Two forces are such that sum of their magnitudes is 18 N and their resultant is 12 N which is perpen

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3-1.Vectors

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Two forces are such that the sum of their magnitudes is $18\; N$ and their resultant is $12\; N$ which is perpendicular to the smaller force. Then the magnitudes of the forces are

A

$12, 5$

B

$14, 4$

C

$5, 13$

D

$10, 8$

(AIEEE-2002)

Solution

(c) Let $P$ be the smaller force and $Q $ be the greater force then according to problem –

$P + Q = 18$……$(i)$

$R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } = 12$…….$(ii)$

$\tan \phi = \frac{{Q\sin \theta }}{{P + Q\cos \theta }} = \tan 90 = \infty $

$P{\rm{ }} + {\rm{ }}Q{\rm{ cos}}\theta = 0$…….$(iii)$

By solving $(i)$, $(ii)$ and $(iii)$ we will get $P = 5,$ and $Q = 13$

Standard 11

Physics

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