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 ,\text{ }R=4\text{ }N $ $ R=4N=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ} $ $ P+Q=4 $ ...(i) When $ \theta \,\text{= }90{}^\circ ,\text{ }R=3\text{ }N $ $ {{P}^{2}}+{{Q}^{2}}=9 $ ?(ii) From Eq. (i) $ {{(P+Q)}^{2}}=16 $ $ \Rightarrow $ $ {{P}^{2}}+{{Q}^{2}}+2PQ=16 $ $ \Rightarrow $ $ 9+2\text{ }PQ=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 $