two forces while acting on a paricle in opposite directions , have the resultant of 10 N . if they act at right to each other , the resultant is found to be 50 N. Find the two forces.

Open in App

Solution

Let A nd B be the two force acting at an angle $θ$ then
Resultant of the two force is given by:

$F = \sqrt{A^{2} + B^{2} + 2 A B \cos θ} w h e n t w o f o r c e s a r e a c t i n g i n o p p o s i t e d i r e c t i o n, θ = 180^{o} a n d r e s u l \tan t f o r c e i s 10 N . t h e r e f o r e 10 = \sqrt{A^{2} + B^{2} + 2 A B \cos 180} 10 = \sqrt{A^{2} + B^{2} - 2 A B} 10 = \sqrt{(A - B)^{2}} 10 = A - B - - - (1) W h e n t w o f o r c e s a r e a c t i n g a t r i g h t a n g l e θ = 90^{o} a n d r e s u l \tan t f o r c e i s 50 N . 50 = \sqrt{A^{2} + B^{2} + 2 A B \cos 90} 50 = \sqrt{A^{2} + B^{2}} 2500 = A^{2} + B^{2} f r o m e q u a t i o n n o . (1) A = 10 + B t h e r e f o r e 2500 = (10 + B)^{2} + B^{2} 2500 = 100 + B^{2} + 20 B + B^{2} B^{2} + 10 B - 1200 = 0 o n s o l v i n g t h i s q u a d r a t i c e q u a t i o n w e g e t B = 30 N o r B = - 40 N u \sin g e q u a t i o n n o (1) w e g e t, t a k i n g p o s i t i v e v a l u e . A = 10 N$

Suggest Corrections

Similar questions

10 N

50 N

10 N

50 N

Join BYJU'S Learning Program