• question_answer A slab is subjected to two forces ${{\vec{F}}_{1}}$and ${{\vec{F}}_{2}}$of same magnitude F as shown in the figure. Force ${{\vec{F}}_{2}}$is in XY-plane while force ${{F}_{1}}$ acts along z-axis at the point $(2\hat{i}+3j).$ The moment of these forces about point 0 will be- [JEE Main 11-Jan-2019 Morning] A) $(3\hat{i}-2j-3\hat{k})F$ B) $(3\hat{i}+2j+3\hat{k})F$ C) $(3\hat{i}-2j+3\hat{k})F$ D) $(3\hat{i}+2j-3\hat{k})F$

Solution :

 [c] The moment of force about point O is given as ${{\vec{r}}_{1}}\times {{\vec{F}}_{1}}\times {{\vec{r}}_{2}}\times {{\vec{F}}_{2}}=(2\hat{i}+3\hat{j})\times (F\hat{k})+(6\hat{j})$$\times [{{F}_{2}}\cos 30(-\hat{j})+{{F}_{2}}\sin 30(-\hat{i})$ $=F[(3\hat{j}-2\hat{j}+3\hat{k})$

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