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′
− ′
+
′
−
′
+
′
−
′
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
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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

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