Chapter 5, Problem 56A

(a)

To determine

The horizontal and vertical components of the given vector.

Answer to Problem 56A

The horizontal and vertical components of the vector E are $E_x=3.0,E_y=4.0$

Explanation of Solution

Given:

A vector E with its length 5. Up and positive directions are assumed positive.

Formula Used:

The horizontal and vertical component of a vector R is given by,

  $\begin{array}{c}{R}_x=R\cos \theta \\ R_y=R\sin \theta \end{array}$

The length or magnitude of a vector is expressed as,

  $R^2=\sqrt{R_x{}^2+R_y{}^2}$

The angle is given by,

  $\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$

Calculation:

For the given magnitude of the vector E which is 5.0 units, the possible perfect square components of the vector are given by,

  $\begin{array}{l}{E}_x=3.0\\ E_y=4.0\end{array}$

The choice of vector components is made so based on the angle made by the vector with the horizontal. The y-component being 4.0 units and the x-component being 3.0 units makes the angle close to 60 degrees which is evident from the given Figure 1.

  

Figure 1

The angle can be calculated as,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)\\ ={\tan }^{-1}\left(\frac{4}{3}\right)\\ =53.13°\end{array}$

Thus, the horizontal and vertical components of the given vector E can be summarized as,

  $\begin{array}{c}{E}_x=5\cos 53.13°\\ =3\\ E_y=5\sin 53.13°\\ =4\end{array}$

Conclusion:

The horizontal and vertical components of the given vector is $E_x=3.0,E_y=4.0$ .

(b)

To determine

The horizontal and vertical components of the given vector.

Answer to Problem 56A

The horizontal and vertical components of the vector F are $F_x=-3.0,F_y=-4.0$

Explanation of Solution

Given:

A vector F with its length -5.0. Up and positive directions are assumed positive.

Formula Used:

The horizontal and vertical component of a vector R is given by,

  $\begin{array}{c}{R}_x=R\cos \theta \\ R_y=R\sin \theta \end{array}$

The length or magnitude of a vector is expressed as,

  $R^2=\sqrt{R_x{}^2+R_y{}^2}$

The angle is given by,

  $\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$

Calculation:

For the given magnitude of the vectorF which is -5.0 units, the possible perfect square components of the vector are given by,

  $\begin{array}{l}{F}_x=-3.0\\ F_y=-4.0\end{array}$

The choice of vector components is made so based on the angle made by the vector with the horizontal. The y-component being -4.0 units and the x-component being -3.0 units makes the angle close to 60 degrees which is evident from the given Figure 1.

  

Figure 2

The angle can be calculated as,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)\\ ={\tan }^{-1}\left(\frac{-4}{-3}\right)\\ =53.13°\end{array}$

Thus, the horizontal and vertical components of the given vector F are,

  $\begin{array}{c}{F}_x=-5\cos 53.13°\\ =-3\\ F_y=-5\sin 53.13°\\ =-4\end{array}$

Conclusion:

The horizontal and vertical components of the given vector are $F_x=-3.0,F_y=-4.0$ .

(c)

To determine

The horizontal and vertical components of the given vector.

Answer to Problem 56A

The horizontal and vertical components of the vector Aare $A_x=-3.0,A_y=0$.

Explanation of Solution

Given:

A vector A with its length -3.0. Up and positive directions are assumed positive.

Formula Used:

The horizontal and vertical component of a vector R is given by,

  $\begin{array}{c}{R}_x=R\cos \theta \\ R_y=R\sin \theta \end{array}$

The length or magnitude of a vector is expressed as,

  $R^2=\sqrt{R_x{}^2+R_y{}^2}$

The angle is given by,

  $\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$

Calculation:

For the given magnitude of the vectorA which is -3.0 units, it is evident from the Figure 3 that, the vector has only a horizontal component, in the negative direction.

  

Figure 3

Thus, the horizontal and vertical components of the given vector A are,

  $\begin{array}{c}{A}_x=-3\\ A_y=0\end{array}$

Conclusion:

The horizontal and vertical components of the given vector are $A_x=-3.0,A_y=0$ .