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 = 0.205 and the coefficient of friction between the ladder and the wall is μ2 = 0.153. Determine the maximum angle with the vertical the ladder can make without falling on the ground.

Answer 1

Lte's start with taking sum of moments at the base of the ladder, ( M = mass of ladder, L = lenegth of ladder , = angle which ladder makes with the wall)

Mg ( L/2) cos - N2sin ( L) - F2cosL =0--------eq(1) ( where N2 is the horizontal force acting on the wall, F2 is the vertical forc arising out of friction between wall and ladder)

Let's apply conditions of vertical and horizontal equilrium

N2 = F1 ( where F1 is the frictional force between the base of ladder and ground )

Mg = N1 + F2 ( where N1 is the vertical normal reaction at the base of ladder )

F2 = 0.153 ( N2)

F1= 0.205 ( N1)

Mg = N1+ 0.153 ( N2)

Mg = F1/ 0.205 + 0.153 ( F1)

F1 = Mg/ 5.03

F2/ 0.153 = Mg/ 5.03

F2= Mg ( 0.0304) ---------eq(2)

N2 = F1=  Mg/ 5.03 -----eq(3)

PLUG eq(2) and eq(3) in eq(1)

Mg /2 cos - N2sin - F2cos =0

Mg/2 cos - Mg/ 5.03 (sin ) - Mg ( 0.0304) cos =0

cos / 2- sin/ 5.03 - cos (0.0304) =0

divide the equatoion by cos

0.5 - tan / 5.03 - (0.0304) =0

tan= 2.362

= tan^-1 ( 2.362) = 67 degree apprx