A 4m long ladder weighing 28kg leans on a frictionless wall its feet rests on the floor 1m away from the wall. Find the reaction forces of the wall and the floor.
Answers
Answer:
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Explanation:
The ladder AB is 4m long, its foot A is at distance AC=1m from the wall. From Pythagons theorem, BC=2
2
m. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces F
1
and F
2
of the wall and the floor respectively. Force F
1
is perpendicular to the wall, since the wall is frictionless. Force F
2
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−F
1
=0 (ii)
For rotational equilibrium, taking the moments of the forces about A,
2
2
F
1
−(1/2)W=0 (iii)
Now W=20g=20×9.8N=196.0N
From (i)N=196.0
From (iii)F
1
=W/4
2
=196.0/4
2
=34.6N
From (ii)F=F
1
=34.6N
F
2
=
F
2
+N
2
=199.0N
The force F
2
makes an angle α with the horizontal,
tanα=N/F=4
2
,α=tan
−1
(4
2
)=80
o
solution
