Question

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

• A. $(2+\sqrt{2})N$ and $(2-\sqrt{2})N$
• B. $(2+\sqrt{3})N$ and $(2-\sqrt{3})N$
• C. $\left(2+\frac{1}{2}\sqrt{2}\right)N$ and $\left(2-\frac{1}{2}\sqrt{2}\right)N$
• D. $\left(2+\frac{1}{2}\sqrt{3}\right)N$ and $\left(2-\frac{1}{2}\sqrt{3}\right)N$

Answer 1

Answer: (C)

Let P and Q are forces. We know that $R=\sqrt{P^2+Q^2+2PQ\cos \theta }$ when $\theta =0{}^{\circ },R=4N$ $R=4N=\sqrt{P^2+Q^2+2PQ}$ $P+Q=4$ ...(i) When $\theta \text{= }90{}^{\circ },R=3N$ $P^2+Q^2=9$ ?(ii) From Eq. (i) ${(P+Q)}^2=16$ $\Rightarrow$ $P^2+Q^2+2PQ=16$ $\Rightarrow$ $9+2PQ=16$ [using Eq. (ii)] $\Rightarrow$ $2PQ=7$ Now, ${(P-Q)}^2=P^2+Q^2-2PQ$ $\Rightarrow$ ${(P-Q)}^2=9-7$ $P-Q=\sqrt{2}$ ...(iii) On solving Eqs. (i) and (iii) $P=\left(2+\frac{1}{2}\sqrt{2}\right)N$ and $Q=\left(2-\frac{1}{2}\sqrt{2}\right)N$