Chapter 3, Problem 30P

(a)

To determine

The $x$ and $y$ component of $\overrightarrow{B}$.

Answer to Problem 30P

The $x$ component of $\overrightarrow{B}$ is $6.9$ and $y$ component of $\overrightarrow{B}$ is $-1.7$.

Explanation of Solution

Since angle is below $x$ axis, take it as negative.

Write the expression for $x$ component of $\overrightarrow{B}$

$B_{\text{x}}=B\cos \theta$ (I)

Here, $B_x$ is the $x$ component of $\overrightarrow{B}$ , $B$ is the magnitude of $\overrightarrow{B}$ and $\theta$ is the angle that $\overrightarrow{B}$ makes below the $x$ axis .

Write the expression for $y$ $x$ component of $\overrightarrow{B}$

$B_{\text{y}}=B\sin \theta$ (II)

Here, $B_y$ is the $y$ component of $\overrightarrow{B}$ and $\theta$ is the angle that $\overrightarrow{B}$ makes below the $x$ axis.

Conclusion:

Substitute $-14°$ for $\theta$ and $7.1$ for $B$ in equation (I) to get $B_x$.

$\begin{array}{c}{B}_{\text{x}}=7.1\cos \left(-14°\right)\\ =6.9\end{array}$

Substitute $-14°$ for $\theta$ and $7.1$ for $B$ in equation (II) to get $B_y$.

$\begin{array}{c}{B}_{\text{y}}=7.1\sin \left(-14°\right)\\ =-1.7\end{array}$

Negative sign for $B_y$ indicate that it lies along negative $y$ axis.

Therefore, the $x$ component of $\overrightarrow{B}$ is $6.9$ and $y$ component of $\overrightarrow{B}$ is $-1.7$.

(b)

To determine

The magnitude and direction of $\overrightarrow{C}$.

Answer to Problem 30P

The magnitude of $\overrightarrow{C}$ is $6.9$ and it is $225°$ counterclockwise from the $+x$ axis.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{C}$.

$C=\sqrt{C_x^2+C_y^2}$ (III)

Here, $C$ is the magnitude of vector $\overrightarrow{C}$ , $C_x$ is the $x$ component of $\overrightarrow{C}$ and $C_y$ is the $y$ component of $\overrightarrow{C}$.

Write the expression for direction of $\overrightarrow{C}$.

${\theta }_{\text{c}}={\tan }^{-1}\left(\frac{C_y}{C_x}\right)$ (IV)

Here, ${\theta }_{\text{c}}$ is the angle than $\overrightarrow{C}$ makes with $x$ axis.

Conclusion:

Substitute $-6.7$ for $C_y$ and $-1.8$ for $C_x$ in equation (III) to get $C$.

$\begin{array}{c}C=\sqrt{{\left(-1.8\right)}^2+{\left(-6.7\right)}^2}\\ =6.9\end{array}$

Substitute $-6.7$ for $C_y$ and $-1.8$ for $C_x$ in equation (IV) to get ${\theta }_{\text{c}}$.

$\begin{array}{c}{\theta }_{\text{c}}={\tan }^{-1}\left(\frac{-6.7}{-1.8}\right)\\ =75°\end{array}$

The $x$ component and $y$ component are negative. Therefore, $\overrightarrow{C}$ lies at $180°+75°=255°$ counterclockwise from the $x$ axis.

Therefore, the magnitude of $\overrightarrow{C}$ is $6.9$ and it is $225°$ counterclockwise from the $+x$ axis.

(c)

To determine

The magnitude and direction of $\overrightarrow{C}+\overrightarrow{B}$.

Answer to Problem 30P

The magnitude of $\overrightarrow{C}+\overrightarrow{B}$ is $9.8$ and it is directed $59°$ clockwise from the $+x$ axis.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{C}+\overrightarrow{B}$.

$\left|\overrightarrow{C}+\overrightarrow{B}\right|=\sqrt{{\left(C_x+B_x\right)}^2+{\left(C_y+B_y\right)}^2}$ (V)

Here, $\left|\overrightarrow{C}+\overrightarrow{B}\right|$ is the magnitude of vector $\overrightarrow{C}+\overrightarrow{B}$ , $C_x$ is the $x$ component of $\overrightarrow{C}$ and $C_y$ is the $y$ component of $\overrightarrow{C}$.

Write the expression for direction of $\overrightarrow{C}+\overrightarrow{B}$.

${\theta }_{\text{C+B}}={\tan }^{-1}\left(\frac{C_y+B_y}{C_x+B_x}\right)$ (VI)

Here, ${\theta }_{\text{C+B}}$ is the angle than $\overrightarrow{C}+\overrightarrow{B}$ makes with $+x$ axis.

Conclusion:

Substitute $6.9$ for $B_x$, $-1.7$ for $B_y$, $-6.7$ for $C_y$ and $-1.8$ for $C_x$ in equation (V) to get $\left|\overrightarrow{C}+\overrightarrow{B}\right|$

$\begin{array}{c}\left|\overrightarrow{C}+\overrightarrow{B}\right|=\sqrt{{\left(-1.8+6.9\right)}^2+{\left(-6.7+-1.7\right)}^2}\\ =9.8\end{array}$

Substitute $6.9$ for $B_x$, $-1.7$ for $B_y$,$-6.7$ for $C_y$ and $-1.8$ for $C_x$ in equation (VI) to get ${\theta }_{\text{C+B}}$.

$\begin{array}{c}{\theta }_{\text{C+B}}={\tan }^{-1}\left(\frac{-6.7+-1.7}{-1.8+6.9}\right)\\ =-59°\end{array}$

Therefore, the magnitude of $\overrightarrow{C}+\overrightarrow{B}$ is $9.8$ and it is directed $59°$ clockwise from the $+x$ axis.

(d)

To determine

The magnitude and direction of $\overrightarrow{C}-\overrightarrow{B}$.

Answer to Problem 30P

The magnitude of $\overrightarrow{C}-\overrightarrow{B}$ is $10$ and it is directed $210°$ counterclockwise from the $+x$ axis.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{C}-\overrightarrow{B}$.

$\left|\overrightarrow{C}-\overrightarrow{B}\right|=\sqrt{{\left(C_x-B_x\right)}^2+{\left(C_y-B_y\right)}^2}$ (V)

Here, $\left|\overrightarrow{C}-\overrightarrow{B}\right|$ is the magnitude of vector $\overrightarrow{C}-\overrightarrow{B}$ , $C_x$ is the $x$ component of $\overrightarrow{C}$ and $C_y$ is the $y$ component of $\overrightarrow{C}$.

Write the expression for direction of $\overrightarrow{C}-\overrightarrow{B}$.

${\theta }_{\text{C-B}}={\tan }^{-1}\left(\frac{C_y-B_y}{C_x-B_x}\right)$ (VI)

Here, ${\theta }_{\text{C}-\text{B}}$ is the angle than $\overrightarrow{C}-\overrightarrow{B}$ makes with $+x$ axis.

Conclusion:

Substitute $6.9$ for $B_x$, $-1.7$ for $B_y$, $-6.7$ for $C_y$ and $-1.8$ for $C_x$ in equation (V) to get $\left|\overrightarrow{C}+\overrightarrow{B}\right|$

$\begin{array}{c}\left|\overrightarrow{C}+\overrightarrow{B}\right|=\sqrt{{\left(-1.8-6.9\right)}^2+{\left(-6.7-\left(-1.7\right)\right)}^2}\\ =10\end{array}$

Substitute $6.9$ for $B_x$, $-1.7$ for $B_y$,$-6.7$ for $C_y$ and $-1.8$ for $C_x$ in equation (VI) to get ${\theta }_{\text{C+B}}$.

$\begin{array}{c}{\theta }_{\text{C+B}}={\tan }^{-1}\left(\frac{-6.7-\left(-1.7\right)}{-1.8-6.9}\right)\\ =30°\end{array}$

Since $x$ component of $\overrightarrow{C}-\overrightarrow{B}$ is negative , it lies at $180°+30°=210°$ counterclockwise from the $+x$ axis.

Therefore, the magnitude of $\overrightarrow{C}-\overrightarrow{B}$ is $10$ and it is directed $210°$ counterclockwise from the $+x$ axis.

(e)

To determine

The $x$ and $y$ components of $\overrightarrow{C}-\overrightarrow{B}$.

Answer to Problem 30P

The $x$ components of $\overrightarrow{C}-\overrightarrow{B}$ is $-8.7$ and $y$ component of $\overrightarrow{C}-\overrightarrow{B}$ is $-5.0$.

Explanation of Solution

Write the expression for the $x$ component of $\overrightarrow{C}-\overrightarrow{B}$.

${\left|\overrightarrow{C}-\overrightarrow{B}\right|}_x=C_x-B_x$ (VII)

Here, ${\left|\overrightarrow{C}-\overrightarrow{B}\right|}_x$ is the x component of vector $\overrightarrow{C}-\overrightarrow{B}$.

Write the expression for the $y$ component of $\overrightarrow{C}-\overrightarrow{B}$.

${\left|\overrightarrow{C}-\overrightarrow{B}\right|}_y=C_y-B_y$ (VII)

Here, ${\left|\overrightarrow{C}-\overrightarrow{B}\right|}_y$ is the $y$ component of vector $\overrightarrow{C}-\overrightarrow{B}$.

Conclusion:

Substitute $6.9$ for $B_x$ and $-1.8$ for $C_x$ in equation (VII) to get ${\left|\overrightarrow{C}+\overrightarrow{B}\right|}_x$.

$\begin{array}{c}{\left|\overrightarrow{C}-\overrightarrow{B}\right|}_x=-1.8-6.9\\ =-8.7\end{array}$

Substitute $-1.7$ for $B_y$ and $-6.7$ for $C_y$ in equation (VI) to get ${\theta }_{\text{C+B}}$.

$\begin{array}{c}{\left|\overrightarrow{C}-\overrightarrow{B}\right|}_y=-6.7-\left(-1.7\right)\\ =-5.0\end{array}$

Therefore, the $x$ components of $\overrightarrow{C}-\overrightarrow{B}$ is $-8.7$ and $y$ component of $\overrightarrow{C}-\overrightarrow{B}$ is $-5.0$.