Applying Newton&s Laws
5-53
5.119.
IDENTIFY:
The analysis is the same as for Problem 5.96.
SET UP:
The speed is related to the period by
22
(
t
a
n
)
/
vR
Th
T
ππ
β
==
, or
2
(tan )/
v
πβ
=
.
EXECUTE:
The maximum and minimum speeds are the same as those found in Problem 5.96,
s
max
s
cos
sin
tan
sin
cos
vg
h
βμ β
βμ
+
=
−
and
s
min
s
cos
sin
tan
sin
cos
h
−
=
+
.
The minimum and maximum values of the period
T
are then
s
min
s
tan
sin
cos
2
cos
sin
h
T
g
ββ
μ
π
−
=
+
and
s
max
s
tan
sin
cos
2
cos
sin
h
T
g
+
=
−
.
EVALUATE:
For
s
0
=
the results for the speeds reduce to
min
max
vv
g
h
.
tan
R
h
=
. The result for
v
then
agrees with the result in Example 5.23, if we take into account that in this problem
is measured from the vertical
whereas in Example 5.23 it is measured relative to the horizontal.
5.120.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to the block and to the wedge.
SET UP:
For both parts, take the
x
-direction to be horizontal and positive to the right, and the
y
-direction to be
vertical and positive upward. The normal force between the block and the wedge is
n
; the normal force between the
wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The
horizontal acceleration of the wedge is
A
, and the components of acceleration of the block are
x
a
and
y
a
.
EXECUTE:
(a)
The equations of motion are then
sin
MA
n
α
= −
,
sin
x
ma
n
=
and
cos
y
ma
n
mg
=−
. Note
that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are
three equations in four unknowns,
A
,
,
x
y
aa
and
n
. Solution is possible with the imposition of the relation between
A
,
x
a
and
y
a
. An observer on the wedge is not in an inertial frame, and should not apply Newton&s laws, but the
kinematic relation between the components of acceleration are not so restricted. To such an observer, the vertical
acceleration of the block is
,
y
a
but the horizontal acceleration of the block is
.
x
aA
−
To this observer, the block
descends at an angle
,
so the relation needed is
tan .
y
x
a
α
−
At this point, algebra is unavoidable. A
possible approach is to eliminate
x
a
by noting that
x
M
m
, using this in the kinematic constraint to eliminate
y
a
and then eliminating
n
. The results are
:
()
t
a
n
(
t
a
n
)
gm
A
Mm
M
−
=
++
t
a
n
(
t
a
n
)
x
gM
a
M
=
(m
)
t
a
n
t
a
n
(
t
a
n
)
y
gM
a
M
−+
=
(b)
When
,
0,
A
>>
→
as expected (the large block won&t move). Also,
2
tan
sin
cos
(1 tan )
tan
1
x
g
ag
g
αα
→=
=
which is the acceleration of the block ( sin
g
in this case),
with the factor of cos
giving the horizontal component. Similarly,
2
sin
y
→−
.
(c)
The trajectory is a spiral.
EVALUATE:
If
mM
>>
, our general results give
0
x
a
=
and
y
= −
. The massive block accelerates straight
downward, as if it were in free-fall.
5.121.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to the block and to the wedge.