Question

A uniform ladder 4 m long weighs 200 N. It is placed against a wall making an angle of 60° with the floor. The coefficient of friction between the wall and the ladder is 0.25 and that between the ground and the ladder is 0.35. The ladder in addition to its own weight, has to support a man of 1000 N at the top at B. Calculate: (i) The horizontal force P to be applied to the ladder at the ground level to prevent slipping. (ii) If the force P is not applied, what should be the minimum inclination of the ladder with the horizontal, so that it does not slip with the man at the top?

Answer 1

Given : length of ladder = 4m

            Weight of ladder = 200 N

           ∠ made by ladder against the wall = 60°

To Find : (i) The horizontal force P to be applied to the ladder at the ground level to prevent slipping.

               (ii) If the force P is not applied, what should be the minimum inclination of the ladder with the horizontal, so that it does not slip with the man at the top

Step-by-step explanation:

i) The horizontal force P to be applied to the ladder at the ground level to prevent slipping:

Resolving forces horizontally,

     $P + F_{8} = N_{w}$

Resolving forces vertically,

$N_{g} + F_{w} = 200 + 1000$

Taking moments about A,

(1000 x 4cos 60°) + (200 x 2cos 60°)

($F_{w}$ x 4cos 60°) + ($N_{w}$ x 4sin 60°)

2200 = 0.25 x $N_{w}$ x 4cos 60° + $N_{w}$ x 4sin 60°

 $N_{w}$= 554.98 N

$F_{w}$ = 0.25 x 554.98

      = 138.75 N

$N_{g}$ = 1061.25 N

$F_{g}$ = 0.35 x 1061.25

    = 371.44 N

P = 554.98 - 371.44

  = 183.54 N

ii) Inclination of the ladder with the horizontal

Consider, the ladder inclined at an angle of theta without any horizontal force acting at the ground level.

Resolving horizontally,

$N_{w}$ = $F_{g}$ = ∪$_{g}$$N_{g}$ = 0.35$N_{g}$