JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • 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) \[(3\hat{i}+2j-3\hat{k})F\]

    Correct Answer: C

    Solution :

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

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