A ladder of length L = 2.7 m and mass m = 21 kg rests on a floor with coefficien

Question

A ladder of length L = 2.7 m and mass m = 21 kg rests on a floor with coefficient of static friction s = 0.56. Assume the wall is frictionless.

1)

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

N

2)

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

°

3)

A person with mass M = 68 kg now stands at the very top of the ladder.

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

N

4)

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

Explanation / Answer

Given that

mass of ladder m=21 kg

length of ladder L=2.7 m

(1) now we find 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 N = 21kg * 9.8m/s² = 205.8 N

(2) now we find the minimum angle the ladder must make with the floor to not slip

At the threshold, Fw= Ff = µmg = 0.56 * 205.8N = 115.25 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:
205.8N * 2.7/2 * cos = 115.25N * 2.7m * sin
277.8N·m = 311.2N·m * tan
= tan^-1(277.8/311.2) = 41.8º

(3) now we find the normal force the floor exerts on the ladder

N2 = 205.8N + 68kg * 9.8m/s² = 872.2N

(4) now we find the minimum angle to keep the ladder from sliding

Ff = 872.2N * 0.56 = 488.4 N = Fw
205.8N * 2.7m/2 * cos + 68kg * 9.8m/s² * 2.7m * cos = 488.4N * 2.7m * sin
277.83N·m + 1799.3N·m = 1318.7N·m * tan
= tan^-1[1318.7/2077.13]=32º

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