Two forces 3,N and 2, N are at an angle theta | vedclass

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Question

(d) $A = 3N$, $B = 2N$ then $R = \sqrt {{A^2} + {B^2} + 2AB\cos 	heta } $

$R = \sqrt {9 + 4 + 12\cos 	heta } $ &hellip;$(i)$

Now $A = 6N$, $B

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(d) $A = 3N$, $B = 2N$ then $R = \sqrt {{A^2} + {B^2} + 2AB\cos 	heta } $

$R = \sqrt {9 + 4 + 12\cos 	heta } $ &hellip;$(i)$

Now $A = 6N$, $B = 2N$ then

$2R = \sqrt {36 + 4 + 24\cos 	heta } $ &hellip;$(ii)$

from $(i)$ and $(ii)$ we get $\cos 	heta = - \frac{1}{2}$

$	herefore $ $	heta = 120^\circ $

Two forces $3\,N$ and $2\, N$ are at an angle $\theta$ such that the resultant is $R$. The first force is now increased to $ 6\,N$ and the resultant become $2R$. The value of is ....... $^o$

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