This preview shows page 1. Sign up to view the full content.
which yields
32
ˆ
(30 k) kg m /s.
t
=⋅
G
A
(f) Again, the acceleration vector is given by
G
at
=−
60
2
.
#
i in SI units, and the net force on
the car is
ma
G
. In a similar argument to that given in the previous part, we have
G
G
G
G
G
τ
=
′
×=
−
−
−
−
−
mr a
t
b
g
b
g
b
g b
gb
g
b
g
ch
e
j
o
k
30
20 0
50
2
..
.
.
#
which yields
2
ˆ
(90 k) N m.
t
G
87. This problem involves the vector cross product of vectors lying in the
xy
plane. For
such vectors, if we write
G
′ =
′′
rx
y
##
i +
j , then (using Eq. 330) we find
G
G
′
′
−
′
rv
x
vy
v
yx
di
#
k.
(a) Here,
G
′
r
points in either the
+
#
i or
the
−
#
i direction (since the particle moves along
the
x
axis). It has no
y
′
or
z
′
components, and neither does
G
v
, so it is clear from the
above expression (or, more simply, from the fact that
ii
=
0
×
) that
G
A
GG
=
′
mr
v
b
g
0
in
this case.
(b) The net force is in the
−
#
i direction (as one finds from differentiating the velocity
expression, yielding the acceleration), so, similar to what we found in part (a), we obtain
=
′
G
G
rF
0.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

Click to edit the document details