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. $(2+\frac{1}{2}\sqrt{2})N$ and $(2-\frac{1}{2}\sqrt{2})N$
  
• D. $(2+\frac{1}{2}\sqrt{3})N$ and $(2-\frac{1}{2}\sqrt{3})N$

Answer 1

The correct option is C

$(2+\frac{1}{2}\sqrt{2})N$ and $(2-\frac{1}{2}\sqrt{2})N$

Let P and Q are forces, we know that $R=\sqrt{P^2+Q^2+2PQ\cos \theta }$
When $\theta =0^0,R=4N=\sqrt{P^2+Q^2+2PQ}$
$P+Q=4$ ...(1)
When $\theta ={90}^0,R=3N$
$P^2+Q^2=9$ ...(2)
From equation (1)
${(P+Q)}^2=16\Rightarrow P^2+Q^2+2PQ=16\Rightarrow 9+2PQ=16\Rightarrow 2PQ=7$
Now ${(P-Q)}^2=P^2+Q^2-2PQ$
$\Rightarrow {(P-Q)}^2=9-7\Rightarrow P-Q=\sqrt{2}$ ...(3)
On solving equation (1) and (3)
$P=(2+\frac{1}{2}\sqrt{2})N$ and $Q=(2-\frac{1}{2}\sqrt{2})N$