The foot of a uniform ladder, of length 2a and weight W, rest on a rough horizontal ground and the top of the ladder rest against a smooth vertical wall. The ladder makes an angle of 60° with the horizontal. A man of weight 4W climbs up the ladder. Given that the ladder begins to slip when the man has climbed two third of the way up the ladder, find the coefficient of friction between the ladder and ground

Let the base of the ladder be point A (on the ground) and the top be point B (against the wall).

• Forces at A (ground):
  • Normal reaction force from the ground, N_A (vertically upwards).
  • Friction force from the ground, F_f (horizontally towards the wall, preventing slip).
• Forces at B (wall):
  • Normal reaction force from the wall, N_B (horizontally away from the wall, since the wall is smooth, there is no vertical friction).
• Weights:
  • Weight of the ladder, W, acting downwards at its center. The center is at a distance 'a' from point A.
  • Weight of the man, 4W, acting downwards at a distance (2/3) × 2a = 4a/3 from point A.
• The ladder makes an angle θ = 60° with the horizontal.

• Sum of vertical forces (ΣF_y = 0): The upward normal force from the ground must balance the downward weights. N_A - W - 4W = 0 N_A = 5W

• Sum of horizontal forces (ΣF_x = 0): The friction force at the ground must balance the normal force from the wall. F_f - N_B = 0 F_f = N_B

• Sum of moments about point A (Στ_A = 0): We choose point A as the pivot to eliminate the moments due to N_A and F_f, as they pass through A. Taking counter-clockwise moments as positive: The normal force from the wall (N_B) creates a counter-clockwise moment. Its lever arm is the vertical height of point B, which is 2a sin(60°). The weight of the ladder (W) creates a clockwise moment. Its lever arm is the horizontal distance from A to its line of action, which is a cos(60°). The weight of the man (4W) creates a clockwise moment. Its lever arm is the horizontal distance from A to its line of action, which is (4a/3) cos(60°).

  N_B (2a sin(60°)) - W (a cos(60°)) - 4W ((4a/3) cos(60°)) = 0

  Substitute sin(60°) = √(3)/2 and cos(60°) = 1/2:
  N_B (2a √(3)/2) - W (a/2) - 4W ((4a/3)/2) = 0
  N_B a √(3) - Wa/2 - 4W (4a/6) = 0
  N_B a √(3) - Wa/2 - 8Wa/3 = 0

  Divide the entire equation by 'a':
  N_B √(3) - W/2 - 8W/3 = 0

  Combine the W terms:
  W/2 + 8W/3 = (3W + 16W)/6 = 19W/6

  So, N_B √(3) = 19W/6
  N_B = (19W) / (6√(3))

When the ladder begins to slip, the friction force F_f reaches its maximum value, which is given by F_f = μ N_A, where μ is the coefficient of friction.

From our horizontal force equilibrium, we know F_f = N_B.
Therefore, N_B = μ N_A

Now, substitute the expressions we found for N_B and N_A:
(19W) / (6√(3)) = μ (5W)

Solve for μ:
μ = [(19W) / (6√(3))] / (5W)
μ = 19 / (6√(3) × 5)
μ = 19 / (30√(3))

To rationalize the denominator, multiply the numerator and denominator by √(3):
μ = (19 × √(3)) / (30√(3) × √(3))
μ = (19√(3)) / (30 × 3)
μ = (19√(3)) / 90