ANSWER
[tex]\begin{equation*} 1.51\text{ }m \end{equation*}[/tex]
EXPLANATION
Parameters given:
The angle of the cable, θ = 30 degrees
Length of the beam, L = 3.00 m
Coefficient of static friction, s = 0.600
Weight of the beam, W = w
First, we have to draw a free-body diagram showing all the forces acting on the beam:
Let us find the components of force in the horizontal and vertical directions.
In the horizontal direction:
[tex]\begin{gathered} \sum F_x=0 \\ \\ N-T\cos\theta=0 \\ \\ N=T\cos\theta \end{gathered}[/tex]
In the vertical direction:
[tex]\begin{gathered} \sum F_y=0 \\ \\ -2w-w+T\sinθ+\mu_sN=0 \end{gathered}[/tex]
Substitute the expression for N into the equation and simplify:
[tex]\begin{gathered} -3w+T\sin\theta+\mu_s(T\cosθ)=0 \\ \\ T(\sin\theta+\mu_s\cosθ)=3w \\ \\ T=\frac{3w}{\sinθ+\mu_s\cosθ} \end{gathered}[/tex]
Substitute the given values into the equation and simplify:
[tex]\begin{gathered} T=\frac{3w}{\sin30+0.600\cos30}=\frac{3w}{0.5+0.5196} \\ \\ T=\frac{3w}{1.0196}=2.942w \end{gathered}[/tex]
Now, take the moments of forces about the left end of the cable. Since the cable is not moving, it is at equilibrium:
[tex]\begin{gathered} \sum M=0 \\ \\ (-2w*x)+(-w*\frac{L}{2})+(\mu_sN*0)+(T\sinθ*L)=0 \end{gathered}[/tex]
Substitute the known values into the equation and solve for x:
[tex]\begin{gathered} (-2wx)-(w*\frac{3.00}{2})+0+(2.942w*\sin30*3.00)=0 \\ \\ -2wx-1.5w+4.413w=0 \\ \\ -2wx=1.5w-4.413w=-3.013w \\ \\ x=\frac{-3.013w}{-2w} \\ \\ x=1.51\text{ }m \end{gathered}[/tex]
That is the minimum distance, x, from point A at which an additional weight, 2w, can be hung.
