Question

Refer to figure (2-E1). Find (a) the magnitude, (b) x and y component and (c) the angle with the X-axis of the resultant of $\overrightarrow{\mathrm{OA}},\overrightarrow{\mathrm{BC}}\mathrm{and}\overrightarrow{\mathrm{DE}}$.

Figure (2-E1)

Answer 1

First, let us find the components of the vectors along the x and y-axes. Then we will find the resultant x and y-components.

x-component of $\overrightarrow{\mathrm{OA}}=2\cos 30°=\sqrt{3}$ m
x-component of $\overrightarrow{\mathrm{BC}}$ =1.5cos120°
$=-\frac{\left(1.5\right)}{2}=-7.5$ m
x-component of $\overrightarrow{\mathrm{DE}}$ = 1cos270°
= 1 × 0 = 0 m

y-component of $\overrightarrow{\mathrm{OA}}$ = 2 sin 30° = 1
y-component of $\overrightarrow{\mathrm{BC}}$ = 1.5 sin 120°
$=\frac{\left(\sqrt{3}\times 1.5\right)}{2}=1.3$
y-component of $\overrightarrow{\mathrm{DE}}$ = 1 sin 270° = −1
x-component of resultant ${\mathrm{R}}_x=\sqrt{3}-0.75+0=0.98\mathrm{m}$

y-component of resultant R_y = 1 + 1.3 − 1 = 1.3 m

$$\begin{gathered} \therefore \mathrm{Resultant},\mathrm{R}=\sqrt{{\left(R_x\right)}^2+{\left(R_y\right)}^2} \\ =\sqrt{{\left(0.98\right)}^2+{\left(1.3\right)}^2} \\ =\sqrt{0.96+1.69} \\ =\sqrt{2.65} \\ =1.6\mathrm{m} \end{gathered}$$

If it makes an angle α with the positive x-axis, then
$$\begin{gathered} \tan \mathrm{\alpha }=\frac{y-component}{x-\mathrm{component}} \\ =\frac{1.3}{0.98}=1.332 \end{gathered}$$
∴ α = tan⁻¹ (1.32)