Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

Answer to Problem 54P

The magnitude of vector is $10.12\text{m}$ and the angle taken counterclockwise from the $+x$ direction is $-32.9°$ , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The $x$ and $y$ components of a displacement vector are $+8.5\text{m}$ and $-5.5\text{m}$ .

Formula used:

Write the expression for $x$ component of the displacement vector.

  $d_x=d\cos \theta$

Here, $d_x$ is the $x$ component of the displacement vector, $d$ is the magnitude of displacement vector and $\theta$ is the angle between the vector and horizontal component.

Rearrange above equation for $d$ .

  $d=\frac{d_x}{\cos \theta }$ ……. (1)

Write the expression for $y$ component of the displacement vector.

  $d_y=d\sin \theta$

Here, $d_y$ is the $y$ component of the displacement vector.

Write the expression for the angle between two vectors.

  $\tan \theta =\frac{d_y}{d_x}$ ……. (2)

Calculation:

Substitute $-5.5\text{m}$ for $d_y$ and $+8.5\text{m}$ for $d_x$ in equation (2).

  $\begin{array}{c}\tan \theta =\frac{-5.5\text{m}}{+8.5\text{m}}\\ \theta ={\tan }^{-1}\left(\frac{-5.5\text{m}}{+8.5\text{m}}\right)\\ =-32.9°\end{array}$

Substitute $-32.9°$ for $\theta$ and $+8.5\text{m}$ for $d_x$ in equation (1).

  $\begin{array}{c}d=\frac{8.5\text{m}}{\cos \left(-32.9°\right)}\\ =10.12\text{m}\end{array}$

Conclusion:

Thus, the magnitude of vector is $10.12\text{m}$ and the angle taken counterclockwise from the $+x$ direction is $-32.9°$ , it is negative because it is taken in counterclockwise direction.

(b)

To determine

The magnitude and angle of velocity vector.

Answer to Problem 54P

The magnitude of vector is $-82.76\text{m/s}$ and the angle taken counterclockwise from the $+x$ direction is $-25.01°$ , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The $x$ and $y$ components of a velocity vector are $-75\text{m}/\text{s}$ and $+35\text{m /s}$ .

Formula used:

Write the expression for $x$ component of the velocity vector.

  $v_x=v\cos \theta$

Here, $v_x$ is the $x$ component of the velocity vector, $v$ is the magnitude of the velocity vector and $\theta$ is the angle between the vector and horizontal component.

Rearrange the above equation for $v$ .

  $v=\frac{v_x}{\cos \theta }$ ……. (3)

Write the expression for the angle between two vectors.

  $\tan \theta =\frac{v_y}{v_x}$ ……. (4)

Here, $v_y$ is the $y$ component of the velocity vector.

Calculation:

Substitute $+35\text{m /s}$ for $v_y$ and $-75\text{m}/\text{s}$ for $v_x$ in equation (4).

  $\begin{array}{c}\tan \theta =\frac{+35\text{m /s}}{-75\text{m}/\text{s}}\\ \theta ={\tan }^{-1}\left(\frac{+35\text{m /s}}{-75\text{m}/\text{s}}\right)\\ =-25.01°\end{array}$

Substitute $-25.01°$ for $\theta$ and $-75\text{m}/\text{s}$ for $v_x$ in equation (3).

  $\begin{array}{c}v=\frac{-75\text{m}/\text{s}}{\cos \left(-25.01°\right)}\\ =-82.76\text{m/s}\end{array}$

Conclusion:

Thus, the magnitude of vector is $-82.76\text{m/s}$ and the angle taken counterclockwise from the $+x$ direction is $-25.01°$ , it is negative because it is taken in counterclockwise direction.

To determine

The magnitude and angle of force vector.

Answer to Problem 54P

The magnitude of vector is $50\text{lb}$ and the angle taken counterclockwise from the $+x$ direction is $-36.87°$ , it is negative because it is in third quadrant.

Explanation of Solution

Given:

The $x$ component of force vector is $40\text{lb}$ and magnitude of force vector $50\text{lb}$ .

Formula used:

Write the expression for $x$ component of the force vector.

  $F_x=F\cos \theta$

Here, $F_x$ is the $x$ component of the force vector, $F$ is the magnitude of the force vector and $\theta$ is the angle between the vector and horizontal component.

Rearrange the above equation for $\theta$ .

  $\begin{array}{c}\cos \theta =\frac{F_x}{F}\\ \theta ={\cos }^{-1}\left(\frac{F_x}{F}\right)\end{array}$ ……. (5)

Calculation:

Substitute $40\text{lb}$ for $F_x$ and $50\text{lb}$ for $F$ in equation (5).

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{F_x}{F}\right)\\ \theta ={\cos }^{-1}\left(\frac{40\text{lb}}{50\text{lb}}\right)\\ =-36.87°\end{array}$

Conclusion:

Thus, the magnitude of vector is $50\text{lb}$ and the angle taken counterclockwise from the $+x$ direction is $-36.87°$ , it is negative because it is in the third quadrant