Question

Two forces P and Q when the angle between them is 90°; their resultant is square root of 10 Newton. But when the angle is reduced to 60° the resultant force is square root of 13 Newton. Calculate the two forces P and Q.

Answer 1

Answer:

A^2 + B^2 = 10     for the right triangle

(A + B cos 60)^2 + (B sin 60)^2 = 13     for the 60 deg angle

(A + B/2)^2 + 3 B^2 / 4 = 13    since cos 60 = 1/2 and sin 60 = 3^1/2 /2

A^2 + A B + B^2 = 13    

A B = 3    since   A^2 + B^2 = 10

A^2 + (3 / A)^2 = 10

A^2 + 9 / A^2 = 10

A^4 -10 A^2 + 9 = 0

(A^2 - 9) * (A^2 - 1) = 0

So A = 3 is one solution with B = 1

Note: A^2 + B^2 = 10

and A^2 + A B + B^2 = 9 + 3 + 1 = 13