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Chapter 5
Page 1
CHAPTER 5  Further Applications of Newton’s Laws
1.
The friction is kinetic, so
F
fr
=
µ
k
F
N
.
With constant velocity,
the acceleration is zero.
Using the force diagram for the crate, we can write
?
F
=
m
a
:
x
component:
F
–
k
F
N
= 0;
y
component:
F
N
–
Mg
= 0.
Thus
F
N
=
Mg
, and
F
=
k
F
N
=
k
Mg
= (0.30)(12.0 kg)(9.80 m/s
2
) =
35 N
.
If
k
= 0, there is
no force
required to maintain constant
speed.
2. (
a
) In general, static friction is given by
F
fr
=
s
F
N
.
Immediately
before the box starts to move, the static friction force reaches
its maximum value:
F
fr,max
=
s
F
N
.
For the instant before the box
starts to move, the acceleration is zero.
Using the force diagram for the box, we can write
?
F
=
m
a
:
x
component:
F
–
s
F
N
= 0;
y
component:
F
N
–
Mg
= 0.
T
h
u
s
F
N
=
Mg
, and
F
=
s
F
N
=
s
Mg
;
25.0 N =
s
(6.0 kg)(9.80 m/s
2
), which gives
s
= 0.43
.
(
b
) When the box accelerates and the friction changes to kinetic, we have
F
–
k
F
N
=
Ma
;
25.0 N –
k
(6.0 kg)(9.80 m/s
2
) = (6.0 kg)(0.50 m/s
2
), which gives
k
= 0.37
.
3. (
a
)
(
b
)
(
c
)
In (
a
) the friction is static and opposes the impending motion down the plane.
In (
b
) the friction is kinetic and opposes the motion down the plane.
In (
c
) the friction is kinetic and opposes the motion up the plane.
4.
If we simplify the forces so that there is one normal force,
we have the diagram shown.
We can write
?
F
=
m
a
:
x
component:
–
F
fr
+
mg
sin
θ
= 0;
y
component:
F
N
–
mg
cos
= 0.
When we combine the two equations, we have
t
a
n
=
F
fr
/
F
N
=
s
.
Thus we have
t
a
n
max
=
s
= 0.8,
max
= 39°
.
M
g
x
y
F
F
fr
F
N
M
g
x
y
F
F
fr
F
N
F
N
m
g
F
fr
F
N
m
g
F
fr
F
N
m
g
F
fr
F
N
m
g
F
fr
x
y
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Page 2
5.
If we simplify the forces so that there is one normal force, we have the
diagram shown.
The friction force provides the acceleration.
We can write
?
F
=
m
a
:
x
component:
F
fr
=
Ma
;
y
component:
F
N
–
Mg
= 0.
Thus we have
a
=
F
fr
/
M
=
µ
s
F
N
/
M
=
s
g
.
The minimum value of
s
is
s,min
=
a
/
g
=
0.20
.
6.
If we simplify the forces so that there is one normal force, we have the
diagram shown.
The friction force provides the acceleration.
We can write
?
F
=
m
a
:
x
component:
F
fr
=
Ma
;
y
component:
F
N
–
Mg
= 0.
Thus we have
a
=
F
fr
/
M
=
s
F
N
/
M
=
s
g
.
The maximum value of
a
is
a
max
=
s
g
= (0.80)(9.80 m/s
2
) =
7.8 m/s
2
.
7.
While the box is sliding down, friction will be up the
plane, opposing the motion.
From the force diagram for
the box, we have ?
F
=
m
a
:
x
component:
mg
sin
θ
–
F
fr
=
ma
;
y
component:
F
N
–
mg
cos
= 0.
From the
x
equation, we have
F
fr
=
mg
sin
–
ma
=
m
(
g
sin
–
a
)
= (15.0 kg)[(9.80 m/s
2
) sin 30° – (0.30 m/s
2
)]
=
69 N
.
Because the friction is kinetic, we have
F
fr
=
k
F
N
=
k
mg
cos
;
69 N =
k
(15.0 kg)(9.80 m/s
2
) cos 30°, which gives
k
= 0.54
.
8.
On the horizontal the only force is the friction force, which
provides the acceleration:
–
F
fr
=
ma
1
.
On the hill, we have
x
component: –
mg
sin
–
F
fr
=
ma
2
,
or
a
2
= –
g
sin
+
a
1
=
–
(
9
.
8
0
m
/
s
2
) sin 13°
+ (– 4.80 m/s
2
) =
– 7.00 m/s
2
.
M
g
x
y
F
fr
F
N
a
M
g
x
y
F
fr
F
N
a
m
g
F
N
x
y
v
F
fr
F
N
m
g
F
fr
x
y
v
Chapter 5
Page 3
9.
For the hanging box we can write
?
F
=
m
a
:
y
component:
m
II
g
–
F
T
= 0.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics, Acceleration, Force, Friction

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