A ladder of length L = 2.1 m and mass m = 18 kg rests on a floor with coefficien

Question

A ladder of length L = 2.1 m and mass m = 18 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 = 65 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?

°

(Survey Question)

5)

Below is some space to write notes on this problem

Explanation / Answer

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

Therefore Fn = 18kg * 9.8m/s² = 176.4 N

2) At the threshold, Fw = Ff = µmg = 0.56 * 176.4N = 98.78 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:

176.4N * 2.1m/2 * cos = 98.78N * 2.1m * sin

185.22N·m = 207.43N·m * tan

185.22 / 207.43 = 0.89

= arctan(0.89) = 41.66º

3) Fn = 176.4N + 65kg * 9.8m/s² = 813 N

4) Ff = 813N * 0.56 = 455 N

= Fw = 176.4N * 2.1m/2 * cos + 65kg * 9.8m/s² * 2.1m * cos = 455N * 2.1m * sin

185.22N·m + 1337.7N·m = 955.5N·m * tan

1522.92N·m = 955.5N·m * tan

1522.92 / 955.5 = 1.59

= arctan(1.59) = 57.83º

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