Unexpected different results from Newton's second law

Question

Please don't ban me. I read through Homework-like questions and I know they should ask about a specific physics concept and show some effort to work through the problem. I hope the question is ok.

I recently came across a mechanic basic problem which I wanted to solve following two different approaches. The odd thing is that the approaches bring different results, so probably there might be some wrong reasoning which I cannot explain. Can anyone explain why results are different?

Here's the figure of the system:

The question is about getting the friction static coefficient between the plane and the bar.

We know the bar weight $P=980N$, and the angle $\theta=60°$, the system is in equilibrium, and the static friction is at its maximum value.

Solution 1 Definition of static friction: $f_{max} =\mu _{s} N\ \rightarrow \ \mu _{s} =\frac{f}{N}$

Newton's second law, forces perpendicular to the plane $y:\ N=P\cdot \cos 60°$

Newton's second law, torques relative to the right edge of the bar: $f\cdot L\sin 60°=P\frac{L}{2} \ \rightarrow \ f=\frac{P}{2\sin 60°}$

Static friction coeff: $\boxed{\mu _{s} =\frac{1}{2\sin 60° \cos 60°}=1.2}$

Solution 2 Definition of static friction: $f_{max} =\mu _{s} N\ \rightarrow \ \mu _{s} =\frac{f}{N}$

Newton's second law, forces perpendicular to the plane $y:\ N=P\cdot \cos 60°$

Newton's second law, torques relative to the left edge of the bar: $T\cdot L\sin 60°=P\cdot \frac{L}{2}\rightarrow T=\frac{P}{2\sin 60°}$

Newton's second law, forces parallel to the plane: $x:\ f+T=P\sin 60°\rightarrow f=P\sin 60°-T=P\sin60°-\frac{P}{2\sin 60°}$

Static friction coeff: $\boxed{\mu _{s} =\frac{\sin 60° -\frac{1}{2\sin 60°}}{\cos 60°}=0.6}$

The solutions for the static friction coeff. are different, in particular solution two is $\frac{1}{2}$ solution 1!!

Answer 1

The first solution is wrong. When you calculate the torques relative of the right edge of the bar you forgot about the torque of force $N$. The correct equation should be $$ f\cdot L \sin\theta + N \cdot L\cos\theta = P \frac{L}{2} $$ and then you'd get the result that agrees with the second solution.

Answer 2

In the section "torques relative to the right edge of the bar:"

The torque from the $N$ should be included...best of luck getting the two ways to match.