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

[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.

 

[SteamKing]

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, ΣM_B = 147.15 * 0.75 - 1.5 * C_v = 0

When you solve for C_v, you obtain the vertical component of T_DE. You still must calculate the horizontal component of T_DE, 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.

 

[DarkAriel7]

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

 

[SteamKing]

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 F_D (You wrote F_C by mistake), you have 1.5 ⋅ F_C 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.

 

[SteamKing]

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.

 

[DarkAriel7]

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

I fixed the missing sin45.

Edit. Nvm

 

[SteamKing]

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.

 

[DarkAriel7]

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

 

[SteamKing]

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.

 

[DarkAriel7]

Thanks. I was really using my fbd wrong

 

[DarkAriel7]

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

 

[SteamKing]

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.

 

[DarkAriel7]

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.