# Homework Help: Vector mechanics for engineers -- Confusion described below

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1. Sep 23, 2015

### MarcusAu314

1. The problem statement, all variables and given/known data
Look I have an exercise about trusses in vector mechanics. It's the following:
https://scontent-lax3-1.xx.fbcdn.net/hphotos-xfp1/v/t1.0-9/12003339_1645511085733429_8210043018491702007_n.jpg?oh=4955534403fb6f3f5f9a40c9c0b8ecbc&oe=569761ED​
I have to determine the forces in every bar and whether they're in tension or compression. Ok, so what I did was calculating the reactions first on the supports. In this case we have one roller and one rough surface (so this means we have the reactions Ey, Ex and Fy). The main confusion here is when I calculated the forces in bars EF and CE I got the same results in the number but opposite condition (instead of tension I've got compression). The following prove my results and I will place the image of the supposed results of the PDF.

2. Relevant equations

For checking this see the "attempt at a solution".

3. The attempt at a solution
FIRST, I need to calculate the reactions on the supports. For joint E there are two reaction forces Ex and Ey while on joint F there is just one reaction force Fy. I calculated those reactions by first calculating the sum of torques on E to find Fy:
$\sum M_E=-900N(2.25m)-900(4.5m)+F_y(3m)=0$
$F_y=2025N$ upwards
$\sum F_x=1800N-E_x=0$
$E_x=1800N$ leftwards
$\sum F_y=2025N-E_y=0$
$E_y=2025N$ downwards

ONCE calculated the reactions I proceeded to calculate the forces on the bars. I proceeded to calculate the forces on bars EF and CE (the ones I previously mentioned) by taking joint E. This was an easy joint since there's not more than one unknown for axes X and Y (I mean, for X axis the only unknown was bar EF and for Y axis the only unknown was bar CE). It's also easy because we already have a known X force and a known Y force (Ex and Ey respectively) which I previously calculated. So to find the force on bars I did the following:
https://scontent-dfw1-1.xx.fbcdn.net/hphotos-xpa1/v/t1.0-9/12002262_1645522482398956_6078775752622087739_n.jpg?oh=6b52fe9217da9f543fc2813644488b73&oe=56A2E38E​
I did the summation on X and Y axes and I got that for EF it was 1800 N rightwards and 2025 N upwards so as EF is going in the bar so it has to be a compression, as the same that CE is going to the bar (I mean it's not leaving the joint outside but going inside) it's a compression. But my book states that's a tension.
https://scontent-dfw1-1.xx.fbcdn.net/hphotos-xpt1/v/t1.0-9/12004140_1645523559065515_2808815127332424313_n.jpg?oh=297521e2fac379aeb5ebe0faae347f99&oe=569DB6A8

Last edited by a moderator: Sep 24, 2015
2. Sep 25, 2015

### PhanthomJay

When you calculated the forces at joint E, the directions of the forces act on the joint. By Newtons 3rd law, the directions of the equal forces on the members are opposite to the joint force directions.

3. Sep 25, 2015

### Staff: Mentor

Yes. EF is pulling to the right on the joint, and CE is pulling upward on the joint. Pulling means that the members are in tension.

Chet

4. Sep 26, 2015

### MarcusAu314

So it is not about whether they go outside or inside the bar, it is whether they leave the joint or not. I thought that in case for a bar that has two joints can be either in compression or tension whether they are going inside the bar or leaving it respectively. By this last thing I refer to this:
This in Spanish means: in the first case they tend to stretch (pulling out) the element and this is in tension and in the second case they tend to compress (pushing in) the element and this is in compression. As you can see in the figure the force EF is pushing in the element with respect to joint E (I know that in joint F it will act the same so this is not my doubt) so how can it be a tension when the force is going in the bar and NOT out?

5. Sep 26, 2015

### Staff: Mentor

EF is pulling to the right on joint E. The joint E is pulling to the left on the bar EF. If you go inside the bar (i.e., put a fictitious break in the bar), the section to the right of the break will be pulling to the right on the section to the left. And the section to the left of the break will be pulling to the left on the section to the right. The key word in all of this is pulling. Pulling means tension. Pushing means compression. Or, if the tail of the arrow is touching the cross section, then you have tension, and, if the head of the arrow is touching the cross section, then you have compression.

Chet