A tetrahedron has edges of length 4 cm. P, Q, R, S are vertices. Find the shortest distance from P to S passing over the midpoint of QR.

To find the shortest path from P to S passing over the edge QR at its midpoint T, we need to unfold the faces $\triangle PQR$ and $\triangle QRS$.
The net consists of two equilateral triangles, $\triangle PQR$ and $\triangle QRS$, sharing the common edge QR. All edges have a length of 4 cm.

1. Draw an equilateral triangle PQR with side length 4 cm.
2. Draw another equilateral triangle QRS, sharing the edge QR with $\triangle PQR$, such that vertex S is on the opposite side of QR from vertex P.
3. Mark the midpoint of the edge QR as T.
4. Draw a straight line segment connecting P to S. This line segment represents the shortest path and will pass through T.

In the net, the path PS is a straight line segment.
Consider $\triangle PQR$. It is an equilateral triangle with side length 4 cm.
The segment PT is the altitude from P to the midpoint T of QR.
The length of an altitude in an equilateral triangle with side $a$ is $\frac{\sqrt{3}}{2}a$.

$$PT = \frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3} \text{ cm}$$

Similarly, consider $\triangle QRS$. It is an equilateral triangle with side length 4 cm.
The segment TS is the altitude from S to the midpoint T of QR.

$$TS = \frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3} \text{ cm}$$

Since P, T, and S are collinear in the unfolded net, the total length of the path PS is the sum of the lengths PT and TS.

$$PS = PT + TS = 2\sqrt{3} + 2\sqrt{3} = 4\sqrt{3} \text{ cm}$$

To measure this length, we can approximate its value:

$$4\sqrt{3} \approx 4 \times 1.73205$$ $$4\sqrt{3} \approx 6.928 \text{ cm}$$

The length of the straight path PS is \boxed{\text{4\sqrt{3} cm}}.