How far up the ladder can you climb before it begins to slip?

[liquidheineken]

A uniform ladder of length 9 m leans against a frictionless vertical wall making an angle of 47° with the ground. The coefficient of static friction between the ladder and the ground is 0.41.

If your mass is 74 kg and the ladder's mass is 33 kg, how far up the ladder can you climb before it begins to slip?

Given:
L = 9 m
M = 74 kg
m = 33 kg
μ = 0.41
θ = 47°
x = unknown

Noted:
F_WL = Force from Wall to Ladder
F_f = Force of Friction
N = Normal ForceI think I've solved it, but I just wanted to be sure. I was trying to get some practice in for my test tomorrow and found this question on these forums unanswered.

Since nothing is moving, the system is in Equilibrium. So net torque = 0, or counter clockwise torques = clockwise torques. Equation for torque; t = Force x Moment Arm

Mg(xL cosθ) + mg(0.5L cosθ) = F_WL(L sinθ)

The L cancels out from both sides, and the g cosθ is factored out of left side.​

g(xM + 0.5m) cosθ = F_WL sinθ

Divide both sides by sinθ​

g(xM + 0.5m) / tanθ = F_WL

Since system is at equilibrium F_WL = F_f = μN = μ(M + m)g​

g(xM + 0.5m) / tanθ = [B]μ(M + m)g[/B]

Multiple both sides by tanθ, and g cancels from both sides​

[B][B][B][B][B][B][B][B][B]xM + 0.5m = [B][B][B][B][B]μ(M + m) tanθ[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]
[B]xM = [B][B][B][B][B][B][B][B][B][B][B][B][B]μ(M + m) tanθ - 0.5m[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]
[B][B][B][B][B][B][B][B][B][B][B][B][B][B]x = [μ(M + m) tanθ - 0.5m] / M = 0.4127...[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]
[B][B][B][B][B][B][B][B][B][B][B][B][B][B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]
[B][B][B][B][B][B][B][B][B][B][B][B][B][B]Max length up ladder = xL ≈ 3.715 m[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]

 

[ehild]

Since system is at equilibrium F_WL = F_f = μN = μ(M + m)g​

Your solution is correct but the sentence above is not.

It should be

Since system is at equilibrium F_WL = F_f ≤ μN = μ(M + m)g

 

[liquidheineken]

can you elaborate on that? I was simply plugging in the equation for Friction at that point. Why is F_f ≤ μN?

 

[BvU]

Matter of careful wording: The system is also at equilibrium when xL = 0 but then $F_f \ne \mu N$

 

[ehild]

can you elaborate on that? I was simply plugging in the equation for Friction at that point. Why is F_f ≤ μN?

The force of static friction is a force that prevents sliding. It does not have a definite value, only a maximum: depending on the surfaces in contact, the maximum value is μ_sN, where μ_s is the coefficient of static friction. If the other forces exceed the static friction, the object starts to slide and is not in equilibrium any more. The friction becomes kinetic, and equal to μ_kN