Chapter 3, Problem 33PQ

(a)

To determine

Component form of vector $\overrightarrow{A}$.

Answer to Problem 33PQ

Component form of vector $\overrightarrow{A}$ is $-4\widehat{i}-2\widehat{j}$.

Explanation of Solution

Write the x-component of vector $\overrightarrow{A}$.

    $A_x=-4.00\text{units}$

Here, $A_x$ is the x-component of vector $\overrightarrow{A}$.

Write the y-component of vector $\overrightarrow{A}$.

    $A_y=-2.00\text{units}$

Here, $A_y$ is the y-component of vector $\overrightarrow{A}$.

Write the expression for component form of vector $\overrightarrow{A}$.

  $A=A_x\widehat{i}+A_y\widehat{j}$

Here, $\widehat{i}$ is the unit normal vector in x-direction and $\widehat{j}$ is the unit normal vector in y-direction.

Conclusion:

Substitute $-4.00\text{units}$ for $A_x$ and $-2.00\text{units}$ for $A_y$ in the above equation for the component form of vector $\overrightarrow{A}$.

    $\begin{array}{c}A=\left(-4.00\text{units}\right)\widehat{i}+\left(-2.00\text{units}\right)\widehat{j}\\ =\left(-4.00\text{units}\right)\widehat{i}-\left(2.00\text{units}\right)\widehat{j}\end{array}$

Therefore, the component form of vector $\overrightarrow{A}$ is $\left(-4.00\text{units}\right)\widehat{i}-\left(2.00\text{units}\right)\widehat{j}$.

(b)

To determine

Magnitude and direction of vector $\overrightarrow{A}$.

Answer to Problem 33PQ

Magnitude is $\sqrt{20}\text{units}$ and directed at $26.56°$ with respect to negative x direction.

Explanation of Solution

Write the expression to find the magnitude of vector $\overrightarrow{A}$.

  $A=\sqrt{A_x^2+A_y^2}$

Here, $A$ is the magnitude of vector $\overrightarrow{A}$.

Since both $A_x$ and $A_y$ are negative, $\overrightarrow{A}$ lies in the third quadrant. Write the equation to find the direction of $\overrightarrow{A}$ with respect to –x axis.

    $\theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right)$

Here, $\theta$ is the direction of $\overrightarrow{A}$ with respect to –x axis.

Conclusion:

Substitute $-4.00\text{units}$ for $A_x$ and $-2.00\text{units}$ for $A_y$ in the equation for $A$.

    $\begin{array}{c}A=\sqrt{{\left(-4.00\text{units}\right)}^2+{\left(-2.00\text{units}\right)}^2}\\ =\sqrt{20}\text{units}\\ =4.47\text{units}\end{array}$

Substitute $-4.00\text{units}$ for $A_x$ and $-2.00\text{units}$ for $A_y$ in the equation for $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-2.00\text{units}}{-4.00\text{units}}\right)\\ ={\tan }^{-1}\left(0.5\right)\\ =26.56°\end{array}$

Thus, the magnitude is $4.47\text{units}$ and directed at $26.56°$ with respect to negative x direction.

The angle from positive $x$ axis given by

  $\begin{array}{c}{\theta }^{\prime }=180°+26.56°\\ =206.56°\\ \approx 207°\end{array}$

Therefore, the magnitude and direction of the vector is $4.47\text{units}$ and $207°$.

(c)

To determine

Identify $\overrightarrow{B}$ such that on addition with $\overrightarrow{A}$ gives a resultant with $6.00\text{units}$ of x-component and no y-component.

Answer to Problem 33PQ

Vector $\overrightarrow{B}$ is $\left(-2.00\text{units}\right)\widehat{i}+\left(2.00\text{units}\right)\widehat{j}$.

Explanation of Solution

Write the relation between $A_x,B_x$ and $C_x$.

  $B_x=C_x-A_x$

Here, $C_x$ is the x-component of $\overrightarrow{C}$ and $B_x$ is the x-component of $\overrightarrow{B}$.

Write the relation between $A_y,B_y$ and $C_y$.

  $B_y=C_y-A_y$

Here, $C_y$ is the y-component of $\overrightarrow{C}$ and $B_y$ is the y-component of $\overrightarrow{B}$.

Write the expression for component form of vector $\overrightarrow{B}$.

  $\overrightarrow{B}=B_x\widehat{i}+B_y\widehat{j}$

Conclusion:

Substitute $-6.00\text{units}$ for $C_x$ and $-4.00\text{units}$ for $A_x$ in equation in the equation for $B_x$.

    $\begin{array}{c}{B}_x=-6.00\text{units}-\left(-4.00\text{units}\right)\\ =-2.00\text{units}\end{array}$

Substitute $0.00\text{units}$ for $C_y$ and $-2.00\text{units}$ for $A_y$ the equation for $B_y$.

  $\begin{array}{c}{B}_y=0.00\text{units}-\left(-2.00\text{units}\right)\\ =2.00\text{units}\end{array}$

Substitute $-2.00\text{units}$ for $B_x$ and $2.00\text{units}$ for $B_y$ in the equation for $\overrightarrow{B}$.

    $\overrightarrow{B}=\left(-2.00\text{units}\right)\widehat{i}+\left(2.00\text{units}\right)\widehat{j}$

Therefore, the vector $\overrightarrow{B}$ is $\left(-2.00\text{units}\right)\widehat{i}+\left(2.00\text{units}\right)\widehat{j}$.