Physics
Mechanics
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Question

Three masses, each equal to M, are placed at the three of side a. The force of attraction on corners of a square unit mass at the fourth corner will be

  1. fraction numerator G M over denominator 3 a squared end fraction
  2. fraction numerator G M over denominator a squared end fraction square root of 3
  3. fraction numerator 3 G M over denominator a squared end fraction
  4. fraction numerator G M over denominator a squared end fraction open square brackets 1 half plus square root of 2 close square brackets

The correct answer is: fraction numerator G M over denominator a squared end fraction open square brackets 1 half plus square root of 2 close square brackets


    F subscript 1 equals F subscript 2 equals fraction numerator G M over denominator a squared end fraction
    Resultant of F subscript 1 and F subscript 2 is square root of 2 fraction numerator G M over denominator a squared end fraction

    Now, F subscript 3 equals fraction numerator G M over denominator left parenthesis square root of 2 a right parenthesis squared end fraction equals fraction numerator G M over denominator 2 a squared end fraction
    Now, fraction numerator square root of 2 G M over denominator a squared end fraction and fraction numerator G M over denominator 2 a squared end fraction act in the same direction.
    Their resultant is  fraction numerator square root of 2 G M over denominator a squared end fraction plus fraction numerator G M over denominator 2 a squared end fraction text  or  end text fraction numerator G M over denominator a squared end fraction open square brackets square root of 2 plus 1 half close square brackets

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