Answer
(a) The magnitude of the larger of the two forces is $5.00N$.
(b) The magnitude of the resultant force is $7.74N$.
Work Step by Step
Given: $F_{1},F_{2}$ are the two forces and $\theta$ is the angle between them.
Let $R$ be the resultant of the two forces,i.e, $R=F_{1}+F_{2}$
(a) (i) At $\theta=0.0^{\circ}$, suppose $F_{1}$ & $F_{2}$ are on the x-axis.
$|R|=\sqrt (R_{x}^2+R_{y}^2)$->(1) and $R_{y}=0$, $R_{x}=F_{1}cos(0^{\circ})+F_{2}cos(0^{\circ})=F_{1}+F_{2}$
From equation (1),we get
$|R|=8.00$
$F_{1}+F_{2}=8.00$->(2)
(ii) At $\theta=90.0^{\circ}$, suppose $F_{1}$ is on the (+)ve x-direction and $F_{2}$ is along the (+)ve y-direction.
Now,
$R_{x}=F_{1}cos(0^{\circ})+F_{2}cos(90.0^{\circ})=F_{1}$
$R_{y}=F_{1}sin(0^{\circ})+F_{2}sin(90.0^{\circ})=F_{2}$
From equation (1), we get
$|R|=\sqrt (F_{1}^2+F_{2}^2)$
$=>5.83^2=(8.00-F_{2})^2+F_{2}^2=64.00+F_{2}^2-16.00F_{2}+F_{2}^2$ ,[using equation (2)]
$=>33.99=2F_{2}^2-16.00F_{2}+64.00$
$=>2F_{2}^2-16.00F_{2}+30.01=0$->(3)
Using quadratic formula in equation (3) gives us,
$F_{2}=\frac{-(-16.00)+-\sqrt ((-16.00)^2-4\times2\times30.01)}{2\times2}$
This gives us $F_{2}=5.00N$ and $F_{1}=3.00N$ or $F_{2}=3.00N$ and $F_{1}=5.00N$
Thus, the magnitude of the larger of the two forces is $5.00N$.
(b) Let $F_{1}=5.00N$ and $F_{2}=3.00N$ with $F_{1}$ along the (+)ve x-direction and $F_{2}$ at an angle of $30.0^{\circ}$ from the (+)ve x-direction.
Again,
$R_{x}=5.00cos(0^{\circ})+3.00cos(30.0^{\circ})=7.598N$
$R_{y}=5.00sin(0^{\circ})+3.00sin(30.0^{\circ})=1.500N$
From equation (1) ,we get
$|R|=\sqrt (7.598^2+1.500^2)=7.74N$
Thus, the magnitude of the resultant force is $7.74N$.
