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: Recall the geometry of an oblique-slip fault.
On a fault plane, the net slip vector is the actual displacement of one block relative to the other. The pitch (or rake) of this vector is the angle it makes with the strike of the fault, measured within the fault plane. The net slip can be split into a strike-slip part, lying along the strike, and a dip-slip part, lying along the dip direction.

Step 2: Split the net slip using the pitch.
For a net slip of magnitude \(N\) and pitch \(\alpha\),
\[ \text{strike-slip component} = N\cos\alpha \]
\[ \text{dip-slip component} = N\sin\alpha \]
With \(N = 100\) m and \(\alpha = 30^{\circ}\),
\[ \text{dip-slip component} = 100 \times \sin 30^{\circ} = 100 \times 0.5 = 50\ \text{m} \]

Step 3: Convert the dip-slip component to heave.
The heave of a fault is the horizontal distance covered by the dip-slip movement, measured perpendicular to the strike. If \(\delta\) is the dip of the fault, the dip-slip displacement resolves into a horizontal part (heave) and a vertical part (throw) as
\[ \text{heave} = (\text{dip-slip component})\cos\delta \]
\[ \text{throw} = (\text{dip-slip component})\sin\delta \]
Here the fault dips at \(\delta = 60^{\circ}\), so
\[ \text{heave} = 50 \times \cos 60^{\circ} = 50 \times 0.5 = 25\ \text{m} \]

Step 4: Final answer.
The heave of this oblique-slip normal fault is 25 m.
\[ \boxed{25} \]