Ladder leaning against a wall friction factor

[Uniquebum]

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.

 

[ehild]

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.

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