Chapter 2, Problem 25P

(a)

To determine

The magnitude and direction of vector.

Answer to Problem 25P

The magnitude of the vector is $9.43\text{cm}$ and the direction is $122°$ from the $+x$ axis.

Explanation of Solution

Write the expression for calculating the magnitude of a vector

  $\text{magnitude}=\sqrt{x^2+y^2}$        (I)

Here, $x$ component, and $y$ component.

Write the expression to calculate the direction,

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{y}{x}\right)\\ d=\theta -180°\end{array}$        (II)

Here, $\theta$ is the angle, and $d$ is the direction

Conclusion:

Substitute $-5.0\text{cm}$ for $x$, and $\text{8}\text{.0cm}$ for $y$ in expression (I)

  $\begin{array}{c}\text{magnitude}=\sqrt{{\left(-5.0\text{cm}\right)}^2+{\left(\text{8}\text{.0cm}\right)}^2}\\ =9.43\text{cm}\end{array}$

Substitute $-5.0\text{cm}$ for $x$, and $\text{8}\text{.0cm}$ for $y$ in expression (II)   

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-5.0\text{cm}}{\text{8}\text{.0cm}}\right)\\ =-57.99°\\ d=\left(-57.99°\right)-180°\\ =122°\text{from the +x axis}\text{.}\end{array}$

The magnitude of the vector is $9.43\text{cm}$ and the direction is $122°$ from the $+x$ axis.

(b)

To determine

The magnitude and direction of vector.

Answer to Problem 25P

The magnitude of the vector is $\left|F\right|=134.16\text{N}$ and the direction is $153.43°$ from the $+x$ axis.

Explanation of Solution

Write the expression for calculating the magnitude of a vector

  $\text{magnitude}=\sqrt{x^2+y^2}$        (I)

Here, $x$ component, and $y$ component.

Write the expression to calculate the direction,

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{y}{x}\right)\\ d=\theta -180°\end{array}$        (II)

Here, $\theta$ is the angle, and $d$ is the direction

Conclusion:

Substitute $120\text{N}$ for $F_x$, and $-60\text{N}$ for $F_y$ in expression (I)

  $\begin{array}{c}\text{magnitude}=\sqrt{{\left(120\text{N}\right)}^2+{\left(-60\text{N}\right)}^2}\\ =134.16\text{N}\end{array}$

Substitute $120\text{N}$ for $F_x$, and $-60\text{N}$ for $F_y$ in expression (I)

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{120\text{N}}{-60\text{N}}\right)\\ =-26.56°\\ d=\left(-26.56°\right)-180°\\ =153.43°\text{from the +x axis}\text{.}\end{array}$

The magnitude of the vector is $\left|F\right|=134.16\text{N}$ and the direction is $153.43°$ from the $+x$ axis.

(c)

To determine

The magnitude and direction of vector.

Answer to Problem 25P

The magnitude of the vector is $\left|V\right|=16.28\text{m/s}$ and the direction is $212.71°$ from the $+x$ axis.

Explanation of Solution

Write the expression for calculating the magnitude of a vector

  $\text{magnitude}=\sqrt{x^2+y^2}$        (I)

Here, $x$ component, and $y$ component.

Write the expression to calculate the direction,

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{y}{x}\right)\\ d=\theta -180°\end{array}$        (II)

Here, $\theta$ is the angle, and $d$ is the direction

Conclusion:

Substitute $-13.7\text{m/s}$ for $V_x$, and $-8.8\text{m/s}$ for $V_y$ in expression (I)

  $\begin{array}{c}\text{magnitude}=\sqrt{{\left(-13.7\text{m/s}\right)}^2+{\left(-8.8\text{m/s}\right)}^2}\\ =16.28\text{m/s}\end{array}$

Substitute $-13.7\text{m/s}$ for $V_x$, and $-8.8\text{m/s}$ for $V_y$ in expression (I)

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-13.7\text{m/s}}{-8.8\text{m/s}}\right)\\ =-32.71°\\ d=\left(-32.71°\right)-180°\\ =212.71°\text{from the +x axis}\text{.}\end{array}$

The magnitude of the vector is $\left|V\right|=16.28\text{m/s}$ and the direction is $212.71°$ from the $+x$ axis.

(d)

To determine

The magnitude and direction of vector.

Answer to Problem 25P

The magnitude of the vector is $\left|a\right|=6.8{\text{m/s}}^2$ and the direction is $1.618°$ from the $+x$ axis.

Explanation of Solution

Write the expression for calculating the magnitude of a vector

  $\text{magnitude}=\sqrt{x^2+y^2}$        (I)

Here, $x$ component, and $y$ component.

Write the expression to calculate the direction,

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{y}{x}\right)\\ d=\theta -180°\end{array}$        (II)

Here, $\theta$ is the angle, and $d$ is the direction

Conclusion:

Substitute $2.3{\text{m/s}}^2$ for $a_x$, and $6.5\times {10}^{-2}{\text{m/s}}^2$ for $a_y$ in expression (I)

  $\begin{array}{c}\text{magnitude}=\sqrt{{\left(2.3{\text{m/s}}^2\right)}^2+{\left(6.5\times {10}^{-2}{\text{m/s}}^2\right)}^2}\\ =2.3{\text{m/s}}^2\end{array}$

Substitute $2.3{\text{m/s}}^2$ for $a_x$, and $6.5\times {10}^{-2}{\text{m/s}}^2$ for $a_y$ in expression (II)

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{6.5\times {10}^{-2}{\text{m/s}}^2}{2.3{\text{m/s}}^2}\right)\\ =1.618°\text{from the +x axis}\text{.}\end{array}$

The magnitude of the vector is $\left|a\right|=6.8{\text{m/s}}^2$ and the direction is $1.618°$ from the $+x$ axis.