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 N, 6 N (b) 13 N, 5 N (c) 10 N, 8 N (d) 16 N, 2 N

Answer: (b). Let the forces be P > Q with P + Q = 18. The resultant is perpendicular to Q, so P² = Q² + R². P² - Q² = 144 ⇒ (P - Q)(18) = 144 ⇒ P - Q = 8 P = 13 N, Q = 5 N So the answer is (b) 13 N, 5 N.

Two forces are such that the sum of their magnitudes is 18 N and… | JEE Main 2002 (AIEEE)

Q25JEE Main 2002 · AIEEEMedium

Equilibrium, Third Law & Frames

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

Show answer and solution

Answer: (b)

Let the forces be $P > Q$ with $P + Q = 18$ . The resultant is perpendicular to $Q$ , so $P^2 = Q^2 + R^2$ .
$P^2 - Q^2 = 144 \Rightarrow (P - Q)(18) = 144 \Rightarrow P - Q = 8$
$P = 13\text{ N}$ , $Q = 5\text{ N}$
So the answer is (b) 13 N, 5 N.

Report a mistake

Equilibrium, Third Law & Frames question from JEE Main 2002 (AIEEE)

More questions on Equilibrium, Third Law & Frames