## Precalculus (6th Edition) Blitzer

a) The resultant force is $-5.62\mathbf{i}+10.01\mathbf{j}$. b) The equilibrium force is $5.62\mathbf{i}-10.01\mathbf{j}.$
(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}