To determine how far the man can walk from the left end before the plank starts to tip, we analyze the torque equilibrium around the right support. The plank begins to tip when the normal force at the left support becomes zero.

1. **Given**:
– Man’s mass: \(80 \, \text{kg}\)
– Plank’s mass: \(10 \, \text{kg}\)
– Plank length: \(5 \, \text{m}\)
– Gravitational acceleration: \(g = 9.8 \, \text{m/s}^2\) (cancels out in equilibrium equations)

2. **Torque Balance**:
– The tipping condition occurs when the torque from the man’s weight equals the torque from the plank’s weight around the right support.
– Let \(x\) be the distance from the left end where the man stands. His distance from the right support is \(5 – x \, \text{m}\).
– The plank’s center of mass is \(2.5 \, \text{m}\) from the right support.

3. **Equation**:
\[
\text{Torque}_{\text{man}} = \text{Torque}_{\text{plank}}
\]
\[
80 \cdot (5 – x) = 10 \cdot 2.5
\]
Solving for \(x\):
\[
400 – 80x = 25 \implies 80x = 375 \implies x = \frac{375}{80} = 4.6875 \, \text{m}
\]

**Conclusion**:
The man can walk up to \(4.6875 \, \text{m}\) from the left end before the plank starts to tip. This position is \(0.3125 \, \text{m}\) from the right support, ensuring equilibrium is maintained until this point.

**Answer**: \(\boxed{4.6875 \, \text{meters}}\)