 N j.
F
+
−
−
−
G
ˆ
(b) The magnitude of
is
G
F
2
2
2
2


( 32.0)
( 20.8)
38.2 N.
xy
FF
F
=+
=
−
+
−
=
G
(c) The angle that
makes with the positive
x
axis is found from
G
F
2
tan
θ
= (
F
2
y
/
F
2
x
) = [(–20.8)/(–32.0)] = 0.656.
Consequently, the angle is either 33.0° or 33.0° + 180° = 213°. Since both the
x
and
y
components are negative, the correct result is 213°. An alternative answer is
.
213
360
147
°−
°=−
°
13. (a) – (c) In all three cases the scale is not accelerating, which means that the two
cords exert forces of equal magnitude on it. The scale reads the magnitude of either of
these forces. In each case the tension force of the cord attached to the salami must be the
same in magnitude as the weight of the salami because the salami is not accelerating.
Thus the scale reading is
mg
, where
m
is the mass of the salami. Its value is (11.0 kg) (9.8
m/s
2
) = 108 N.
19. (a) Since the acceleration of the block is zero, the components of the Newton’s
second law equation yield
T
–
mg
sin
= 0
F
N
–
mg
cos
= 0.
Solving the first equation for the tension in the string, we find
Tm
g
==
°
=
sin
.
.
85
9 8
30
42
2
kg
m / s
N .
ch
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View Full Document(b) We solve the second equation in part (a) for the normal force
F
N
:
( )( )
2
cos
8.5 kg
9.8 m/s
cos 30
72 N .
N
Fm
g
θ
==
°
=
(c) When the string is cut, it no longer exerts a force on the block and the block
accelerates. The
x
component of the second law becomes –
mg
sin
=
ma
, so the
acceleration becomes
2
sin
9.8 sin 30
4.9 m/s .
ag
=−
°= −
The negative sign indicates the acceleration is down the plane. The magnitude of the
acceleration is 4.9 m/s
2
.
25. In terms of magnitudes, Newton’s second law is
F
=
ma
, where
F
=
G
F
net
,

aa
=
G
,
and
m
is the (always positive) mass. The magnitude of the acceleration can be found
using constant acceleration kinematics (Table 21). Solving
v
=
v
0
+
at
for the case where
it starts from rest, we have
a
=
v
/
t
(which we interpret in terms of magnitudes, making
specification of coordinate directions unnecessary). The velocity is
v
= (1600 km/h)
(1000 m/km)/(3600 s/h) = 444 m/s, so
F
500
444
18
12
10
5
kg
ms
s
N.
bg
.
.
×
29. The acceleration of the electron is vertical and for all practical purposes the only force
acting on it is the electric force. The force of gravity is negligible. We take the +
x
axis to
be in the direction of the initial velocity and the +
y
axis to be in the direction of the
electrical force, and place the origin at the initial position of the electron. Since the force
and acceleration are constant, we use the equations from Table 21:
x
=
v
0
t
and
ya
t
F
m
t
F
H
G
I
K
J
1
2
1
2
22
.
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 Spring '08
 drty
 mechanics, Force, kg

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