Statics. Determine the smallest angle $θ$ The coefficient of friction at A and B is $0.5$

Question

Determine the smallest angle θ at which the uniform triangular plate of weight W can remain at rest. The coefficient of static friction at $ A $ and $B $ is $0.5.$

Based on my Free body diagram which I think is correct, I wonder why I can't get the answer $θ$ = $36.9$ so I know that Friction Force is, $Ffa=(0.5)Na$ and $Ffb=(0.5)Nb$ So I tried to perform Moment at A. and do the summation of forces from

$X$-axis which is Ffa = Nb

and summation of forces at

$Y$-axis which is W=Na + Ffb.

I can derive the weight of the triangle by moment from A or B, assuming at A so it will be $W\frac{2}{3}(270)cosθ$ = $(Na)(270)sinθ$+$(Ffb)(270)cosθ$ and then i can just perform derivation however I can't get the angle $θ=36.9$ I used $\frac{2}{3}$ since W is located at the centroid of the triangle. If you may please show your step by step solution.

Answer 1

The moment arm at $A$ of the weight is the distance from $A$ to the centroid. If we place a coordinate system with $A$ as the origin and the $270$ side along the positive $x$-axis, then the centroid is located at $(\frac 2 3(270), \frac 13(180)) = (180, 60)$. The moment arm at $A$ is therefore $\sqrt{180^2 + 60^2} = 60\sqrt{10}$, not the $180$ that you are using.

Answer 2

you forgot the other coordinate of the W which is W(180/3)sinθ, and this rotates CCW at A.

Also, don't forget these:

Fa=(0.5)Na     Fb=(0.5)Nb

Horizontal Forces:

  0.5Na = Nb

Vertical Forces:

  0.5Nb + Na = W

So, W = 2.5 Nb

So, the moment at A: CW=CCW

    Wcosθ(2/3)(270) = (Nb)(270)sinθ + (Ffb)(270)cosθ + Wsinθ(180/3)
   
   (2.5Nb)cosθ(180) = Nb(270)sinθ + 0.5Nb(270cosθ) + (2.5Nb)sinθ(60)

then cancel all Nb's

    450cosθ = 270sinθ + 135cosθ + 150sinθ

    315cosθ = 420sinθ

    tanθ = 0.75,

    θ = 36.9