Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. 3-35, points A and B coincided before the rock in the foreground slid down to the right. The net displacement $\overrightarrow{\mathbf{A}\mathbf{B}}$is along the plane of the fault.The horizontal component of is the strike-slip AC. The component $\overrightarrow{\mathbf{A}\mathbf{B}}$of that is directed down the plane of the fault is the dip-slip AD.(a) What is the magnitude of the net displacement$\overrightarrow{\mathbf{A}\mathbf{B}}$if the strike-slip is 22.0 mand the dip-slip is 17.0 m? (b) If the plane of the fault is inclined at angle$\mathit{\varphi }\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{0}\mathbf{°}$to the horizontal, what is the vertical component of$\overrightarrow{\mathbf{A}\mathbf{B}}$?

Strike-slip $AC=22.0m$and dip-slip$AD=17.0\mathrm{m}$

The inclination angle of the fault plane to the horizontal is$\varphi =52°$

The vector addition theorem can be used to find the net displacement and its vertical component can be found by using trigonometry and the information given in the figure.

Formula:

$\overrightarrow{AD}=\overrightarrow{AD}+\overrightarrow{AD}\dots \dots \dots \left(\mathrm{i}\right)$

$s\mathrm{in}\varphi =\frac{\mathrm{opposite}\mathrm{side}}{\mathrm{hypotenuse}}\dots \dots \dots .\left(\mathrm{ii}\right)$

$\overrightarrow{AB}$and$\overrightarrow{AD}$ are the horizontal and vertical component of$\overrightarrow{AB}$

According to the triangle vector addition law,

$\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{AD}$

$$\begin{gathered} A{B}^2=A{C}^2+A{D}^2 \\ AB=\sqrt{A{C}^2+A{D}^2} \\ =\sqrt{{22}^3+{17}^2} \\ =27.8\mathrm{m} \end{gathered}$$

Hence, the magnitude of net displacement is 27.8 m

Now the fault is inclined at an angle$\varphi$ to the horizontal, hence, the vertical component of$\overrightarrow{AB}$ is the opposite side of that angle$\varphi$

The vertical component of $\overrightarrow{AB}$,

$$\begin{gathered} A{B}_v=AD\sin \varphi \\ =17\sin 52 \\ =13.4\mathrm{m} \end{gathered}$$

Therefore, the vertical component of$\overrightarrow{AB}=13.4\mathrm{m}$ .