Horizontal Force in Bottom Member of a Truss

  • #1
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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  • #2
SteamKing
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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!
Have you tried drawing a free body diagram of the structure?
 
  • #3
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 cant tell if i am over thinking everything or what.
 
  • #4
SteamKing
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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 cant 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°) ?
 
  • #5
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..
 
  • #6
SteamKing
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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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