The sum of two force is 9 N. Their resultant which is perpendicular to the smaller force is of 6N. Find the magnitude of the forces.
Answers
Answer:
Since R is given to be perpendicular to A, then I shall take A along the (+ x) axis and R will be along the (+ y) axis. Vector A has no component in the (+ y) direction. Clearly, B must be in the Second Quadrant and A must cancel with the x-component of B and R would equal to the y-component of B. We can set up the following:
x-component of B; (B cos k) = magnitude of A. But A = 9 - B
B (cos k) = 9 - B
B (sin k) = 6
B is the hypotenuse of the triangle with leg (B-y) and base (B-x). Then using Pythagorean: B^2 = [B (cos k)]^2 + [B (sin k)]^2 = (9 - B)^2 + 36
Removing brackets and using cos-square + sin-square = unity, and solving for B, we obtain: 18 B = 117; → B = 6.5 N and A = 2.5 N.
Explanation:
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