Ladder leaning against a wall friction factor

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SUMMARY

The discussion centers on calculating the minimum angle at which a ladder can lean against a wall without slipping, given a friction factor K between the ladder and the ground. The correct formula derived is tan(θ) = 1/(2K), which ensures stability without slipping. A common misconception arises from an alternative formula tan(θ) = 2K, which is incorrect. The introduction of a friction factor between the wall and the ladder complicates the equation, leading to potential negative angles, indicating a misunderstanding in the application of friction forces.

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  • Understanding of static equilibrium in physics
  • Knowledge of trigonometric functions and their applications
  • Familiarity with friction concepts in mechanics
  • Basic principles of forces acting on inclined planes
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  • Learn about the role of friction in inclined planes
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This discussion is beneficial for physics students, mechanical engineers, and anyone interested in understanding the mechanics of inclined structures and friction forces.

Uniquebum
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Ladder of mass M and length L leans against a vertical wall. The friction factor between the ladder and ground is K. Calculate the minimum angle at which the ladder can stay in position without slipping off ignoring the friction between the wall and the ladder.

Calculating momentum equilibrium
N_2*L*sin(\theta) = 0.5*G*L*cos(\theta)
(N_2 = force between wall and ladder = KMg)
(G = Mg)
Which leads to
tan(\theta) = \frac{1}{2K}

Anyhow, is this correct or am i missing something? I found a website giving an answer of
tan(\theta) = 2K

What if i put a friction factor between the wall and the ladder? To this i get an answer of
tan(\theta) = \frac{2K_2-1}{2K_1}
which feels wrong as it might result in a negative angle.
 
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Uniquebum said:
Which leads to
tan(\theta) = \frac{1}{2K}

Anyhow, is this correct or am i missing something? I found a website giving an answer of
tan(\theta) = 2K

It is correct. Probably the other website denoted the other angle by theta.

Uniquebum said:
What if i put a friction factor between the wall and the ladder? To this i get an answer of
tan(\theta) = \frac{2K_2-1}{2K_1}
which feels wrong as it might result in a negative angle.

It really is not right. How did you get it?


ehild
 

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