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

  • question_answer
    When forces \[{{F}_{1}},{{F}_{2}},{{F}_{3}}\] are acting on a particle of mass m such that \[{{F}_{1}}\] and \[{{F}_{2}}\] are mutually perpendicular, then the particle remains stationary. If the force \[{{F}_{1}}\] is now removed, then the acceleration of the particle is   [AIEEE 2002]

    A) \[{{F}_{1}}/m\]

    B) \[{{F}_{2}}{{F}_{3}}/m{{F}_{1}}\]

    C) \[({{F}_{2}}-{{F}_{3}})/m\]

    D) \[{{F}_{2}}/m\]

    Correct Answer: A

    Solution :

    [a] In this question, we have to think about that "a body or system to be in equilibrium, net force acting on it should be zero.” The particle is remains stationary under the action of three forces \[{{F}_{1}},{{F}_{2}}\] and \[{{F}_{3}}\], it means resultant force is zero.
    Since, in second case, \[{{F}_{1}}\] is removed (in terms of magnitude we are talking now), the forces acting are \[{{F}_{2}}\] and \[{{F}_{3}}\] the resultant of which has the magnitude as \[{{F}_{1}}\], so acceleration of particle is \[\frac{{{F}_{1}}}{m}\] in the direction opposite to that of \[{{F}_{1}}\].

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