James10492
Junior Member
- Joined
- May 17, 2020
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'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?