Q 11-04-013NEETJEE MainEAMCET 2012 (Engineering)Easy
The sum of the magnitudes of two forces acting at a point is $16$ N. If their resultant is normal to the smaller force and has a magnitude of $8$ N, then the forces are
Answer: (D) $6$ N and $10$ N
Let the smaller force be $F_1$ and the larger $F_2$. The resultant is perpendicular to $F_1$, so $F_1$, $R$ and $F_2$ form a right triangle with $F_2$ as the hypotenuse:
$$F_2^2 - F_1^2 = R^2 = 64 \;\Rightarrow\; (F_2 - F_1)(F_2 + F_1) = 64$$
With $F_1 + F_2 = 16$: $F_2 - F_1 = 4$. So $F_1 = 6$ N and $F_2 = 10$ N.
Solution by Sreeraj P, M.Sc Physics