Question

A uniform ladder 3 m long weighing 20 kg leans against a frictionless wall. its foot rest on a rough floor 1m from the wall. the reaction forces of the wall and floor are

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

o

Answer 2

Answer:

Required force is 196 N.

Explanation:

The ladder AB is 3 m long,its foot A is at distance AC= 1 m from the wall.

By Pythagoras theorem, BC=$2 \sqrt{2} \: m$

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 respectively.Force$F_1$is perpendicular to the wall is frictionless.Force

$F_2$is resolved into two components, The normal reaction N and force of friction F.

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

$N-W=0......(1)$Taking the forces in the horizontal direction,$F-F_1=0.......(2)$

For rotational equilibrium, taking the moments of the forces about A,$2 \sqrt{2} F_1 - \frac{1}{2}W =0.....(3)$

Now,$W=20 g=20 \times 9.8 N=196 N$

From equation (1),

$N=196$

$F_2= \sqrt{ {F}^{2} + {N}^{2} }$=199.0 N,The force$F_2$makes an angle $\alpha$with the horizontal,

$\tan( \alpha ) =$N/F=$4 \sqrt{2}$

So,

$\alpha = {tan}^{ - 1} (4 \sqrt{2} ) = {80}^{o}$