# A Force of 140 Kn Passes Through Point C (-6,2,2) and Goes to Point B (6,6,8). Calculate Moment of Force About Origin. - Engineering Mechanics

A force of 140 kN passes through point C (-6,2,2) and goes to point B (6,6,8). Calculate moment of force about origin.

#### Solution

Given : C (-6,2,2)
B (6,6,8)
To find : Moment of force about origin
Assume πΜ and πΜ be the position vectors of points B and C respectively w.r.t O (0,0,0)
overline(OB)=overlineb=6overlineI+6overlineJ+8overlineK
overline(OC)=6overlineb+2overlinej+2overlinek
overline(CB)=(6overlineI+6overlineJ+8overlineK)-(-6overlineI+2overlineJ+overlineK)
=2(6overlineI+2overlineJ+3overlineK)
overline(|CB|)=2xsqrt(6^2+2^2+3^2)
=14
Unit vector along overline(CB)=(CB)/(|CB|)=(6i+2j+3k)/7
Force along overline(CB)=overlineF=140x(6i+2j+3k)/7
=120overlineI+40 overlineJ+60overlineK
Moment of πΉΜ about O =overline(OB)xoverlineF

 π π π 6 6 8 120 40 60

=40overlineI+600overlineJ-480K
Moment of F about C is 40overlineI+600overlineJ-480overlinek kNm

Concept: Moment of force about a point
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