Proof involving static rigid body

[James10492]

'A uniform ladder rests in limiting equilibrium with one end on rough horizontal ground and the other against a rough vertical wall. The coefficient of friction between the ladder and the ground is m1 and the coefficient of friction between the ladder and the wall is m2. Given that the ladder makes an angle x with the ground,

show that tan x = (1-m1m2)/2m1'


(I will refer to the symbols given to the forces in my diagram):
(*edit sorry I forgot, of course W is the weight acting at the centre of the uniform rod)

Taking moments:

W cos x = 2 H sinx

W = 2 H tan x


Resolving the forces vertically,

R + F₂ = W

Horizontally,

F₁ = H


F₁ = Rm₁ and F₂ = H m₂


tan x = W / 2 H

= R + F₂ / 2H

= R + F₂ / 2Rm₁

= R + Hm₂ / 2Rm₁

= R + Rm₁m₂ / 2Rm₁

= 1 +m₁m₂/2m₁

which looks great, and makes sense to me, but is not the right answer. What did I do wrong?

 

[skeeter]

rotational equilibrium about the foot of the ladder ...

[imath]W \cos{x} = 2H\sin{x} + 2F_2\cos{x} = 2H(\sin{x} + m_2\cos{x})[/imath]

try again ...

 

[skeeter]

[imath] (R+F_2)\cos{x} = 2H(\sin{x} + m_2 \cos{x} ) [/imath]

[imath] R+m_2 H = 2H(\tan{x}+m_2)[/imath]

[imath] \dfrac{F_1}{m_1} + m_2 F_1 = 2F_1(\tan{x} + m_2)[/imath]

[imath] \dfrac{1}{2m_1} + \dfrac{ m_2}{2} = \tan{x} + m_2[/imath]

[imath] \tan{x} = \dfrac{1}{2m_1} + \dfrac{m_1m_2}{2m_1} - \dfrac{2m_1m_2}{2m_1} = \dfrac{1-m_1m_2}{2m_1}[/imath]