Taking torques about the contact point on the floor. The angle with the wall is \(60°\), so angle with floor is \(30°\). Normal from wall (horizontal) \(\times L\sin30° = mg\times\frac{L}{2}\cos30°\). Wait — using torque about foot: \(N_w\cdot L\sin30° = mg\cdot\frac{L}{2}\cos30°\Rightarrow N_w = \frac{mg\cos30°}{2\sin30°} = \frac{20\times10\times\frac{\sqrt{3}}{2}}{2\times\frac{1}{2}}=100\sqrt{3}\). Friction \(= N_w = 100\sqrt{3}\approx173\ \text{N}\). Answer: \(100\sqrt{3}\ \text{N}\).