Chapter 6, Problem 14CT

To determine

To find: The direction and the magnitude of the resultant of these forces.

Answer to Problem 14CT

The magnitude of the resultant force is $14.75°$ and the angle it makes with the positive $x$ axis.

Explanation of Solution

Given:

The magnitude of the forces are $250\text{pounds}$ and $130\text{pounds}$ , the angle at which these two forces act are $45°$ and $-60°$ respectively with the positive $x$ axis.

Calculation:

Consider the force $u=250$ acts at $45°$ and the force $v=-130$ acts at . $-60°$ .

The component form of $u$ is,

  $\begin{array}{c}u=250\cos 45°i+250\sin 45°j\\ =176.77i+176.77j\\ =176\left(i+j\right)\end{array}$

The component form of the vector $v$ is,

  $\begin{array}{c}v=130\cos \left(-60\right)°i+130\sin \left(-60\right)j\\ =65i-113.1j\end{array}$

Consider the vector sum of the resultant force is,

  $\begin{array}{c}u+v=176.77i+176.77j+65i-113.1j\\ =241.77i+63.67j\end{array}$

Then, the magnitude of the resultant force is,

  $\begin{array}{c}\left|u+v\right|=\sqrt{{\left(241.77\right)}^2+{\left(241.77\right)}^2}\\ =250.013\end{array}$

The direction of the angle of force is,

  $\begin{array}{l}\tan \theta =\frac{63.67}{241.77}\\ \arctan \theta =\arctan \left(\frac{63.67}{241.77}\right)\\ \theta =14.75°\end{array}$