p05_079.pdf
(
81 KB
)
Pobierz
Chapter 5 - 5.79
79. Since (
x
0
,y
0
)=(0
,
0) and
v
0
=6
.
0i, we have from Eq. 2-15
x
=(6
.
0)
t
+
1
2
a
x
t
2
1
2
a
y
t
2
.
y
=
These equations express uniform acceleration along each axis; the
x
axis points east and the
y
axis
presumably points north (the assumption is that the figure shown in the problem is a view
from above
).
Lengths are in meters, time is in seconds, and force is in newtons.
Examination of any non-zero (
x, y
) point will su.ce, though it is certainly a good idea to check results by
examining more than one. Here we will look at the
t
=4
.
0 s point, at (8
.
0
,
8
.
0). The
x
equation becomes
8
.
0=(6
.
0)(4
.
0) +
2
a
x
(4
.
0)
2
. Therefore,
a
x
=
−
2
.
0m
/
s
2
.The
y
equation becomes 8
.
0=
2
a
y
(4
.
0)
2
.
Thus,
a
y
=1
.
0m
/
s
2
. The force, then, is
F
=
ma
=
24i+12j
−→
(27
153
◦
)
where the vector has been expressed in unit-vector and then magnitude-angle notation. Thus, the force
has magnitude 27 N and is directed 63
◦
west of north (or, equivalently, 27
◦
north of west).
−
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