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liquidheineken
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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:
FWL = Force from Wall to Ladder
Ff = 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) = FWL(L sinθ)
xM = μ(M + m) tanθ - 0.5m
x = [μ(M + m) tanθ - 0.5m] / M = 0.4127...
Max length up ladder = xL ≈ 3.715 m
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:
FWL = Force from Wall to Ladder
Ff = 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) = FWL(L sinθ)
The L cancels out from both sides, and the g cosθ is factored out of left side.
g(xM + 0.5m) cosθ = FWL sinθDivide both sides by sinθ
g(xM + 0.5m) / tanθ = FWLSince system is at equilibrium FWL = Ff = μN = μ(M + m)g
g(xM + 0.5m) / tanθ = μ(M + m)gMultiple both sides by tanθ, and g cancels from both sides
xM + 0.5m = μ(M + m) tanθxM = μ(M + m) tanθ - 0.5m
x = [μ(M + m) tanθ - 0.5m] / M = 0.4127...
Max length up ladder = xL ≈ 3.715 m