How to solve this problem while taking torques about the lower end of the ladder?

This is a problem from my introductory physics textbook:

The ladder shown in the figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder at the middle. The angle between two legs is $$60^{\circ}$$. The fat person sitting on the ladder has a mass of $$80$$ kg. Find the tension in the crossbar.

This is the solution from the book:

Since the ladder is in vertical equilibrium, $$2\text{ N}=800N\rightarrow \text{N} = 400N$$ Next consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, $$\text{N} (2\text{ m}) \tan 30^{\circ}=\text{T}(1 \text{ m})$$ This gives, $$\text{ T} = (2 / \sqrt {3}) \times 400 = 460N$$

I would like to solve the latter half of the problem (using the fact that the left half of the ladder is in rotational equilibrium) by taking torques about the lower end of the left leg of the ladder. I assumed that the weight of the man would be evenly distributed between both legs, leading to a weight that is numerically equal to the normal force. However, the torques from the weight and the tension force line up in the same direction (into the plane of your screen) and do not cancel.

What is wrong with my thinking here?

There's a missing force you didn't consider. The 2 arms of the ladders exert horizontal forces on each other at their point of contact.

• Yes, thanks. Since that is numerically equal to the tension force for horizontal equilibrium and in the opposite direction, the torques finally cancel. Nov 15, 2020 at 6:34

. I assumed that the weight of the man would be evenly distributed between both legs, leading to a weight that is numerically equal to the normal force.

So far so good

However, the torques from the weight and the tension force line up in the same direction (into the plane of your screen) and do not cancel

This doesn't sound right. Remember that torque is perpendicular distance to force from point of reference times the force and not directly the distance.

If you consider rotational equilibrium about this point you get that normal must be half of the weight. Consider the torque balance,

$$\vec{\tau} = \sum r_{\perp} F= (1) T - (1) T+ xN- \frac{x}{2} W$$ With $$x$$ is the distance between legs, Solve considering $$\vec{\tau} = 0$$

You get,

$$2N=W$$

The key point here is that the torque about any point should give same since sum of net force is zero (Proof). There is a person called John Alexiou who has written really detailed answers about torque many times, I suggest you start reading those starting from (here)