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A ladder is positioned against a wall, it forms a $30^\circ$ angle with the floor. A person that was holding the ladder releases it. The ladder starts accelerating down and to the right. The wall is frictionless but the floor is not.

I did the $\mathrm{FBD}$:

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Where $F_g$ is the ladder's weight, $N_w$ the normal force from the wall, $N_f$ the normal force from the floor, and $F_f$ the frictional force exerted by the floor on the ladder.

The statement of the problem ranks the forces in the following way:

$$F_g>N_f>N_w>F_f$$

As the ladder is accelerating down and to the right, it is easy to demonstrate that $F_g>N_f$ and $N_w>F_f$ by Newton's Second Law.

I have doubts regarding how to complete the ranking, this is what I've tried:

Establishing the center of mass of the ladder as the axis of rotation, the ladder is rotating in the counterclockwise direction, which means the direction of Net Torque is counterclockwise.

$$\tau_{net}=\tau_{N_f}-\tau_{F_f}+\tau_{N_w}$$

Since it rotates in the counterclockwise direction.

$$\tau_{N_f}>\tau_{F_f}+\tau_{N_w}$$

$$\frac{L}{2}N_f \cos(\theta)>\frac{L}{2}F_f \sin(\theta)+\frac{L}{2}N_w \sin(\theta)$$

Where $\theta$ is the angle between the ladder and the floor.

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Well, considering vertical motion with acceleration $a_y$ (in +ve Y-direction) then $$\sum F_y=ma_y\iff N_f-F_g=-ma_y<0\iff F_g>N_f$$ Similarly, for horizontal motion towards right, $$\sum F_x=ma_x\iff N_w-F_f=ma_x>0\iff N_w>F_f$$ Considering rotation of ladder anticlockwise with angular acceleration $\alpha$ about C.M. in plane of paper $$\sum T_{net}=I\alpha $$$$N_f\cdot \frac{L}{2}\cos30^\circ-N_w\cdot \frac{L}{2}\sin30^\circ-F_f\cdot \frac{L}{2}\sin30^\circ=I\alpha>0\iff N_f>N_w$$ $$\therefore F_g>N_f>N_w>F_f$$

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