ILLUSTRATIVE EXAMPLES 



69 



Let e be this angle of friction, and let a be the inclination of ladder to the 

 horizontal. Then, from the geometry of the triangle ACP, we have 



AC _ AP 

 sine 



and since APB is a right angle, 



AP - AB cos 



Thus 



lit \ 



( e a]- 



\2 / 



A C = AP sin e sec a = AB sin e sin (e + a) sec a. 



Thus slipping will begin as soon as the man has climbed a height equal to 

 sin e sin (e + a) sec a times the whole height. 



The condition that the man can reach the top without slipping is that 

 sin e sin (e + a) sec a shall be greater than unity, or that 



sin e sin (e + a) > cos [(e + a) e] 



> sin e sin (e + a) + cos e cos (e + ) 



Thus for the condition to be satisfied cos e cos (e + a) must be negative ; i.e. 

 e + a must be greater than 90. Thus the angle between the ladder and the 

 vertical must be less than the angle of friction. This is also clear from the 

 figure, for when the man reaches B, two of the forces, namely the reaction at B 

 and the weight of the man, both pass through B, so that the third force must 

 also pass through B ; i.e. the reaction at A must have AB for its line of action, 

 and if the ladder just slips here, the angle between AB and the vertical 

 must be e. 



4. I/, in the last problem, the man has 

 ascended to some point C without the ladder 

 slipping, what are the reactions at A and B ? 



Here it is not known what angles the 

 reactions make with the normals : all that 

 is known is that these angles are less than 

 the angle of friction. 



Let us resolve the reaction at A into two 

 components NI, FI, and that at B into two 

 components N%, F%, these being horizontal 

 and vertical as in the figure. Then the 

 forces acting on the system composed of 

 the man and ladder are the five forces 

 NI, FI, N 2 , F 2 , and W. 



Resolving vertically, W NI Fz = Q. 



Resolving horizontally, FI N% = 0. 



Taking moments about A, 



W- AC cos a - F 2 AB cos a - N 2 AB sin a = 0. 



(a) 



