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Resultant of two forces P and Q is `sqrt(3) Q`. The resultant is inclined at `30^@` with P . Show that, either P =Q or P=2Q.

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The resultant of two forces P and Q is 5 sqrt(P^2+Q^2) when the angle between them is alpha . When the angle changes to (90^@-alpha) , the resultant becomes 3sqrt(P^2+Q^2) . Show that, tan alpha =1/3 .

Knowledge Check

  • The resultant of two forces vecF_1 and vecF_2 is P. If vecF_2 is reversed, the resultant is Q, then

    A
    `F_1^2+F_2^2=2(P^2+Q^2)`
    B
    `F_1^2-F_2^2=2(P^2+Q^2)`
    C
    `P^2+Q^2=2(F_1^2+F_2^2)`
    D
    `P^2-Q^2=2(F_1^2-F_2^2)`
  • If the resultant of two forces of magnitudes P and Q acting at a point at an angle of 60^@ is sqrt7 Q, then P/Q is :

    A
    1
    B
    44230
    C
    2
    D
    4
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