JEE Main · 2004 · 24 Apr 2004 · Offline · Vector Algebra

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With two forces acting at a point, the maximum effect is obtained when their resultant is 4 N . If they act at right angles, then their resultant is 3 N . Then the forces are

(A) $(2 + 2)$ N and $(2 - 2)$ N
(B) $(2 + 3)$ N and $(2 - 3)$ N
(C) $(2 + \frac{1}{2} 2)$ N and $(2 - \frac{1}{2} 2)$ N✓ Correct
(D) $(2 + \frac{1}{2} 3)$ N and $(2 - \frac{1}{2} 3)$ N

Answer: (C)

Solution

Let the forces be

F_{1}

and

F_{2}

. The maximum resultant occurs when they are parallel, giving

F_{1} + F_{2} = 4

, and at right angles the resultant is

F_{1}^{2} + F_{2}^{2} = 3

, so

F_{1}^{2} + F_{2}^{2} = 9

. From

(F_{1} + F_{2})^{2} = 16

we get

2 F_{1} F_{2} = 16 - 9 = 7

, so

F_{1} F_{2} = \frac{7}{2}

. The forces are the roots of

t^{2} - 4 t + \frac{7}{2} = 0

, namely

t = \frac{4 \pm 2}{2} = 2 \pm \frac{1}{2} 2

, so they are

(2 + \frac{1}{2} 2)

N and

(2 - \frac{1}{2} 2)

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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JEE Main · 2004 · 24 Apr 2004 · Offline · Vector Algebra

Solution

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