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A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall. Find the reaction forces of the wall and the floor. (1) 34.6 N, 199 N (2) 30 N, 150 N (3) 24 N, 146 N (4) 75 N, 24 N

A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall. Find the reaction forces of the wall and the floor.
(1) 34.6 N, 199 N
(2) 30 N, 150 N
(3) 24 N, 146 N
(4) 75 N, 24 N

Grade:12th pass

2 Answers

Saurabh Kumar
askIITians Faculty 2400 Points
9 years ago
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13 Points
5 years ago
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

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