Min. angle at which ladder will not slip

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SUMMARY

The discussion focuses on determining the minimum angle, θ, at which a uniform ladder of mass m and length l will not slip against a frictionless wall, given a static friction coefficient, μ_s. The participant outlines their approach using free body diagrams (FBD) and equations of equilibrium, including SFx and SFy. However, they receive feedback indicating that their torque equation is flawed, specifically regarding the lever arm for the wall force moment, which needs to be expressed as a function of L and θ.

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  • Understanding of static equilibrium concepts in physics
  • Familiarity with free body diagrams (FBD)
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  • Basic grasp of friction and its coefficients
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Homework Statement


A uniform ladder of mass, m, and length, l, leans at an angle, \theta, against a frictionless wall. If the coefficient of static friction between the ladder and the ground is \mu_{s}, what is the minimum angle at which the ladder will not slip?



Homework Equations





The Attempt at a Solution


I've attached my FBD.

SFx = mg - N = 0
SFy = \mumg - Fw = 0
St = FwL - mg(L/2) = 0

mg = N
\mumg = Fw
FwL = mg(L/2)
L/2 = \mu

Where do I go from here?
 

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Your FBD is good, and your sum of forces = 0 equations are good, but your sum of torques equation is incorrect. L, for example, is not the lever arm for the wall force moment about the base. You need to calculate that arm as a function of L and theta. L is the length of the ladder as measured along the incline.
 

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