′
− ′
+
′
−
′
+
′
−
′
yF zF
zF
xF
yF
zy
xz
y x
di
bg
##
#
.
ijk
(a) Here,
G
G
′ =
r
r
where
ˆˆ
ˆ
3.0i 2.0j 4.0k,
r
=−+
G
and
G
G
FF
=
1
. Thus, dropping the prime in
the above expression, we set (with SI units understood)
x
= 3.0,
y
= –2.0,
z
= 4.0,
F
x
= 3.0,
F
y
= –4.0 and
F
z
= 5.0. Then we obtain
G
G
G
τ
=× =
−
−
⋅
rF
1
60
30
.
#
.
#
.
#
ij
k
N
m
.
e
j
(b) This is like part (a) but with
G
G
=
2
. We plug in
F
x
= –3.0,
F
y
= –4.0 and
F
z
=
–5.0
and obtain
G
G
G
+
−
⋅
2
26
18
#
.
k
N
m
.
e
j
(c) We can proceed in either of two ways. We can add (vectorially) the answers from
parts (a) and (b), or we can first add the two force vectors and then compute
G
G
G
G
=× +
F
12
(these total force components are computed in the next part). The result
is
()
32i 24k N m.
F
=
−
⋅
GG
(d) Now
G
G
G
′
=−
rr
r
o
where
o
ˆ
3.0i 2.0j 4.0k.
r
=++
G
Therefore, in the above expression, we
set
0,
4.0,
0,
xy
z
′′
′
==
−
=
and
3.0 3.0 0
4.0 4.0
8.0
5.0 5.0 0.
x
y
z
F
F
F
=−=
−
We get
G
G
G
G
=
′
×+
=
F
0
.
22. If we write
G
′ =
′
+
′
+ ′
rx
y
z
###
i
j
k, then (using Eq. 330) we find
G
G
′ ×
r
F
is equal to
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View Full Document30. (a) The acceleration vector is obtained by dividing the force vector by the (scalar)
mass:
a
→
=
F
→
/
m
= (3.00 m/s
2
)i
^
– (4.00 m/s
2
)j
^
+ (2.00 m/s
2
)k
^
.
(b) Use of Eq. 1118 leads directly to
L
→
=
(42.0 kg
.
m
2
/s)i
^
+ (24.0 kg
.
m
2
/s)j
^
+ (60.0 kg
.
m
2
/s)k
^
.
(c) Similarly, the torque is
rF
τ
=×
G
GG
= (–8.00 N
.
m)i
^
– (26.0 N
.
m)j
^
– (40.0 N
.
m)k
^
.
(d) We note (using the Pythagorean theorem) that the magnitude of the velocity vector is
7.35 m/s and that of the force is 10.8 N.
The dot product of these two vectors is
v
→
.
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 Fall '07
 silvera
 mechanics, Angular Momentum, τ

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