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.