fisica1-youn-e-freedman-exercicios-resolvidos

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 a
x
and a
y
. The equations of motion are then
MA = −n
sin α
ma
ma
x
y
= n sin α
= n cosα − 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, a , a and n. Solution
is possible with the imposition of the relation between A, a
x
and a
y
.
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 a
y,
but the
horizontal acceleration of the block is a x
− A.
To this observer, the block descends at an
angle α , so the relation needed is
ay
= − tan α.
ax
− A
At this point, algebra is unavoidable. Symbolic-manipulation programs may save some
M
solution time. A possible approach is to eliminate a
x
by noting that ax
= − A (a result
m
that anticipates conservation of momentum), using this in the kinematic constraint to
eliminate a
y
and then eliminating n. The results are:
− gm
A =
( M + m)
tan α + ( M
a
x
gM
=
( M + m)
tan α + ( M
tan α)
tan α)
− g(
M + m) tan α
ay
=
( M + m)
tan α + ( M tan α)
(b) When M >> m, A → 0,
as expected (the large block won’t move). Also,
g
tanα
ax →
tanα
+ (1 tanα
)
= g 2 = g sin α cosα,
which is the acceleration of the block
tan α + 1
( g sin α in this case), with the factor of cos α giving the horizontal component. Similarly,
a y
→ −g sin 2 α.
(c) The trajectory is a spiral.
x
y