Chapter 10: Simple and Harmonic Motion and Elasticity.
Selected Conceptual Questions, pp.312.
Problems pp. 313: # 10, 13, 24, 34, 39, 49, 57, 66
Chapter 16: Waves and Sound.
Selected Conceptual Questions, pp.499.
Problems pp. 500: # 6, 16, 20, 23, 27, 42, 60, 70, 88, 104
Chapter 17: The Principle of Linear Superposition and Interference
Phenomena
.
Selected Conceptual Questions, pp.527.
Problems pp. 528: # 21, 37, 39, 43, 51, 52
Chapter 10: Simple and Harmonic Motion and Elasticity
10.
REASONING
The free-body diagram shows the
magnitudes and directions of the forces acting on the
block. The weight
mg
acts downward. The
maximum force of static friction
f
s
max
acts upward
just before the block begins to slip. The force from
the spring
F
x
Applied
=
kx
(Equation 10.1) is directed
to the right. The normal force
F
N
from the wall
points to the left. The magnitude of the maximum
force of static friction is related
to the
magnitude of the normal force according to
Equation 4.7
(
29
max
s
s
N
f
F
μ
=
, where
μ
s
is the coefficient of static friction. Since the block is at
equilibrium just before it begins to slip, the forces in the
x
direction must balance and the
forces in the
y
direction must balance. The balance of forces in the two directions will provide
two equations, from which we will determine the coefficient of static friction.
SOLUTION
Since the forces in the
x
direction and in the
y
direction must balance, we have
N
s
N
and
F
kx
mg
F
=
=
Substituting the first equation into the second equation gives
(
29
(
29
(
29
(
29
(
29
2
s
N
s
s
1.6 kg 9.80 m/s
or
0.79
510 N/m 0.039 m
mg
mg
F
kx
kx
=
=
=
=
=
+
y
+
x
F
N
F
s
max
=
μ
s
F
N
mg
F
x
Applied
=
kx
1