Equilibrium in 2D: Finding Tension and Support Reactions in a Cable System

• DarkAriel7
In summary, the task is to find the tension in the cables and the support reactions at points A, B, and E for a pole supporting a pedestrian and traffic light. Neglecting the weight of the cables, the masses of the traffic signal, pedestrian signal, beam BCD, beam EGH, and pole AF are given. The equations of equilibrium in 2D are used to solve for the tensions and reactions. A mistake is made in assuming the vertical component of the tension in line DE is the same as the force in the moment equation. The vertical and horizontal components of the tension must be separately calculated. The use of moments in the equation also requires the inclusion of the horizontal component of tension. Further errors are made in the
DarkAriel7

Homework Statement

A pole supporting a pedestrian and traffic light is shown. Neglecting the weight of the cables. Find the tension in the cables and and the support reactions at A,B,E.

Mass of traffic signal 20kg
Mass of pedestrian signal 10kg
Mass of beam BCD 5kg center of gravity C
Mass of beam EGH 8kg center of gravity G
Mass of pole AF center of gravity midpoint A and F

The visual is in the uploaded image. I am very sorry that for whatever reason they came out backwards. I do not know how to fix this.

Homework Equations

Sum of forces = 0
Equations of Equilibrium in 2d

The Attempt at a Solution

I left this in the attached image.

DarkAriel7 said:

Homework Statement

A pole supporting a pedestrian and traffic light is shown. Neglecting the weight of the cables. Find the tension in the cables and and the support reactions at A,B,E.

Mass of traffic signal 20kg
Mass of pedestrian signal 10kg
Mass of beam BCD 5kg center of gravity C
Mass of beam EGH 8kg center of gravity G
Mass of pole AF center of gravity midpoint A and F

The visual is in the uploaded image. I am very sorry that for whatever reason they came out backwards. I do not know how to fix this.

Homework Equations

Sum of forces = 0
Equations of Equilibrium in 2d

The Attempt at a Solution

I left this in the attached image.
View attachment 92638
View attachment 92639
You've made some mistakes right off with the calculation of the tension in Line DE which holds up the Pedestrian Light.

You have assumed that the tension in line DE is the same force as the vertical component of DE in your moment equation, namely 73.57N.

Taking moments about B, ΣMB = 147.15 * 0.75 - 1.5 * Cv = 0

When you solve for Cv, you obtain the vertical component of TDE. You still must calculate the horizontal component of TDE, which in this case is trivial, since the acute angle between the line DE and the bar DB is 45°.

I haven't checked the rest of your calculations. I think it would be easier to go step-by-step with such a long calculation.

In that equation. Moments about B. Do I need to include Ex(the supporting force along the x-axis at E)?

DarkAriel7 said:
In that equation. Moments about B. Do I need to include Ex(the supporting force along the x-axis at E)?
Not according to your moment equilibrium equation. For the moment produced by FD (You wrote FC by mistake), you have 1.5 ⋅ FC but there is no sin (45°) factor like there was in the vertical force equations.

You may have written this moment equation thinking that the full tension is being used, but it's only the vertical component of the tension which is holding up the pedestrian light and its support pole.

DarkAriel7 said:
In that equation. Moments about B. Do I need to include Ex(the supporting force along the x-axis at E)?
No. You make a free body of just the arm BD (which includes the pedestrian light) and a portion of the line DE which is connected to BD.

If I am trying to find out moments about B went do I not need Ex?

I fixed the missing sin45.

Edit. Nvm

DarkAriel7 said:
If I am trying to find out moments about B went do I not need Ex?

I fixed the missing sin45.

Edit. Nvm
No. See Post #5 above.

So that means that also do not need E in my sum of X and Y equations right? Because I left E out of the diagram

DarkAriel7 said:
So that means that also do not need E in my sum of X and Y equations right? Because I left E out of the diagram
Once you have figured out the tension which keeps the pedestrian signal and its arm in equilibrium, the finding the load at E is trivial: (it's just the tension in the line DE).
You can resolve the tension at E into components if you wish.

Thanks. I was really using my fbd wrong

DE 104.051
Bx. 73.57
By. 73.57
GF. 701.94
Ex. 554.26
Ey. 34.34
Ax. 0
Ay. 421.83
Ma. 1136.49

I think I got them all

DarkAriel7 said:
DE 104.051
Bx. 73.57
By. 73.57
GF. 701.94
Ex. 554.26
Ey. 34.34
Ax. 0
Ay. 421.83
Ma. 1136.49

I think I got them all
I haven't checked your calculations for the traffic signal.

For the reaction at point A, Ay is going to be the weight of the pole and all the lights hung off of it. You don't include the vertical components of the tensions from the light arms in this calculation, since these serve only to keep the lights and their supports in equilibrium.

After all, you can't lift yourself up by your own bootstraps.

The same with calculating the moment at A. You calculate how much moment is created by hanging the lights to one side of the pole. That is the moment which the base of the pole must react against.

Also, for your reaction calculations, it is important to specify the units for each quantity and the direction in which these forces and moments act. In other words, you should always make clear what is a positive force in the horizontal and vertical directions, and whether a positive moment acts clockwise or counterclockwise.

Thanks. I did not include anything except the weights for A. I realized that same thing you said. You can't push on the ground more or less than what you weight.

I left the units line that because this teacher lines it like that. He cares mostly about the number. He checks the fbd for the direction.

1. What is equilibrium in a 2D problem?

Equilibrium in a 2D problem refers to a state in which all the forces acting on an object are balanced, resulting in no net force or acceleration.

2. How is equilibrium in a 2D problem determined?

To determine equilibrium in a 2D problem, the vector sum of all the forces acting on an object must be equal to zero. This means that the forces in the x-direction and y-direction must be balanced.

3. What is the difference between stable and unstable equilibrium in a 2D problem?

In a stable equilibrium, the object will return to its original position if it is slightly displaced from its equilibrium position. In an unstable equilibrium, the object will move further away from its equilibrium position if it is slightly displaced.

4. Can an object be in equilibrium if it is moving in a 2D problem?

Yes, an object can be in equilibrium even if it is moving in a 2D problem. This is known as dynamic equilibrium, where the object is experiencing balanced forces but is still moving at a constant velocity.

5. How does the center of mass play a role in equilibrium in a 2D problem?

The center of mass is the point at which an object's mass is evenly distributed. In equilibrium, the center of mass must remain stationary or move at a constant velocity. This means that the forces acting on the object must be balanced around the center of mass.

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