Physics
Mechanics
Easy

Question

A uniform rod of length L (in between the supports) and mass m is placed on two supports A and B. The rod breaks suddenly at length L/10 from the support B. Find the reaction at support A immediately after the rod breaks.

  1. 9 over 40 m g
  2. 19 over 40 m g
  3. fraction numerator m g over denominator 2 end fraction
  4. 9 over 20 m g

The correct answer is: 9 over 40 m g


    table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell text  Torque  end text equals tau equals 9 over 10 m g open parentheses 9 over 20 L close parentheses end cell row cell equals I alpha equals m over 3 open parentheses 9 over 10 L close parentheses squared alpha semicolon space of 1em a equals fraction numerator 3 g over denominator 2 L end fraction end cell end table
table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell text  Acceleration  end text a subscript C M end subscript equals alpha left parenthesis A C right parenthesis equals fraction numerator 3 g over denominator 2 L end fraction open parentheses fraction numerator 9 L over denominator 20 end fraction close parentheses equals fraction numerator 27 g over denominator 40 end fraction end cell row cell N subscript A equals 9 over 40 m g end cell end table

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