Hey AUDREY, good to see you again. a) Find the value of $\mu$, the coefficient of friction between the ladder and the floor. Step 1: Identify forces and set up equilibrium equations. Let $R_A$ be the normal reaction force at the floor (point A). Let $F_A$ be the friction force at the floor (point A). Let $R_B$ be the normal reaction force at the wall (point B). The weight of the ladder $W$ acts at its center, $a$ from A, as it is uniform and has length $2a$. The ladder makes an angle of $45^\circ$ with the horizontal. For vertical equilibrium (sum of vertical forces = 0): $$R_A - W = 0$$ $$R_A = W$$ For horizontal equilibrium (sum of horizontal forces = 0): $$F_A - R_B = 0$$ $$F_A = R_B$$ Step 2: Take moments about point A. Taking moments about point A eliminates $R_A$ and $F_A$ from the equation. The weight $W$ creates a clockwise moment. The perpendicular distance from A to the line of action of $W$ is $a \cos 45^\circ$. The normal force $R_B$ creates an anticlockwise moment. The perpendicular distance from A to the line of action of $R_B$ is $2a \sin 45^\circ$. Sum of moments about A = 0: $$R_B (2a \sin 45^\circ) - W (a \cos 45^\circ) = 0$$ Since $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$: $$R_B (2a) \frac{1}{\sqrt{2}} - W (a) \frac{1}{\sqrt{2}} = 0$$ $$2a R_B - Wa = 0$$ $$2a R_B = Wa$$ $$R_B = \frac{Wa}{2a}$$ $$R_B = \frac{W}{2}$$ Step 3: Calculate the coefficient of friction $\mu$. From horizontal equilibrium, $F_A = R_B = \frac{W}{2}$. For limiting equilibrium, the friction force is $F_A = \mu R_A$. Substitute the values of $F_A$ and $R_A$: $$\frac{W}{2} = \mu W$$ $$\mu = \frac{W}{2W}$$ $$\mu = \frac{1}{2}$$ The value of $\mu$ is $\boxed{\frac{1}{2}}$. b) Find how far up the ladder a boy of weight $2W$ can climb before the ladder begins to slip. Step 1: Identify new forces and conditions. Let the boy of weight $2W$ be at a distance $x$ from A along the ladder. The ladder is on the verge of slipping, so the friction force $F_A$ is at its maximum value, $F_A = \mu R_A$. We use $\mu = \frac{1}{2}$ from part (a). Step 2: Apply equilibrium conditions for forces. For vertical equilibrium: $$R_A - W - 2W = 0$$ $$R_A = 3W$$ For horizontal equilibrium: $$F_A - R_B = 0$$ $$F_A = R_B$$ Step 3: Calculate the maximum friction force and $R_B$. Since the ladder is about to slip, $F_A = \mu R_A$. $$F_A = \frac{1}{2} (3W)$$ $$F_A = \frac{3W}{2}$$ From horizontal equilibrium, $R_B = F_A = \frac{3W}{2}$. Step 4: Take moments about point A. The weight of the ladder $W$ creates a clockwise moment. Perpendicular distance is $a \cos 45^\circ$. The weight of the boy $2W$ creates a clockwise moment. Perpendicular distance is $x \cos 45^\circ$. The normal force $R_B$ creates an anticlockwise moment. Perpendicular distance is $2a \sin 45^\circ$. Sum of moments about A = 0: $$R_B (2a \sin 45^\circ) - W (a \cos 45^\circ) - 2W (x \cos 45^\circ) = 0$$ Since $\sin 45^\circ = \cos 45^\circ$: $$R_B (2a) - W(a) - 2W(x) = 0$$ Substitute $R_B = \frac{3W}{2}$: $$\frac{3W}{2} (2a) - Wa - 2Wx = 0$$ $$3Wa - Wa - 2Wx = 0$$ $$2Wa - 2Wx = 0$$ $$2Wa = 2Wx$$ $$x = a$$ The boy can climb a distance of $\boxed{a}$ from end A. Send me the next one 📸