Horizontal Force in Bottom Member of a Truss

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Discussion Overview

The discussion revolves around a problem from the Statics section of a review for the FE exam, specifically focusing on determining the horizontal force in the bottom member of a truss. Participants explore the application of trigonometric functions to resolve forces in a structural analysis context.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Exploratory

Main Points Raised

  • One participant expresses confusion about the division of cos(60) by cos(30) in the context of calculating forces in the truss members.
  • Another participant suggests drawing a free body diagram (FBD) to clarify the situation.
  • There is a discussion about the distribution of a 10 kN vertical force between two members, with participants questioning how this affects the forces in members AC and BC.
  • One participant identifies that the expression 5 / cos(30°) could represent the vertical component of force in member AC when considering symmetry in the frame.
  • Another participant confirms the calculation of force in member AC as Fac=5/cos(30), leading to a numerical value of 5.77 kN, and subsequently calculates the horizontal force in member AB as Fab=2.88 kN.
  • There is acknowledgment of overthinking the problem, with a suggestion that visualizing a force triangle could help clarify the relationships between the forces.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the application of trigonometric functions in this context. While some clarify aspects of the problem, there is no consensus on the initial confusion surrounding the division of cos(60) by cos(30).

Contextual Notes

Participants reference specific angles and forces without fully resolving the implications of these relationships. The discussion includes assumptions about symmetry and the distribution of forces that may not be universally accepted.

CSawyer717
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Hello,
I've been studying for the FE, and this question in the Statics section of the review I am going through has really tripped me up for some reason. I understand Statics, and how to determine forces in X and Y directions, determining X and Y components using trig, etc, etc..

Maybe I am overthinking this whole thing, but for some reason this problem has me awfully confused. it's in two parts; first part (not shown) just asks what rxns at A and B are (Ra=Rb=5 kN)
second part (in attached image) asks for the horizontal force in the bottom component. I have the solution there, as well - I just have no idea how the solution was reached. what confuses me, is why do you divide cos(60)/cos(30)? I understand where the 60 and 30 degree angles are from - just not sure why you're dividing the two to get the answer.
any help would be greatly appreciated!
 

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CSawyer717 said:
Hello,
I've been studying for the FE, and this question in the Statics section of the review I am going through has really tripped me up for some reason. I understand Statics, and how to determine forces in X and Y directions, determining X and Y components using trig, etc, etc..

Maybe I am overthinking this whole thing, but for some reason this problem has me awfully confused. it's in two parts; first part (not shown) just asks what rxns at A and B are (Ra=Rb=5 kN)
second part (in attached image) asks for the horizontal force in the bottom component. I have the solution there, as well - I just have no idea how the solution was reached. what confuses me, is why do you divide cos(60)/cos(30)? I understand where the 60 and 30 degree angles are from - just not sure why you're dividing the two to get the answer.
any help would be greatly appreciated!
Have you tried drawing a free body diagram of the structure?
 
I have, and I'm still not understanding why you need to divide cos(60)/cos(30). Also, is that 10/2 from the one 10kN vertical force being split between the two members AC and BC? so when you draw your FBD about point A, you're only seeing half of that 10 kN force? i understand the cos(60) is the X component in member AC. I just get hung up where the cos(30) comes into play. do you need to determine the force in member AC from both, point A and point C? and that's where the 30 degree angle comes into play?

very frustrating; at first glance, it seemed like a simple problem - i can't tell if i am over thinking everything or what.
 
CSawyer717 said:
I have, and I'm still not understanding why you need to divide cos(60)/cos(30). Also, is that 10/2 from the one 10kN vertical force being split between the two members AC and BC? so when you draw your FBD about point A, you're only seeing half of that 10 kN force? i understand the cos(60) is the X component in member AC. I just get hung up where the cos(30) comes into play. do you need to determine the force in member AC from both, point A and point C? and that's where the 30 degree angle comes into play?

very frustrating; at first glance, it seemed like a simple problem - i can't tell if i am over thinking everything or what.
You need to lay out what the individual parts of the expression are telling you. I think you are just confused seeing several parts rolled up into one expression.

You have correctly discerned that (10/2) comes from applying symmetry to the frame.

What force is given by the expression 5 / cos (30°) ?
 
SteamKing said:
You need to lay out what the individual parts of the expression are telling you. I think you are just confused seeing several parts rolled up into one expression.

You have correctly discerned that (10/2) comes from applying symmetry to the frame.

What force is given by the expression 5 / cos (30°) ?

would that be the vertical component of AC if you made a FBD around point C? So, since you can just apply symmetry, you're able to assume only a 5kN downward force at C, and a force in member AC. then solving for Fac, you come up with Fac=5/cos(30), resulting in Fac=5.77kN. then summing horizontal forces about point A, you get Fab=5.77cos(60), Fab=2.88

oh, wow. you're a magician. thank you very much for the help! there was a ton of overthinking and confusion from just seeing the solution there..
 
CSawyer717 said:
would that be the vertical component of AC if you made a FBD around point C? So, since you can just apply symmetry, you're able to assume only a 5kN downward force at C, and a force in member AC. then solving for Fac, you come up with Fac=5/cos(30), resulting in Fac=5.77kN. then summing horizontal forces about point A, you get Fab=5.77cos(60), Fab=2.88

oh, wow. you're a magician. thank you very much for the help! there was a ton of overthinking and confusion from just seeing the solution there..
Sometimes, just laying out a force triangle can answer many questions.
 

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