Question

Three displacement vectors of a croquet ball are shown in Figure P3.38, where $\mathbf{\left|}\overrightarrow{\mathbf{A}}\mathbf{\right|}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathit{u}\mathit{n}\mathit{i}\mathit{t}\mathit{s}\mathbf{,}\mathbf{}\mathbf{\left|}\overrightarrow{\mathbf{B}}\mathbf{\right|}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathit{u}\mathit{n}\mathit{i}\mathit{t}\mathit{s}\mathbf{,}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{\left|}\overrightarrow{\mathbf{C}}\mathbf{\right|}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathit{u}\mathit{n}\mathit{i}\mathit{t}\mathit{s}\mathbf{.}$Find

(a) the resultant in unit vector notation and (b) the magnitude and direction of the resultant displacement.

Short answer

Nothing

Step-by-step solution

Definition of vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

Understanding the figure

Solving for part (a)

First, we find the resultant vector in x and y diection

$\begin{array}{rcl}{R}_x& =& B\cos {45}^o+C\cos {45}^o\\ \square & =& 40\cos \cos {45}^o+30\cos {45}^o\\ \square & =& 49.5\\ R_x& =& A+B\sin {45}^o-C\cos {45}^o\\ \square & =& 20+40\sin {45}^o-30\cos {45}^o\\ \square & =& 27.1\end{array}$

Then, we write the resultant vector in unit-vector notation:

$$\begin{gathered} \overrightarrow{R}=R_{x}i+R_{y}j \\ \square =49.5i+27.1j \end{gathered}$$

Solving for part (b)

The magnitde of $\overrightarrow{R}$is found by:

$$\begin{gathered} \left|R\right|=\sqrt{R_x^2+R_y^2} \\ \square =\sqrt{49.5^2+27.1^2} \\ \square =56.4 \end{gathered}$$

The direction of $\vec{R}$ is found by:

$$\begin{gathered} \theta ={\tan }^{-1}\frac{R_y}{R_x} \\ \square ={\tan }^{-1}\frac{27.1}{49.5} \\ \square =28.7^o \end{gathered}$$

Final solution

$$\begin{gathered} \left(a\right)\overrightarrow{R}=49.5i+27.1j \\ \left(b\right)\left|R\right|=56.4at\theta =28.7^o \end{gathered}$$