- #1

thegreengineer

- 54

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## Homework Statement

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.## Homework Equations

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

## 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:

[itex]\sum M_E=-900N(2.25m)-900(4.5m)+F_y(3m)=0[/itex]

[itex]F_y=2025N[/itex]

**upwards**

[itex]\sum F_x=1800N-E_x=0[/itex]

[itex]E_x=1800N[/itex]

**leftwards**

[itex]\sum F_y=2025N-E_y=0[/itex]

[itex]E_y=2025N[/itex]

**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

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