Question

F A B Find the moment of the force, F = 19N shown above Figure with respect to the origin and with respect to point C when A(4, -2, 6), B(4, 3, 2), and C( - 2,5,5) m. + j+ Mo=( Mc =( k)N.m k)N.m i + j + Submit Question

Answer 1

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a = 4i-2; +6K To = 44+3j+2K rc = -2i+ 5j +5k Co-ordinates A-> (4-2,6) B 4,3,2) C(-2,5,5) TÊi = 190 Solution:- Moment vector can be found using the concept of Cross product of a vector r ice from the point to any on the line of action of the force) and the force vector. * First finding force vector :- 8 = LEICÊ unit vector along Since A and B and is directed passes through B to Ê IŠBA For A FA- FB TBAL FRA) where c) 14 A: from final-initial Magnitude Ê 541 A (4i-2j+6)=(4 i + 3j+2x) to Ba| – 5j + 4k =) - 5; +4k 152 +42 1917 -5 jt 4 k7 √41 141 14-833 + 11.87K :. É. Db

* Moment about origin rBox Hi +3j+26] = Cross product of position vector from origin to any point along the force vector and the force vector itself. (I am considering point B, since it lies on the force vector's line of action) (ro-ro) x [Teo (4 € + 3j+ 2x) * (-14.83+ 11.87k) theag* * Solving cross product using determinant. 요 j 3) i(3x11.87- 26-14-83)) j (4x11-87 - 2x0) + K (4*(-14-63)) 65-27 8 - 47.48 9 - 59.32 * Moment about point c. first findinding finding position vector directed from point e to B rb- ře 3 (i + 3; + 2K)-(-2; + S; +Sk) K А 3 2 0 -14-83 11.87 IMO 8 BC 3 6i-2;-3k Mc = řec x j к (61-23-3x)* (-14.83% + 11.87k) Solving cross product using determinat ¿ (-2]x11-87-(-376-14-83]) -j (6*11.81-0) 0 -14.83 1:87 tk ( 6x[-14.83]-0) Mc 7 -68-23i - 71.22; -88.98k 6 -2 -3