A ladder weighing W rests with one end against arough vertical wall and the other end on a rough floor. It is inclined at an angle of 45^(@) with the horizontal. Coefficient of friction of the ladder with the floor and the wall are mu and mu' respectively. Show that the minimum horizontal force that can move the base of the ladder towards the wall is (W(1+2mu-mumu'))/(2(1-mu')).

A ladder weighing W rests with one end against arough vertical wall an

1.	A ladder weighing W rests with one end against arough vertical wall and the other end on a rough floor. It is inclined at an angle of 45^(@) with the horizontal. Coefficient of friction of the ladder with the floor and the wall are mu and mu' respectively. Show that the minimum horizontal force that can move the base of the ladder towards the wall is (W(1+2mu-mumu'))/(2(1-mu')).
Answer» Solution :A horizontal force F is applied at point B of the ladder AB Fig. such that the ladder is at the limiting stage of equilibrium. The figure shows the FORCES ACTING at this stage. At equilibrium the horizontal components give, F= R.+ `mu R ""cdots (1)` and the vertical components give, W +`mu.R.=R "" cdots (2)` `because ANGLEABO= 45^(@)`, we have OA =OB=2L (say). At equilibrium the moments of the forces about B give, WL +`mu.R..2L=R..2L` :. W =2R.(1-`mu`) or, `R. = (W)/(2(1-mu.))""cdots(3)` `:.` From (2) we get, R =W `[1+(mu.)/(2(1-mu.))] ""cdots(3)` and from (1) we get, `F=W [(1)/(2(1-mu.))+mu+(mu.mu)/(2(1-mu.))]` =`W((1+2mu-mumu.))/(2(1-mu.))`