A `3m` along ladder weighing `20 kg` leans on a frictionless wall. Its feet rest on the floor `1 m` from the wall. Find the reaction forces of the wall and the floor.

Options

A. 25`sqrt 2` N, 203 N
B. 50`sqrt 2` N, 230 N
C. 203 N, 25`sqrt 2` N
D. `230 N, 50sqrt 2` N

Answer

A

Explanation

Let AB be ladder.
`therefore` AB=3m
Its foot A is at distance 1m
from the wall,
`therefore` AC=1 and
BC `=sqrt((AB)^(2)-(AC)^(2))`
`=sqrt((3)^(2)-(1)^(2))=2sqrt(2)`m
The various forces acting on the ladder are
i) Weight W acting at its center of gravity G.
ii) Since the wall is fictionless, reaction force `R_(1)` of the wall acting perpendicular to the wall.
iii) Reaction force `R_(2)` of the floor. This force can be resolved into two components, the normal reaction N and the force of friction f.
For translatory equilibrium in the horizontal direction,
`f-R_(1)=0" or" f=R_(1)`
For translatory equilibrium in the vertical direction,
`N=W=20g = 20 xx 10= 200N`...................(ii)
For rotational equilibrium, taking moment of the forces about A, we get
`R_(1)(2sqrt(2))-W(1/2)=0`
`R_(1)=W/(4sqrt(2))= 200/(4sqrt(2))= 25 sqrt(2)N`..............(iii)
`R_(2)=sqrt((N^(2)+f^(2)) = sqrt((200N)^(2)+(25sqrt(2)N)^(2)))=203N`