Question

A ladder of length L = 2.5 m and mass m = 20 kg rests on a floor with coefficient of static friction μ_s = 0.52. Assume the wall is frictionless.

1)What is the normal force the floor exerts on the ladder? 196.2N

2) What is the minimum angle the ladder must make with the floor to not slip?

3)

A person with mass M = 63 kg now stands at the very top of the ladder.

What is the normal force the floor exerts on the ladder?

4)What is the minimum angle to keep the ladder from sliding?

Answer 1

given data
length of the ladder L=2.5m
mass m=20
μs = 0.52

1)What is the normal force the floor exerts on the ladder?

Since the wall is frictionless, the normal force on the floor supports the full weight of the ladder.

Therefore Fn = 20kg * 9.8m/s² = 196 N

2) What is the minimum angle the ladder must make with the floor to not slip?

At the threshold, Fw = Ff = µmg = 0.52 * 196N = 196 N
where Ff is the friction force at the floor and Fw is the opposing force at the wall.
Sum the moments about the base of the ladder:
196N * 2.5m/2 * cosΘ = 102N * 2.5m * sinΘ
245N·m = 255N·m * tanΘ
Θ = 43.85º

3.A person with mass M = 63 kg now stands at the very top of the ladder.

What is the normal force the floor exerts on the ladder?

Fn = 196N + 63kg * 9.8m/s² = 813.4 N

4)What is the minimum angle to keep the ladder from sliding?

Ff = 813.4N * 0.52 = 423 N = Fw
196N * 2.5m/2 * cosΘ + 63kg * 9.8m/s² * 2.5m * cosΘ = 423N * 2.5m * sinΘ
(196/2)+63*9.8 = 423 *tan(Θ )

Θ = 59.4º