**Problem 12:** Consider a ladder weighing 200 N resting against a smooth wall such that it makes an angle 60^{0} with the horizontal. Find the reactions on the ladder due to the wall and ground.

**Solution**

Given data

Length

of the ladder = AB = L

AC = CB

= L/2

Weight of the

ladder = W = 200 N

Θ = 60^{0}

Required

- Reaction of the wall
- Reaction of the ground

Assume

Normal

Reaction of the wall = R

Reaction

force of the ground = F

Angle of

F and horizontal = φ

Now

apply 1^{st} condition of equilibrium

Apply 2^{nd}

condition of equilibrium with A as point of rotation, clockwise torque as

negative and anti-clockwise as positive,

Now

consider the ∆AEB,

Similarly,

in ∆ACD

Substitute

the values from (iv) and (v) in (iii)

As θ =

60^{0}, therefore,

**Magnitude of R**

Now to

find the magnitude and direction of F, we use equations (i) and (ii)

And

This is

the reaction of the ground on the ladder.

**Direction**

This is

the angle of the reaction of the ground on the ladder.

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