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

`\alpha =29.60`

Step-by-step 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= ((u1)m*g)/([1 + (u1)(u2)])` [3]

Torques about the point where the ladder meets the ground:

`m*g((L)/(2))sin\alpha= (N_2)(L)cos\alpha+(u2)(N_2)(L)sin\alpha`

`((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 =
`(2*(u1))/(1-u1*u2)`

`\alpha =tan^(-1)*((2*0.265)/(1-0.265*0.253))`

`\alpha =tan^(-1)*(0.568)`

`\alpha =29.60`