JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Equilibrium of Forces

  • question_answer
    Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12 N. Then the magnitudes of the forces are                     [AIEEE 2002]

    A)             12 N, 6 N

    B)               13 N, 5N

    C)             10 N, 8 N

    D)               16 N, 2 N

    Correct Answer: B

    Solution :

                    \[A+B=18\]                            ?(i)             \[12=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]               ?(ii)             \[\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\tan 90{}^\circ \] Þ \[\cos \theta =-\frac{A}{B}\]            ?(iii)               By solving (i), (ii) and (iii), \[A=13N\] and \[B=5N\]


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