Chapter 3, Problem 17P

Solution

(a)

To Determine:The two possible values for x -component.

Solution: $\pm 62.25$

Explanation:

Given:

The resultant magnitude is $\left|\overrightarrow{V}\right|=90$ .

The y-component of $V_y$ vector $\overrightarrow{V}$ is -65.0.

Formula used:

The magnitude of vector can be obtained by:

  $\left|\overrightarrow{V}\right|=\sqrt{\left(V_x{}^2\right)+{\left(V_y\right)}^2}$

And, the direction θ is defined as:

  $\theta ={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$

Calculation:

Substitute the values and solve:

  $\begin{array}{l}\left|\overrightarrow{V}\right|=\sqrt{\left(V_x{}^2\right)+{\left(V_y\right)}^2}\\ 90=\sqrt{{\left(V_x\right)}^2+{\left(-65\right)}^2}\\ V_x=\pm \sqrt{\left({90}^2-{\left(-65\right)}^2\right)}\\ V_x=\pm 62.25\end{array}$

Conclusion:

Thus, the two possible values of x-component are $\pm 62.25$ .

(b)

To Determine: A vector that can be added to the positive x -component to give resultant of 80 units in negative x -direction.

Solution:A vector $\overrightarrow{U}$ can be added which has a magnitude of 156.4 and directed ${155.5}^{\text{o}}$ counter clockwise from positive x -axis.

Given:

The resultant vector is $\overrightarrow{V}+\overrightarrow{U}=-80\widehat{i}$ .

The x -component of vector $\overrightarrow{V}$ is $V_x=62.25$ .

The y-component of $V_y$ vector $\overrightarrow{V}$ is -65.0.

Formula used:

The magnitude of vector can be obtained by:

  $\left|\overrightarrow{V}\right|=\sqrt{\left(V_x{}^2\right)+{\left(V_y\right)}^2}$

And, the direction θ is defined as:

  $\theta ={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$

Calculation:

Consider the vector to be added: $\overrightarrow{U}$

Then,

  $\begin{array}{l}{V}_y+U_y=0\\ -65+U_y=0\\ U_y=65\end{array}$

And

  $\begin{array}{l}{V}_x+U_x=-80\\ 62.25+U_x=-80\\ U_x=-142.25\end{array}$

Substitute the values and solve to find the resultant of vector $\overrightarrow{U}$

  $\begin{array}{c}\left|\overrightarrow{U}\right|=\sqrt{\left(U_x{}^2\right)+{\left(U_y\right)}^2}\\ =\sqrt{{\left(-142.25\right)}^2+{\left(65\right)}^2}\\ =156.4\end{array}$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{U_y}{U_x}\right)\\ ={\tan }^{-1}\left(\frac{65}{-142.25}\right)\\ =-{24.6}^{\text{o}}\\ =180-24.6={155.5}^{\text{o}}\end{array}$

Conclusion:

Thus, a vector $\overrightarrow{U}$ can be added which has a magnitude of 156.4 and directed ${155.5}^{\text{o}}$ counter clockwise from positive x -axis.