A Force of 100 N acts at a point P(-2,3,5)m has its line of action passing through Q(10,3,4)m. Calculate moment of this force about origin (0,0,0).

#### Solution

Given: O = (0,0,0)

P= (4.5, -2) Q= (-3,1,6) A= (3,2,0) F=100 N To find : Moment of the force about origin

Solution: Let 𝑝̅ and 𝑞 ̅be the position vectors of points P and Q with respect to the origin 0

∴` baro9=bar(-2l)+3J+5k`

∴` bar oQ=10barl+3barJ+4k`

∴ `bar(PQ)=bar(oQ)-bar(oP)=(10barl+3barJ4bark)-(-2barl-3barJ+5k)`

=`12barl-bark`

∴ `|pQ| I=sqrt(12^2+(-1)^2)=sqrt145`

Unit vector along`PQ= bar(PQ)=bar(PQ)/|bar(PQ)|=(12barl-bark)/sqrt145`

Force along `PQ=barf=100xx(12barl-bark)/sqrt145`

Moment of F about O = `bar(OP)xxbarf`

`barl bar J bark`

`=100/sqrt145 xx -2 3 5 `

` 12 0 -1`

=`8.3045 (-3bari+58barj-36-bark)`

=`-24.9135barj+481.661.66barj-298.962bark Nm`

Moment of the force =`-24.9135barj+481.661.66barj-298.962bark Nm`