This problem involves a uniform ladder $AB$ with mass $M = 30 \, \text{kg}$ and length $L = 4 \, \text{m}$, leaning against a frictionless wall at an angle $\alpha$ with the floor, which has friction. We need to derive the formula for the minimum coefficient of static friction $\mu_s$ required between the ladder and the floor to keep the ladder in equilibrium.

Solution Outline and Calculation

1. Free-Body Diagram and Forces:

   • Weight of the ladder ($W = Mg = 30 \cdot 9.8 = 294 \, \text{N}$), acting downward at the ladder's center of gravity, located at $L/2$ from point $A$.
   • Normal force from the wall ($F_w$) acting horizontally at point $A$, since the wall is frictionless.
   • Normal force from the floor ($N$) acting vertically upward at point $B$.
   • Frictional force at the floor ($f = \mu_s N$) acting horizontally at point $B$.

2. Conditions for Equilibrium:

   • Horizontal equilibrium: The horizontal components must balance. $F_w = f = \mu_s N$
   • Vertical equilibrium: The vertical components must balance. $N = Mg$

3. Torque Equilibrium: Taking torques around point $B$ (where the ladder touches the floor) eliminates $f$ and $N$, simplifying calculations:

   • Torque due to $Mg$: $\tau_{Mg} = Mg \cdot \frac{L}{2} \cos \alpha$
   • Torque due to $F_w$: $\tau_{F_w} = F_w \cdot L \sin \alpha$

   Setting the torques equal for rotational equilibrium: $Mg \cdot \frac{L}{2} \cos \alpha = F_w \cdot L \sin \alpha$ Simplifying, we get: $F_w = \frac{Mg}{2} \cot \alpha$

4. Substitute $F_w = \mu_s N$: Since $F_w = \mu_s N$ and $N = Mg$, substitute these values: $\mu_s Mg = \frac{Mg}{2} \cot \alpha$ Dividing both sides by $Mg$, we get: $\mu_s = \frac{1}{2} \cot \alpha$

Final Answer:

The minimum coefficient of static friction $\mu_s$ required is: $\mu_s = \frac{1}{2} \cot \alpha$

This formula gives the minimum friction needed between the ladder and the floor as a function of the angle $\alpha$ to prevent slipping.

Would you like more details on each step or have any questions?

Additional Questions:

1. How would the minimum coefficient of friction change if the ladder's length or mass changed?
2. What happens to $\mu_s$ if the angle $\alpha$ becomes very small or very large?
3. How would the solution change if the wall also had friction?
4. Can you calculate $\mu_s$ for specific values of $\alpha$, for example, $\alpha = 45^\circ$?
5. What are the implications of a frictionless wall on the equilibrium conditions?

Tip:

In problems involving static equilibrium, taking torque around a point where multiple unknown forces act can simplify calculations by eliminating those forces from the torque equation.