Answer
a) The resultant force is $-5.62\mathbf{i}+10.01\mathbf{j}$.
b) The equilibrium force is $5.62\mathbf{i}-10.01\mathbf{j}.$
Work Step by Step
(a)
The force $\mathbf{F}$ in vector form is given as
$\mathbf{F}=\left\| \left. \mathbf{F} \right\| \right.\cos \theta \mathbf{i}+\left\| \left. \mathbf{F} \right\| \right.\sin \theta \mathbf{j}$
Substitute the value $\theta =70{}^\circ $ and $\left\| \left. {{\mathbf{F}}_{1}} \right\| \right.=8$.
The force ${{\mathbf{F}}_{\mathbf{1}}}$ is given as:
$\begin{align}
& {{\mathbf{F}}_{1}}=\left\| \left. {{\mathbf{F}}_{1}} \right\| \right.\cos 70{}^\circ \mathbf{i}+\left\| \left. {{\mathbf{F}}_{1}} \right\| \right.\sin 70{}^\circ \mathbf{j} \\
& =8\cos 70{}^\circ \mathbf{i}+8\sin 70{}^\circ \mathbf{j} \\
& =2.74\mathbf{i}+7.52\mathbf{j}
\end{align}$
Substitute the value $\theta =140{}^\circ $ and $\left\| \left. {{\mathbf{F}}_{2}} \right\| \right.=6$.
The force ${{\mathbf{F}}_{2}}$ is given as
$\begin{align}
& {{\mathbf{F}}_{2}}=\left\| \left. {{\mathbf{F}}_{2}} \right\| \right.\cos 140{}^\circ \mathbf{i}+\left\| \left. {{\mathbf{F}}_{2}} \right\| \right.\sin 140{}^\circ \mathbf{j} \\
& =6\cos 140{}^\circ \mathbf{i}+6\sin 140{}^\circ \mathbf{j} \\
& =-4.60\mathbf{i}+3.86\mathbf{j}
\end{align}$
Substitute the value $\theta =200{}^\circ $ and $\left\| \left. {{\mathbf{F}}_{3}} \right\| \right.=4$.
The force ${{\mathbf{F}}_{3}}$ is given as
$\begin{align}
& {{\mathbf{F}}_{3}}=\left\| \left. {{\mathbf{F}}_{3}} \right\| \right.\cos 200{}^\circ \mathbf{i}+\left\| \left. {{\mathbf{F}}_{3}} \right\| \right.\sin 200{}^\circ \mathbf{j} \\
& =4\cos 200{}^\circ \mathbf{i}+4\sin 200{}^\circ \mathbf{j} \\
& =-3.76\mathbf{i}-1.37\mathbf{j}
\end{align}$
The resultant force is given by $\mathbf{F}$.
$\begin{align}
& \mathbf{F}={{\mathbf{F}}_{1}}+{{\mathbf{F}}_{2}}+{{\mathbf{F}}_{3}} \\
& {{\mathbf{F}}_{1}}+{{\mathbf{F}}_{2}}+{{\mathbf{F}}_{3}}=\left( 2.74-4.60-3.76 \right)\mathbf{i}+\left( 7.52+3.86-1.37 \right)\mathbf{j} \\
& =-5.62\mathbf{i}+10.01\mathbf{j}
\end{align}$
(b)
The equilibrium force is when net force is 0.
Let the equilibrium force be given by $\mathbf{G}$.
Then,
$\begin{align}
& \mathbf{F}+\mathbf{G}\text{=0} \\
& \mathbf{G}=\mathbf{-F} \\
\end{align}$
Substitute the value of $\mathbf{F}=-5.62\mathbf{i}+10.01\mathbf{j}$ in the above equation to get
$\begin{align}
& \mathbf{G}=-\left( -5.62\mathbf{i}+10.01\mathbf{j} \right) \\
& =5.62\mathbf{i}-10.01\mathbf{j}
\end{align}$
