Question

A 3m long ladder weight 20 kg leans on a frictionless wall . its feet rest on the floor 1 m from the wall . show in fig . find. the reaction force of the wall and the floor ​

Answer 1

The ladder AB is 3m long, its foot A is at distance AC=1m from the wall. From Pythagons theorem, BC=22m. The forces on the ladder are its weight Wacting at its centre of gravity D, reaction forces F1 and F2 of the wall and the floor respectively. Force F1 is perpendicular to the wall, since the wall is frictionless. Force F2 is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.

For translational equilibrium, taking the forces in the vertical direction,

N−W=0                                       (i)

Taking the forces in the horizontal direction,

F−F1=0                                    (ii)

For rotational equilibrium, taking the moments of the forces about A, 

22F1−(1/2)W=0                                          (iii)

Now W=20g=20×9.8N=196.0N

From (i)N=196.0

From (iii)F1=W/42=196.0/42=34.6N

From (ii)F=F1=34.6N

F2=F2+N2=199.0N

The force F2 makes an angle α with the horizontal,

tanα=N/F=42,α=tan−1(42)=80o