TWO INDIVIDUAL FORCES OF MAGNITUDE F1 AND F2 ACTS ON A BODY OF MASS 1 KG. IF THE PRODUCT OF NET MAGNITUDE OF FORCES PRODUCED IS 27 N THEN FIND THE MAGNITUDE OF FORCES F1 AND F2.(ASSUME F1>F2)

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Solution

$S u p p o s e t w o f o r c e s F_{1} a n d F_{2} a c t o n a b o d y a t a n g l e θ . T h e r e s u l \tan t f o r c e F i s g i v e n b y, {F^{2}}^{} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ, b u t F^{2} = 27 N . T h e r e f o r e, 27 = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ - - - (1) o r 27 = F_{1}^{2} + F_{2}^{2} - 2 F_{1} F_{2} \cos θ - - - - (2) . N o w s u b t r a c t i n g e q u a t i o n (2) f r o m e q u a t i o n (1), w e g e t, 0 = 2 F_{1} F_{2} \cos θ o r \cos θ = 0 . T h e r e f o r e e q u a t i o n (1) a n d e q u a t i o n (2) u n d e r t h i s c o n d i t i o n b e c o m e s, 27 = F_{1}^{2} + F_{2}^{2}, a n d i n c a s e F_{1} = F_{2}, t h e n 27 = 2 F_{1}^{2} o r F_{1} = F_{2} = \frac{27}{\sqrt{2}} N . O t h e r p o s s i b i l i t i e s a r e n o t p o s s i b l e .$

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