# Tension/Compression in Joints

1. Jun 5, 2012

### Hyperfluxe

Statics - Forces in a Truss

1. The problem statement, all variables and given/known data
http://i.imgur.com/NbZlc.png

2. Relevant equations
Equations of equilibrium (sum of the forces in all directions = 0, sum of the moments about an point = 0)

3. The attempt at a solution
My attempt: http://i.imgur.com/qlLCY.jpg
Hopefully you can read it. What I did is that I did the FBD of the entire system neglecting the internal forces, and I found the reaction forces. Then I did a FBD for joint A alone and solved for the two forces. Me and my friends are getting different answers so I just want to make sure. What I'm unsure about is my signs and the whole tension vs. compression thing. I don't understand it at all. Let me know where I went wrong (if I went wrong). Thanks!

EDIT: I'm not really sure why this was moved to the engineering section, this problem is from my Statics course which is an introductory physics course (right?).

Last edited: Jun 5, 2012
2. Jun 5, 2012

### SammyS

Staff Emeritus
Re: Statics - Forces in a Truss

Your solution looks fine to me. I didn't check the numbers, but they appear to be in the right ballpark.

From my experience, a Statics course is generally Engineering, even if it's taught by the physics department. (No, I wasn't the one who moved it.)

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3. Jun 5, 2012

### Hyperfluxe

Thank you :)
A lot of people in my class seem to be getting 23.7N for the axial force in AE, and they found that the joint is under tension. I don't see anything wrong with my solution so I'm keeping it.

I'm confused about one thing: how exactly do you know if a member is under compression or tension? For example, for my free-body diagram at joint A, both axial forces point towards the joint, so I call that compression. Is that how it works? Also, when you get a negative answer for a force or reaction force, do you carry that negative sign in the rest of your equations? I know that you have to change the direction on the FBD after.

4. Jun 6, 2012

### Ziv7

In calculations compression units are negative and tension are positive

if you have 2 forces going away from the beam you would draw arrows inside the beams going in [ -> <- ] (opposite to reaction) - that beam would be in tension

this is how i was taught to do it, hope it helps.

5. Jun 6, 2012

### SammyS

Staff Emeritus
They would get 23.7N by adding 17 + 17cos(60°) . But those forces are in opposite direction, so you're solution is correct.

6. Jun 6, 2012

### PhanthomJay

Always, when a force of a member on a joint points toward the joint, it is in compression; if the force in a member on a joint points away from the joint, it is in tension. As Sammy S has noted, your answer is correct. And yes, if you get a negative answer, you assumed the wrong direction, and must reverse the direction on your FBD.

7. Jun 6, 2012

### Hyperfluxe

Thanks so much guys, I appreciate it =)