Given, F1=2F(−i^)+2F3(−j^) r1=0i^+6j^ Torque due to F1 force τF1=r1×F1=6j^×(2F(−i^)+2F3(−j^))=3 F(k^) Torque due to F2 force τF2=(2i^+3j^)×τnet=τF1+τF2Fk^=3 Fi^+2 F(−j^)=3Fi+2 F(−j^)+3 F(k^)=(3i^−2j^+3k^)F
A slab is subjected to two forces F1 and F2 of same magnitude F as shown in the figure. Force F2 is in XY- plane while force F1 acts along z -axis at the point (2i+3j). The moment of these forces about point O will be: 
Held on 11 Jan 2019 · Verified 6 Jul 2026.
(3i^−2j^+3k^) F
(3i^−2j^−3k^)F
(3i^+2j^−3k^)F
(3i^+2j^+3k^)F
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