22. If we write
r
= x
ˆi+ y
ˆj + z
ˆ
k, then (using Eq. 3-30) we find
r
× F is equal to
(y
F
z
− z
F
y
)ˆi + (z
F
x
− x
F
z
)ˆj + (x
F
y
− y
F
x
) ˆ
k .
(a) Here,
r
=
r where
r = 3ˆi
− 2ˆj + 4 ˆk, and F = F
1
. Thus, dropping the primes in the above
expression, we set (with SI units understood) x = 3, y =
−2, z = 4, F
x
= 3, F
y
=
−4 and F
z
= 5.
Then we obtain
τ =
r
× F
1
=
6.0ˆi
− 3.0ˆj− 6.0 ˆk
N
·m.
(b) This is like part (a) but with
F =
F
2
. We plug in F
x
=
−3, F
y
=
−4 and F
z
=
−5 and obtain
τ =
r
× F
2
=
26ˆi + 3.0ˆj
− 18 ˆk
N
·m.
(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
τ =
r
×
F
1
+
F
2
(these total force
components are computed in the next part). The result is
32ˆi
− 24 ˆk
N
·m.
(d) Now
r
=
r
−r
o
where
r
o
= 3ˆi+2ˆj+ 4 ˆ
k. Therefore, in the above expression, we set x
= 0, y
=
−4,
z
= 0, F
x
= 3
− 3 = 0, F
y
=
−4 − 4 = −8 and F
z
= 5
− 5 = 0. We get τ = r
×
F
1
+
F
2
= 0.