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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 perpendicular to the smaller force and magnitude of resultant is $12\, N$. Then the magnitudes of the forces are
A
$12\, N, 6 \,N$
B
$13\, N, 5\,N$
C
$10\, N, 8 \,N$
D
$16\, N, 2\, N$
Solution
(b) $A + B = 18$…(i)
$12 = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $…(ii)
$\tan \alpha = \frac{{B\sin \theta }}{{A + B\cos \theta }} = \tan 90^\circ $
$⇒$ $\cos \theta = – \frac{A}{B}$…(iii)
By solving (i), (ii) and (iii),
$A = 13\,N$ and $B = 5\,N$
Standard 11
Physics
