The two vectors A and B in the figure have equal magnitudes of 13.5 m and the angles are θ1 = 33° and θ2 = 110°. How to find (a) the x component and (b) the y component of their vector sum R , (c) the magnitude of R, and (d) the angle R ?

Sep 22, 2015

Here's what I got.

Explanation:

I don't wave a good way of drawing you a diagram, so I'll try to walk you through the steps as they come along.

So, the idea here is that you can find the $x$-component and the $y$-component of the vector sum, $R$, by adding the $x$-components and $y$-components, respectively, of the $\vec{a}$ and $\vec{b}$ vectors.

For vector $\vec{a}$, things are pretty straighforward. The $x$-component will be the projection of the vector on the $x$-axis, which is equal to

${a}_{x} = a \cdot \cos \left({\theta}_{1}\right)$

Likewise, the $y$-component will be the projection of the vector on the $y$-axis

${a}_{y} = a \cdot \sin \left({\theta}_{1}\right)$

For vector $\vec{b}$, things are a little more complicated. More specifically, finding the corresponding angles will be a little tricky.

The angle between $\vec{a}$ and $\vec{b}$ is

${\theta}_{3} = {180}^{\circ} - {\theta}_{2} = {180}^{\circ} - {110}^{\circ} = {70}^{\circ}$

Draw a parallel line to the $x$-axis that intersects the point where the tail of $\vec{b}$ and head of $\vec{a}$ meet.

In your case, line $m$ will be the $x$-axis and line $a$ the parallel line you draw.

In this drawing, $\angle 6$ is ${\theta}_{1}$. You know that $\angle 6$ is equal to $\angle 3$, $\angle 2$, and $\angle 7$.

The angle between $\vec{b}$ and the $x$-axis will be equal to

${180}^{\circ} - \left({\theta}_{1} + {\theta}_{2}\right) = {180}^{\circ} - {143}^{\circ} = {37}^{\circ}$

This means that the $x$-component of vector $\vec{b}$ will be

${b}_{x} = b \cdot \cos \left({37}^{\circ}\right)$

Now, because the angle between the $x$-component and the $y$-component of a vector is equal to ${90}^{\circ}$, it follows that the angle for the $y$-component of $\vec{b}$ will be

${90}^{\circ} - {37}^{\circ} = {53}^{\circ}$

The $y$-component will thus be

${b}_{y} = b \cdot \sin \left({37}^{\circ}\right)$

Now, keep in mind that the $x$-component of $\vec{b}$ is oriented in the opposite direction of the $x$-component of $\vec{a}$. This means that the $x$-component of $\vec{R}$ will be

${R}_{x} = {a}_{x} + {b}_{x}$

${R}_{x} = 13.5 \cdot \cos \left({33}^{\circ}\right) - 13.5 \cdot \cos \left({37}^{\circ}\right)$

${R}_{x} = 13.5 \cdot 0.04 = \textcolor{g r e e n}{\text{0.54 m}}$

The $y$-components are oriented in the same direction, so you have

${R}_{y} = {a}_{y} + {b}_{y}$

${R}_{y} = 13.5 \cdot \left[\sin \left({110}^{\circ}\right) + \sin \left({37}^{\circ}\right)\right]$

${R}_{y} = 13.5 \cdot 1.542 = \textcolor{g r e e n}{\text{20.82 m}}$

The magnitude of $\vec{R}$ will be

${R}^{2} = {R}_{x}^{2} + {R}_{y}^{2}$

$R = \sqrt{0.54 \text{^2 + 20.82""^2)" m" = color(green)("20.83 m}}$

To get the angle of $\vec{R}$, simply use

$\tan \left({\theta}_{R}\right) = {R}_{y} / {R}_{x} \implies {\theta}_{R} = \arctan \left({R}_{y} / {R}_{x}\right)$

theta_R = arctan((20.82color(red)(cancel(color(black)("m"))))/(0.54color(red)(cancel(color(black)("m"))))) = color(green)(88.6""^@)