Question

A 5m long ladder of mass 20 kg rests against a frictionless wall. Its lower end rests on a
floor 3-m from the wall. Calculate the force needed to prevent the ladder from sliding away
from the wall.​

Answer 1

Given:

⇒ Length of ladder, say AB = 3m

⇒ Weight of ladder = 20Kg

⇒ Distance of ladder from wall, say AE = 1m

To find:

⇒ Reaction forces of the wall = ?

⇒ Reaction forces of the floor = ?

Solution:

⇒ Solving by Pythagoras theorem, 

⇒ (BE)² = 3² - 1²

⇒ (BE)² = 9 - 1

⇒ (BE)² = 8

⇒ BE = √8

⇒ BE = 2.83m

⇒ Let us understand the various forces acting are:

⇒ The weight, mg of the ladder acting vertically downward at C

⇒ The horizontal force, F due to the wall at B.

⇒ Reaction of the floor f acting along AB. This is the force exerted by the ground on the ladder. It is the resultant of fx and fy.

⇒ Resolving the forces, to find solution to the problem:

⇒ Resolving the horizontal components of force, we get F = fx which is the force exerted by the wall on the ladder.

⇒ Resolving the vertical components of force, we get mg = fy 

⇒ i.e 20 × 9.8 = fy 

⇒ ∴ fy = 196N

⇒ We know that the algebraic sum of moment of forces about A = 0

⇒ Hence, F × BE = mg × AD

⇒ F × (2.83) = 196 × (0.5)

⇒ F =  (196 × 0.5)/ 2.83

⇒ F = 98/ 2.83

⇒ ∴F = 34.63N

 ⇒ ∴F = 34.63 N which is also fx which is the force exerted by the wall on the ladder.

⇒ So the resultant is f = √(fx² + fy² + fx.fy.cosθ)

⇒ f =  √[ (34.63² + (196)² + 34.63 × 196 × Cos 90°] 

⇒ f = √( 1199.2369 + 38416 + 0) (∵since cos 90° = 0, fx.fy.cosθ = 0)

⇒ f = √39615.2369

⇒ f = 199.03

⇒∴f ≈ 199N

⇒ f =  199N which is the force exerted by the ground on the ladder.

Answer 2

Given: Length of ladder $l$ $=5m$

          Distance of lower end of the  ladder from the wall $=3m$

          mass of ladder $m$ $=20kg$

Now, solving the Pythagoras theorem, we can calculate the vertical distance from the top of the ladder to the floor, (say h)

                  $5^{2}= 3^{2}+h^{2}$

                 $h=\sqrt{16}$

                   $h=4$            

In the case of the frictionless wall, the vertical and horizontal forces exerted on the wall, so

          $mg=20\times 9.8$

          $F_{Vertical} =196N$

And the algebraic sum of moment of forces is

               $F\times 4=F_{vertical} \times 1.5$

                $F=\frac{196\times 1.5}{4}$

                 $F=73.5N$

Hence the force needed to prevent the ladder from sliding will be 73.5Newton

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