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.
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.