Problem 12: Consider a ladder weighing 200 N resting against a smooth wall such that it makes an angle 600 with the horizontal. Find the reactions on the ladder due to the wall and ground.
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num46.png)
Solution
Given data
Length
of the ladder = AB = L
AC = CB
= L/2
Weight of the
ladder = W = 200 N
Θ = 600
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 1st condition of equilibrium
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num47.png)
Apply 2nd
condition of equilibrium with A as point of rotation, clockwise torque as
negative and anti-clockwise as positive,
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num48.png)
Now
consider the ∆AEB,
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num49.png)
Similarly,
in ∆ACD
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num50.png)
Substitute
the values from (iv) and (v) in (iii)
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num51.png)
As θ =
600, therefore,
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num52.png)
Magnitude of R
Now to
find the magnitude and direction of F, we use equations (i) and (ii)
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num53.png)
And
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num54.png)
This is
the reaction of the ground on the ladder.
Direction
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num55.png)
This is
the angle of the reaction of the ground on the ladder.
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