How to find forces in wires at equilibrium

In summary, the conversation was about finding tensions in suspension wires on a picture using equilibrium equations for moments and forces. The matrix obtained had a rank of two and further assumptions needed to be made about the deformation of the wires and beam. The suggested solution involved taking the beam to be completely straight and rigid, and considering all wires to have the same modulus of elasticity. This led to a solution with forces of 0.1, 0.2, 0.3, and 0.4.
  • #1
harmyder
33
1
I need to find tensions in suspensions wires on the picture:
fZruy.png


I wrote 4 equilibrium equations for moments in A, B, C, D and equilibrium about y-axis, but i got matrix with rank two.

[tex]\begin{pmatrix}
3 & 2 & 1 & 0 \\
0 & 1 & 2 & 3 \\
-1 & 0 & 1 & 2 \\
-2 & -1 & 0 & 1 \\
1 & 1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
F_A \\
F_B \\
F_C \\
F_D
\end{pmatrix}
=
\begin{pmatrix}
1 \\
2 \\
1 \\
0 \\
1
\end{pmatrix}[/tex]

How to find the forces?
 
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  • #2
Yes, there ard only two equations to be had from the usual linear and rotational force balances. To get further, you will need to make assumptions about how the wires and beam deform under loads.
 
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  • #3
haruspex said:
Yes, there ard only two equations to be had from the usual linear and rotational force balances. To get further, you will need to make assumptions about how the wires and beam deform under loads.
... Specifically, I would suggest taking the beam to be completely straight and rigid, but allow the wires all the same modulus of elasticity. That is enough to get a solution.
 
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  • #4
[tex]\delta_A = \delta_C - 2x[/tex][tex]\delta_B = \delta_C - x[/tex][tex]\delta_D = \delta_C + x[/tex]
Thus,
[tex]F_A = \frac{\delta_A SE}{L} = \frac{\delta_C SE}{L} - 2x\frac{SE}{L}[/tex]

And further,
[tex]F_B - F_A = F_C - F_B = F_D - F_C.[/tex]

Now matrix looks like this:
[tex]\begin{pmatrix}
-1 & 2 & -1 & 0 \\
0 & -1 & 2 & -1 \\
3 & 2 & 1 & 0 \\
-1 & 0 & 1 & 2
\end{pmatrix}
\begin{pmatrix}
F_A \\
F_B \\
F_C \\
F_D
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
1 \\
1
\end{pmatrix}[/tex]

It has solution 0.1, 0.2, 0.3, 0.4.
 

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  • #5
harmyder said:
[tex]\delta_A = \delta_C - 2x[/tex]
[tex]\delta_B = \delta_C - x[/tex]
[tex]\delta_D = \delta_C + x[/tex]
Thus,
[tex]F_A = \frac{\delta_A SE}{L} = \frac{\delta_C SE}{L} - 2x\frac{SE}{L}[/tex]

And further,
[tex]F_B - F_A = F_C - F_B = F_D - F_C.[/tex]

Now matrix looks like this:
[tex]\begin{pmatrix}
-1 & 2 & -1 & 0 \\
0 & -1 & 2 & -1 \\
3 & 2 & 1 & 0 \\
-1 & 0 & 1 & 2
\end{pmatrix}
\begin{pmatrix}
F_A \\
F_B \\
F_C \\
F_D
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
1 \\
1
\end{pmatrix}[/tex]

It has solution 0.1, 0.2, 0.3, 0.4.
That's the answer I get.
 
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1. How do I determine the net force on a wire at equilibrium?

At equilibrium, the net force on a wire is equal to zero. This means that the vector sum of all the forces acting on the wire must be zero. To determine the net force, you can use vector addition to calculate the sum of all the individual forces acting on the wire.

2. What types of forces can act on a wire at equilibrium?

There are two types of forces that can act on a wire at equilibrium: tension and gravitational force. Tension is the pulling force exerted by the wire, while gravitational force is the downward force exerted by the Earth.

3. How do I calculate the tension in a wire at equilibrium?

To calculate the tension in a wire at equilibrium, you can use the equation T = mg, where T is the tension, m is the mass of the object attached to the wire, and g is the acceleration due to gravity. In this case, the tension is equal to the weight of the object.

4. Can the angle of a wire affect the forces acting on it at equilibrium?

Yes, the angle of the wire can affect the forces acting on it at equilibrium. As the angle of the wire increases, the component of the gravitational force acting on the wire also increases. This means that the tension in the wire must also increase to keep the wire in equilibrium.

5. How does the length of a wire affect the forces at equilibrium?

The length of a wire does not affect the forces acting on it at equilibrium, as long as the wire is not stretched beyond its elastic limit. This means that the tension and gravitational force will remain the same regardless of the length of the wire, as long as it is at equilibrium.

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