Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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=

Sagot :

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.

View image LealaL296698
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.