Question 2.5: The link in Fig. 2-18a is subjected to two forces F1 and F2.......
The link in Fig. 2-18a is subjected to two forces \mathbf{F}_{1} and \mathbf{F}_{2}. Determine the magnitude and direction of the resultant force.
I
Scalar Notation. First we resolve each force into its x and y components, Fig. 2-18 b, then we sum these components algebraically.
\begin{array}{rlrl} \stackrel{十}{\rightarrow}\left(F_{R}\right)_{x} \quad =\Sigma F_{x} ; \quad\left(F_{R}\right)_{x} & =600 \cos 30^{\circ} \mathrm{N}-400 \sin 45^{\circ} \mathrm{N} \\ & =236.8 \mathrm{~N} \rightarrow \\ +\uparrow\left(F_{R}\right)_{y}=\Sigma F_{y} ; \quad \left(F_{R}\right)_{y} & =600 \sin 30^{\circ} \mathrm{N}+400 \cos 45^{\circ} \mathrm{N} \\ & =582.8 \mathrm{~N} \uparrow \end{array}
The resultant force, shown in Fig. 2-18 c, has a magnitude of
\begin{aligned} F_{R} & =\sqrt{(236.8 \mathrm{~N})^{2}+(582.8 \mathrm{~N})^{2}} \\ & =629 \mathrm{~N} \end{aligned}
From the vector addition,
\theta=\tan ^{-1}\left(\frac{582.8 \mathrm{~N}}{236.8 \mathrm{~N}}\right)=67.9^{\circ}
II
Cartesian Vector Notation. From Fig. 2-18b, each force is first expressed as a Cartesian vector.
\begin{aligned} & \mathbf{F}_{1}=\left\{600 \cos 30^{\circ} \mathbf{i}+600 \sin 30^{\circ} \mathbf{j}\right\} \mathbf{N} \\ & \mathbf{F}_{2}=\left\{-400 \sin 45^{\circ} \mathbf{i}+400 \cos 45^{\circ} \mathbf{j}\right\} \mathbf{N} \end{aligned}
Then,
\begin{aligned} \mathbf{F}_{R}=\mathbf{F}_{1}+\mathbf{F}_{2}= & \left(600 \cos 30^{\circ} \mathrm{N}-400 \sin 45^{\circ} \mathrm{N}\right) \mathbf{i} \\ & +\left(600 \sin 30^{\circ} \mathrm{N}+400 \cos 45^{\circ} \mathrm{N}\right) \mathbf{j} \\ = & \{236.8 \mathbf{i}+582.8 \mathbf{j}\} \mathrm{N} \end{aligned}
The magnitude and direction of \mathbf{F}_{R} are determined in the same manner as before.
NOTE: Comparing the two methods of solution, notice that the use of scalar notation is more efficient since the components can be found directly, without first having to express each force as a Cartesian vector before adding the components. Later, however, we will show that Cartesian vector analysis is very beneficial for solving three-dimensional problems.