A 4.0 m long uniform ladder with the weight of 120 N leans against a wall making it 70° above a cement floor as shown in Figure. Assuming the wall is frictionless, but the floor is not, determine the forces exerted on the ladder by the floor and by the wall. ( Ans. 122N) 



Given:

Let the ladder ground contact point is A and to the wall be B then the Length of ladder `\overline {AB}` (Figure-2) = L = 4.0 m
Weight of ladder = W = 120 N
Center of gravity of rod =`\ frac {L}{2}` = 2.0 m
Angle 𝛳 = 70°
 

To Find:

The force of Reaction on the ladder due to wall = Fw = ?
The force of Reaction on the ladder due to ground = F = ?

Solution:

A figure-2 is constructed to explain the Figure 1 of the question, in which we have resolved the ladder length L into its components which are L cos𝛳 and L sin 𝛳 x-component and y-component respectively. The clock wise and anti clockwise torques are also marked in the figure-2 according to the given conditions in the question.  

First, we will apply the first condition of equilibrium, which is mathematically represented as

Figure-2
Σ F = 0

In x-axis (Horizontal direction)

Σ Fₓ = 0

Fₓ - Fw = 0

or

Fₓ = Fw ------------(1)

In y-axis (vertical direction)

Σ Fy = 0 Fy - W = 0 or Fy = W = 120 N

We will find the value of Fₓ by applying the second condition of equilibrium. So, let's consider the ladder moments at about point A. The second condition of equilibrium is mathematically written as


Σ T = 0

or

Σ T Anti-clockwise =  Σ T Clockwise 

As shown in figure-2, the anti-clockwise torque is due to the Force of Reaction on the ladder due to the wall that is  T = Fw L sin 70° and clockwise torque is due to the weight of the ladder W that is  T = W `\overline {AG}` cos 70° thus

Fw L sin 70° = W `\overline {AG}` cos 70°

as `\overline {AB}` = L so, `\overline {AG}` = `\frac {L}{2}` = `\frac {4}{2}` = 2

by putting values

Fw ｘ 4 ｘ0 .940 = 120 ｘ 2 ｘ 0.342

Fw ｘ 3.759 = 82.085

Fw = `\frac {82.085}{3.759}`

Fw = 21.8 N

by equation (1)

Fw = Fₓ = 21.8 N -----------Ans. 1

This is the reaction force on the ladder due to the wall

Now the reaction force on the ladder due to the ground is given by (using Pythagoras theorem)

F = `sqrt {Fx^2 + Fy^2}`

by putting values

F = `sqrt {(21.8 N)^2 + (120 N)^2}`

F = `sqrt {475.24 N² + 1,4400 N²}`

F = `sqrt {14,875.24 N²}`

F = 121.96 N

F = 122 N ------------Ans. 2
This is the reaction force on the ladder due to ground


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