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As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of = 30°, the length of the beam is L = 3.00 m, the coefficient of static friction between the wall and the beam is s = 0.600, and the weight of the beam is represented by w. Determine the minimum distance x from point A at which an additional weight 2w (twice the weight of the rod) can be hung without causing the rod to slip at point A. x =

As Shown In The Figure Below A Uniform Beam Is Supported By A Cable At One End And The Force Of Friction At The Other End The Cable Makes An Angle Of 30 The Len class=

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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.

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