Laws Of Motion Question 81
Question: Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12 N. Then the magnitudes of the forces are [AIEEE 2002]
Options:
A) 12 N, 6 N
B) 13 N, 5N
C) 10 N, 8 N
D) 16 N, 2 N
Show Answer
Answer:
Correct Answer: B
Solution:
$ A+B=18 $ -(i)
$ 12=\sqrt{A^{2}+B^{2}+2AB\cos \theta } $ -(ii)
$ \tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\tan 90{}^\circ $
therefore $ \cos \theta =-\frac{A}{B} $ -(iii) By solving (i), (ii) and (iii), $ A=13N $
and $ B=5N $
BETA
