# Torque and dynamics of a slipping ladder [closed]

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 $$\mathrm{FBD}$$:

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.

• since the center of mass as well as falling is moving towards the right, then $N_w > F_f.$ – Nelson Vanegas A. May 24 '20 at 22:05
• – John Alexiou May 25 '20 at 1:20

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 $$\sum T_{net}=I\alpha$$$$N_f\cdot \frac{L}{2}\cos30^\circ-N_w\cdot \frac{L}{2}\sin30^\circ-F_f\cdot \frac{L}{2}\sin30^\circ=I\alpha>0\iff N_f>N_w$$ $$\therefore F_g>N_f>N_w>F_f$$