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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Self Evaluation Test - Laws of Motion

  • question_answer
    An overweight acrobat, "weighing" in at 115 kg, wants to perform a single hand stand. He tries to cheat by resting one foot against a smooth frictionless vertical wall. The horizontal force there is 130 N. What is the magnitude of the force exerted by the floor on his hand? Answer in N.

    A) 1134    

    B) 1257

    C) 997      

    D) 1119

    Correct Answer: A

    Solution :

    [a] The acrobat has a force acting on his hand that we resolve into two perpendicular components:  the vertical one is the reaction to the weight \[(115\times 9.8N=1127N)\]and the horizontal one balances the 130N force from the wall. These two forces give a resultant force F of \[F=\sqrt{{{1127}^{2}}+{{130}^{2}}}=1134N\]


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