Question

A ladder with a length of 5.0 m and a weight of 600 N is placed so its base is on the ground 4.0 m from a vertical frictionless wall, and its tip rests 3.0 m up the wall. The ladder remains in this position only because of the static friction force between the ladder and the ground. You, with a weight of 500 N, are standing on the ladder at a horizontal distance of 2.10 m from the wall. Assume the mass of the ladder is uniformly distributed.

(a) Find the magnitude of the upward normal force applied by the ground on the ladder.
(b) Find the magnitude of the horizontal normal force applied by the wall on the ladder.
(c) Determine the minimum possible coefficient of static friction for the ladder-ground interaction, for the ladder to remain at rest.

Answer 1

Answer:

(a) N₁= 1100 N

(b) N₂ = 716.66 N

(c) μs =0.65

Explanation:

We apply Newton's first law:

∑Fx = 0  

∑Fy =0

∑F : algebraic sum of the forces in Newton (N)

∑MA=0

∑MA : algebraic sum of moments at point A.(N*m)  

Look at the free body diagram in the attached graph

(a) Find the magnitude of the upward normal force applied by the ground on the ladder (N₁)

∑Fy = 0  positive (+) is upward

N₁-500-600=0

N₁= 500+600 = 1100 N

(b) Find the magnitude of the horizontal normal force (N₂) applied by the wall on the ladder.

$H= \sqrt{5^{2}-4^{2}  }$

H= 3m

∑MA=0 :positive moment (+) counterclockwise

Ff(3)-N₁(4)+500(2.1)+600(2) = 0   ,Ff: friction force

(3)Ff = 1100(4)-500(2.1)-600(2)

$F_{f} = \frac{1100(4)-500(2.1)-600(2)}{3}$

$F_{f} = 716.66 N$

∑Fx = 0  ,positive (+) to the right

Ff-N₂ = 0

Ff=N₂

N₂ = 716.66 N

(c) Determine the minimum possible coefficient of static friction (μs) for the ladder-ground interaction, for the ladder to remain at rest.

Ff = μs *N₁

μs = Ff /N₁

μs =716.66 / 1100

μs =0.65