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

• A. $25\sqrt{2}N,203N$
• B. $50\sqrt{2}N,203N$
• C. $203N,25\sqrt{2}N$
• D. $203N,50\sqrt{2}N$

Answer 1

Answer: (A)

$25\sqrt{2}N,203N$

Answer 2

Let AB be ladder

$\therefore$AB = 3m

Its foot A is a distance 1m from the wall.

$\therefore$AC = 1m

And $BC = \sqrt{(AB)^{2} - (AC)^{2}} = \sqrt{(3)^{2} - (1)^{2}} = 2\sqrt{2}m$

The various force acting on the ladder are

(i) Weight W acting at its center of gravity G.

(ii) Reactions force $R_{1}$of the wall acting Perpendicular to the wall ($\because$the wall is frictionless).

(iii) Reactions force $R_{2}$of the floor. This force can be resolved into two components, the normal reactions N and the force of frictions f.

For translator equilibrium in the horizontal directions,

$f - R_{1} = 0orf = R_{1}$ …..(i)

For translator equilibrium in the vertical directions,

N – W = 20g = 20 × 10 = 200 N ….. (ii)

For rotational equilibrium, taking moment of the forces about A, we get

$R_{1}\left( 2\sqrt{2} \right) - W\left( \frac{1}{2} \right) = 0$

$R_{1} = \frac{W}{4\sqrt{2}} = \frac{200}{4\sqrt{2}} = 25\sqrt{2}N$ ….. (iii)

From (ii), $f = R_{1} = 25\sqrt{2}N$

$R_{2} = \sqrt{N^{2} + f^{2}} = \sqrt{(200N)^{2} + (25\sqrt{2}N)^{2}} = 203N$