Question

a uniform ladder 3m long weights 200N.it is placed against a wall making an angle of 60° with the floor as shown in figure
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Answer 1

Answer:

The ladder AB is 3m 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

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