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5.120:
a) There are many ways to do these sorts of problems; the method presented is
fairly straightforward in terms of application of Newton’s laws, but involves a good deal
of algebra. 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
. The equations of motion are then
.
cos
sin
sin
mg
n
ma
n
ma
n
MA
y
x

α
=
α
=
α

=
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
,
y
x
a
a
,
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
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