Question

An oblique-slip normal fault dipping \(60^{\circ}\)E shows a net slip of 100 m. If the net slip vector has a pitch of \(30^{\circ}\), then the heave of the fault is m (answer in integer).

Hint:

Split the net slip into dip-slip and strike-slip using the pitch angle first, then project the dip-slip component onto the horizontal using the dip angle to get the heave.

Answer 1

Correct Answer: 25

Step 1: Break the slip into two directions.
On the fault surface the net slip $N = 100$ m does not point purely along strike or purely along dip. Its pitch of $30^{\circ}$ tells us how far it leans away from the strike direction, toward the dip direction. Along-strike and along-dip components of this slip are
\[ N\cos(\text{pitch}) \quad \text{and} \quad N\sin(\text{pitch}) \]

Step 2: Find the along-dip part first.
\[ N\sin 30^{\circ} = 100 \times 0.5 = 50\ \text{m} \]
This 50 m is how far the hanging wall has moved measured straight down the dip of the fault surface.

Step 3: Project the dip-slip onto the horizontal.
Heave is defined as the horizontal separation caused by the fault, at right angles to the strike. Since the fault plane itself dips at $60^{\circ}$ from horizontal, the along-dip displacement of 50 m has a horizontal shadow of
\[ \text{heave} = 50\cos 60^{\circ} \]
because $\cos(\text{dip})$ gives the horizontal share of a displacement measured along the dipping plane.

Step 4: Work out the number.
\[ \text{heave} = 50 \times 0.5 = 25\ \text{m} \]

Step 5: State the result.
The fault's heave comes out to 25 m.
\[ \boxed{25} \]