Problem 1 - Entrance Test

A uniform ladder of mass M and length L leans against a smooth vertical wall. The base of the ladder rests on a rough horizontal floor. The ladder makes an angle θ with the horizontal. A person of mass m stands at the very top of the ladder (at a distance L from the base). If the coefficient of static friction between the ladder and the floor is μs, what is the minimum angle θ (with respect to the horizontal) at which the ladder can stand without slipping?

A. A) tan θ = (M + 2m) / (2μs M)B. B) tan θ = (M + m) / (2μs (M + 2m))C. C) tan θ = (M + m) / (2μs (M + m))D. D) tan θ = (M + 2m) / (2μs (M + m))E. E) tan θ = (M + 2m) / (μs (M + m))

Correct: D

Let's analyze the forces acting on the ladder: 1. Weight of the ladder (Mg) acting downwards at its center of mass (L/2 from the base). 2. Weight of the person (mg) acting downwards at the top (L from the base). 3. Normal force from the floor (N_f) acting upwards at the base. 4. Static friction force from the floor (f_s) acting horizontally towards the wall at the base. 5. Normal force from the wall (N_w) acting horizontally away from the wall at the top. For equilibrium: Sum of forces in the x-direction: ΣFx = 0 => f_s - N_w = 0 => f_s = N_w (Eq. 1) Sum of forces in the y-direction: ΣFy = 0 => N_f - Mg - mg = 0 => N_f = (M + m)g (Eq. 2) For the ladder to be on the verge of slipping, the static friction reaches its maximum value: f_s_max = μs * N_f. So, f_s = μs * (M + m)g (Eq. 3) From (Eq. 1) and (Eq. 3): N_w = μs * (M + m)g (Eq. 4) Now, sum of torques about a convenient pivot point. Let's choose the base of the ladder (point of contact with the floor) to eliminate N_f and f_s from the torque equation. Στ = 0 (clockwise torques = counter-clockwise torques) Clockwise torques: - Due to ladder's weight: Mg * (L/2) * cos θ - Due to person's weight: mg * L * cos θ Total clockwise torque = (Mg * L/2 + mg * L) * cos θ = (M/2 + m)gL cos θ Counter-clockwise torques: - Due to wall's normal force: N_w * L * sin θ Equating torques: (M/2 + m)gL cos θ = N_w * L * sin θ (M/2 + m)g cos θ = N_w * sin θ Substitute N_w from (Eq. 4): (M/2 + m)g cos θ = [μs * (M + m)g] * sin θ Divide both sides by g and cos θ: (M/2 + m) = μs * (M + m) * (sin θ / cos θ) (M/2 + m) = μs * (M + m) * tan θ tan θ = (M/2 + m) / (μs * (M + m)) tan θ = ( (M + 2m) / 2 ) / (μs * (M + m)) tan θ = (M + 2m) / (2μs (M + m)) The correct choice is D.