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:


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.


Since it rotates in the counterclockwise direction.


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

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


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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