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Laws Of Motion

Question
CBSEENPH11020752

Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is

  • 0.5 N

  • 1.5 N

  • fraction numerator square root of 3 over denominator 4 end fraction space straight N
  • square root of 3 space straight N

Solution

B.

1.5 N

Minimum additional force needed
F = - (Fresultant)x
straight F subscript resultant space equals space left square bracket left parenthesis 4 minus 2 right parenthesis left parenthesis cos space 30 space bold j with hat on top space minus space sin space 30 space bold i with hat on top right parenthesis space plus space left parenthesis cos space 60 space bold i with hat on top space plus sin space 60 space bold j with hat on top right parenthesis
space equals space open square brackets 2 open parentheses fraction numerator square root of 3 over denominator 2 end fraction bold j with hat on top minus 1 half bold i with hat on top close parentheses space plus open parentheses 1 half bold i with hat on top space plus fraction numerator square root of 3 over denominator 2 end fraction bold j with hat on top close parentheses close square brackets
equals open square brackets open parentheses square root of 3 space plus fraction numerator square root of 3 over denominator 2 end fraction close parentheses bold j with hat on top space plus open parentheses negative bold i with bold hat on top space plus fraction numerator bold i with bold hat on top over denominator 2 end fraction close parentheses close square brackets
equals open square brackets negative 1 half bold i with bold hat on top space plus fraction numerator 3 square root of 3 over denominator 2 end fraction bold j with hat on top close square brackets
equals fraction numerator bold i with hat on top over denominator 2 end fraction plus fraction numerator 3 square root of 3 over denominator 2 end fraction space bold j with hat on top
therefore comma
straight F space equals space minus space open parentheses fraction numerator bold i with hat on top over denominator 2 end fraction close parentheses space equals space 1 half bold i with bold hat on top
Hence comma space space vertical line straight F vertical line space equals space 0.5 space straight N