Higher Engineering Mathematics, Sixth Edition

282 Higher Engineering Mathematics

Hence

p×( 2 r× 3 q)=( 4 i+j− 2 k)

×(− 24 i− 42 j− 12 k)

=

∣

∣

∣

∣∣

∣

∣

ijk

41 − 2

− 24 − 42 − 12

∣

∣

∣

∣∣

∣

∣

=i(− 12 − 84 )−j(− 48 − 48 )

+k(− 168 + 24 )

=− 96 i+ 96 j− 144 k

or− 48 ( 2 i− 2 j+ 3 k)

Practical applications of vector products

Problem 9. Find the moment and the magnitude

of the moment of a force of(i+ 2 j− 3 k)newtons

about pointBhaving co-ordinates (0, 1, 1), when

the force acts on a line throughAwhose

co-ordinates are (1, 3, 4).

The momentMabout pointBof a force vectorFwhich

has a position vector ofrfromAis given by:

M=r×F

ris the vector fromBtoA,i.e.r=BA.

But BA=BO+OA=OA−OB (see Problem 13,

page 262), that is:

r=(i+ 3 j+ 4 k)−(j+k)

=i+ 2 j+ 3 k

Moment,

M=r×F=(i+ 2 j+ 3 k)×(i+ 2 j− 3 k)

=

∣ ∣ ∣ ∣ ∣ ∣ ∣

ij k

12 3

12 − 3

∣ ∣ ∣ ∣ ∣ ∣ ∣

=i(− 6 − 6 )−j(− 3 − 3 )

+k( 2 − 2 )

=− 12 i+ 6 jNm

The magnitude ofM,

|M|=|r×F|

=

√

[(r•r)(F•F)−(r•F)^2 ]

r•r=( 1 )( 1 )+( 2 )( 2 )+( 3 )( 3 )= 14

F•F=( 1 )( 1 )+( 2 )( 2 )+(− 3 )(− 3 )= 14

r•F=( 1 )( 1 )+( 2 )( 2 )+( 3 )(− 3 )=− 4

|M|=

√

[14× 14 −(− 4 )^2 ]

=

√

180Nm=13.42Nm

Problem 10. The axis of a circular cylinder

coincides with thez-axis and it rotates with an

angular velocity of( 2 i− 5 j+ 7 k)rad/s. Determine

the tangential velocity at a pointPon the cylinder,

whose co-ordinates are(j+ 3 k)metres, and also

determine the magnitude of the tangential velocity.

ThevelocityvofpointPonabodyrotatingwithangular

velocityωabout a fixed axis is given by:

v=ω×r,

whereris the point on vectorP.

Thus v=( 2 i− 5 j+ 7 k)×(j+ 3 k)

=

∣ ∣ ∣ ∣ ∣ ∣ ∣

ijk

2 − 57

013

∣ ∣ ∣ ∣ ∣ ∣ ∣

=i(− 15 − 7 )−j( 6 − 0 )+k( 2 − 0 )

=(− 22 i− 6 j+ 2 k)m/s

Themagnitude ofv,

|v|=

√

[(ω•ω)(r•r)−(r•ω)^2 ]

ω•ω=( 2 )( 2 )+(− 5 )(− 5 )+( 7 )( 7 )= 78

r•r=( 0 )( 0 )+( 1 )( 1 )+( 3 )( 3 )= 10

ω•r=( 2 )( 0 )+(− 5 )( 1 )+( 7 )( 3 )= 16

Hence,

|v|=

√

( 78 × 10 − 162 )

=

√

524m/s=22.89m/s