Answer:
[tex]\alpha =29.60[/tex]
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:
[tex]N_{2}= (u1)(N_{1})[/tex] [1]
Vertical forces:
[tex]N_1 + (u2)(N_{2})=m*g[/tex] [2]
Substitute [2] into [1]:
[tex]N_2 = (u1)*[m*g - (u2)(N_2)][/tex]
[tex]N_2= \frac{(u1)m*g}{[1 + (u1)(u2)]}[/tex] [3]
Torques about the point where the ladder meets the ground:
[tex]m*g(\frac{L}{2})sin\alpha= (N_2)(L)cos\alpha+(u2)(N_2)(L)sin\alpha[/tex]
[tex](\frac{1}{2})*m*g=(N_2)*cot\alpha+(u2)(N_2)[/tex]
[tex][1 + (u1)(u2) - 2(u2)(u1)]/2 [1 + (u1)(u2)]= [(u1)/[1 + (u1)(u2)]]cot\alpha[/tex]
tanα = [tex]\frac{2*(u1)}{1-u1*u2}[/tex]
[tex]\alpha =tan^{-1}*(\frac{2*0.265}{1-0.265*0.253})[/tex]
[tex]\alpha =tan^{-1}*(0.568)[/tex]
[tex]\alpha =29.60[/tex]
