find the magnitude of the two forces such that it they are at right angles, their resultant is 10N, But if they act an 60 degree ,their resultant is √13 N
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Let resultant at right angle be p(= √15N) and at 60 degrees be q(= √18N.
Now, p = √(x2 + y2) = √15,
q = √(x2 + y2 +2xycos60) = √18.
So x2 + y2 = 15,
x2 + y2 + xy = 18,
therefore xy = 3 or y = 3/x.
Now, x2 + (3/x)2 = 15,
x4 + 9 = 15x2,
x4 - 15x2 + 9 = 0.
Now find the roots of this quadratic equation, the two values of x will correspond to the magnitudes of the two vectors.
Now, p = √(x2 + y2) = √15,
q = √(x2 + y2 +2xycos60) = √18.
So x2 + y2 = 15,
x2 + y2 + xy = 18,
therefore xy = 3 or y = 3/x.
Now, x2 + (3/x)2 = 15,
x4 + 9 = 15x2,
x4 - 15x2 + 9 = 0.
Now find the roots of this quadratic equation, the two values of x will correspond to the magnitudes of the two vectors.
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