Two forces of magnitude $P$ & $Q$ acting at a point have resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. Angle between the forces is :
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$
Two forces ${F_1} = 1\,N$ and ${F_2} = 2\,N$ act along the lines $x = 0$ and $y = 0$ respectively. Then the resultant of forces would be
Two forces having magnitude $A$ and $\frac{ A }{2}$ are perpendicular to each other. The magnitude of their resultant is
Following sets of three forces act on a body. Whose resultant cannot be zero
The resultant of $\vec A$ and $\vec B$ makes an angle $\alpha $ with $\vec A$ and $\beta $ with $\vec B$,
When $n$ vectors of different magnitudes are added, we get a null vector. Then the value of $n$ cannot be