Ladder problem (statics) with 2 sources of friction

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SUMMARY

The discussion centers on calculating the maximum angle θ that a uniform ladder of mass m and length L can make with a wall without slipping, given equal coefficients of static friction (μ) between the ladder and the floor and wall. The solution involves applying static equilibrium equations, particularly focusing on torque and forces in both the x and y directions. The final expression for θ is derived as θ = arctan(μ/(1-(1+μ²)/2)). The participants confirm the validity of the calculations and discuss the implications of μ values greater than 1.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with torque calculations
  • Knowledge of trigonometric functions, particularly inverse functions
  • Basic concepts of friction and normal forces
NEXT STEPS
  • Study the derivation of static equilibrium equations in physics
  • Explore the implications of varying coefficients of friction in real-world applications
  • Learn about the behavior of materials with μ > 1 and their practical significance
  • Investigate advanced torque problems involving multiple forces and angles
USEFUL FOR

Students studying physics, particularly those focusing on statics, engineers involved in structural analysis, and anyone interested in the mechanics of friction and equilibrium in physical systems.

natasha13100
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Homework Statement


A uniform ladder of mass m and length L rests against the wall as shown. The coefficients of static friction between the floor and the ladder and between the wall and the ladder are equal to each other (μ). What is the maximum value of angle θ that the ladder can make with the wall without sliding? (Use m for m, L for L, mu for μ, and arcsin, arccos, and arctan for the inverse trig functions.)

Homework Equations


f(friction)≤μN(normal force)
t(torque)=r(moment arm)Fsinθ
G=mg

The Attempt at a Solution


FBD is attached

because the ladder is not moving
t=-mgLsinθ/2+NfLsinθ-μNfLcosθ=0
Fx=Nw-ff=0
Fy=Nf+fw-mg=0

Solve t for Nf
NfLsinθ-μNfLcosθ=mgLsinθ/2
Nftanθ-μNf=mgtanθ/2
Nf=mgtanθ/(2(tanθ-μ))

When θ is at its maximum without the ladder slipping, f=μN.
ff=μmgtanθ/(2(tanθ-μ))
Solve Fxfor Nw
Nw=μmgtanθ/(2(tanθ-μ))
Also fw2mgtanθ/(2(tanθ-μ))

Solve Fy for θ.
mgtanθ/(2(tanθ-μ))+μ2mgtanθ/(2(tanθ-μ))=mg
tanθ/(2(tanθ-μ))+μ2tanθ/(2(tanθ-μ))=1
tanθ=2(tanθ-μ)/(1+μ2)
(tanθ-μ)/tanθ=(1+μ2)/2
1-μ/tanθ=(1+μ2)/2
μ/tanθ=1-(1+μ2)/2
tanθ=μ/(1-(1+μ2)/2)
θ=arctan(μ/(1-(1+μ2)/2))
 

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Your work looks good to me. You can simplify your answer a bit.

[Would there be a solution if ##\mu > 1##? For most substances ##\mu < 1##, but there exist substances for which ##\mu > 1##.]
 
Last edited:
Oh yes it is. I finally figured out where my mistake was. I've been trying to simplify the answer and it would come out wrong. I would get an answer like 2/(1+μ). A simple algebra mistake. Thanks. :) By the way, if I am thinking correctly, the angle would exist but it would be extremely small.
 

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