Question

Two forces 3N and 2 N are at an angle θ such that the resultant is R. The first force is now increased to 6N and the resultant become 2R. The value of θ is-

• A. $30^{0}$
• B. $60^{0}$
• C. $90^{0}$
• D. $120^{0}$

Answer 1

Answer: (D)

$120^{0}$

Answer 2

$A = 3N$, $B = 2N$ then

$R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}$

$R = \sqrt{9 + 4 + 12\cos\theta}$ …(i)

Now $A = 6N$, $B = 2N$ then

$2R = \sqrt{36 + 4 + 24\cos\theta}$ …(ii)

from (i) and (ii) we get $\cos\theta = - \frac{1}{2}\therefore$ $\theta = 120{^\circ}$