Question

A ladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. The coefficient of friction between the ladder and the horizontal surface is μ1 = 0.265 and the coefficient of friction between the ladder and the wall is μ2 = 0.253. Determine the maximum angle with the vertical the ladder can make without falling on the ground.

Answer 1

Answer:

$\alpha =29.60$

Explanation:

Let the normal force of the wall on the ladder be N2 and the normal force of the ground on the ladder be N1.

Horizontal forces:

$N_{2}= (u1)(N_{1})$ [1]

Vertical forces:

$N_1 + (u2)(N_{2})=m*g$ [2]

Substitute [2] into [1]:

$N_2 = (u1)*[m*g - (u2)(N_2)]$

$N_2= \frac{(u1)m*g}{[1 + (u1)(u2)]}$ [3]

Torques about the point where the ladder meets the ground:

$m*g(\frac{L}{2})sin\alpha= (N_2)(L)cos\alpha+(u2)(N_2)(L)sin\alpha$

$(\frac{1}{2})*m*g=(N_2)*cot\alpha+(u2)(N_2)$

$[1 + (u1)(u2) - 2(u2)(u1)]/2 [1 + (u1)(u2)]= [(u1)/[1 + (u1)(u2)]]cot\alpha$

tanα = $\frac{2*(u1)}{1-u1*u2}$

$\alpha =tan^{-1}*(\frac{2*0.265}{1-0.265*0.253})$

$\alpha =tan^{-1}*(0.568)$

$\alpha =29.60$