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
–
µ
k
F
N
= 0;
y
-component:
F
N
–
Mg
= 0.
Thus
F
N
=
Mg
, and
F
=
µ
k
F
N
=
µ
k
Mg
= (0.30)(12.0 kg)(9.80 m/s
2
) =
35 N
.
If
µ
k
= 0, there is
no force
required to maintain constant
speed.
2.
(
a
) In general, static friction is given by
F
fr
=
µ
s
F
N
.
Immediately
before the box starts to move, the static friction force reaches
its maximum value:
F
fr,max
=
µ
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
–
µ
s
F
N
= 0;
y
-component:
F
N
–
Mg
= 0.
Thus
F
N
=
Mg
, and
F
=
µ
s
F
N
=
µ
s
Mg
;
25.0 N =
µ
s
(6.0 kg)(9.80 m/s
2
), which gives
µ
s
= 0.43
.
(
b
)
When the box accelerates and the friction changes to kinetic, we have
F
–
µ
k
F
N
=
Ma
;
25.0 N –
µ
k
(6.0 kg)(9.80 m/s
2
) = (6.0 kg)(0.50 m/s
2
), which gives
µ
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
θ
= 0.
When we combine the two equations, we have
tan
θ
=
F
fr
/
F
N
=
µ
s
.
Thus we have
tan
θ
max
=
µ
s
= 0.8,
θ
max
= 39°
.
M
g
x
y
F
F
fr
F
N
M
g
x
y
F
F
fr
F
N
F
N
m
g
θ
F
fr
F
N
m
g
θ
F
fr
F
N
m
g
θ
F
fr
F
N
m
g
θ
F
fr
x
y

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