One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the floor. A man weighing 700 N could climb up to 7.0 m before slipping. What is the coefficient of static friction between the floor and the ladder

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

which statement about this sum is true? −245+2.7

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