# Mechanics forces problem

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#1
I wasn't sure which was best forum to post this problem into, but here goes.

I have attached an image of a relatively simple (they say!) problem in which I have to find the forces in each truss. But I'm at a complete loss at how they get to their answers.

I know the assembly is in equilibrium, so all the forces in X and Y directions = 0. I need to set where my positive directions are before I start don't I? So I set mine as left and up as positive.

The problem I have is that I find Fcd = 20KN. So that means my original free body diagram was correct in assuming the force of 20KN was acting upward. But then when we move on to point D the answers then show the 20KN force acting down. But I just calculated it as +ve acting up.

I'm at a complete loss with these problems. And how do you tell if a truss is under compression or tension? It seems random to me. There's no explanation.

(P1 = 20KN P2 = 10 KN) (I tried to insert pic but it kept putting it in upside down for some reason)
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1 year ago
#2
(Original post by Darwinion)
...
Don't normally visit the physics forum, so my terminiology may not be 100%. Hope this clears up your issues.

It helps to distinguish between forces acting on a truss, and forces acting in a truss.

Consider CD.

We have 20kN acting downwards on the truss at C.

Since we have static equilibrium, there must be a force in the truss of 20kN acting upwards at C.

Again since the truss is in equilibrium and there is a force of 20kN in the truss acting upwards at C, there must be a force of 20kN in the truss acting downwards at D.

I.e. the force acts in opposite directions at either end of the truss.

Compression/tension.

Example. Take a metal cylinder and try and compress it lengthwise by hand. The cylinder resists. Forces act outwards stopping your hands. Cylinder is under compression.
I.e. When the forces at each end act outwards the object is under compression.

And when they act inwards, it's under tension. It's resisting you pulling it apart.

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#3
Thanks. It's taken me ages to realise that when I'm calculating the forces in the beams I'm working out the REACTION forces. If a reaction force is in direct opposition to an external force then it meets it "head on" as such and has an equal reaction force at the other end like <-------> hence compression (opposite to what you naturally think as these are reaction forces).
If the reaction force is in same direction as an external force then the reaction force meets half way with it's counterpart in the opposite direction like -----><----- and that's tension as the reaction forces are pulling back on the beam being stretched.

I struggled with this so much but thank you.
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1 year ago
#4
(Original post by Darwinion)
Thanks. It's taken me ages to realise that when I'm calculating the forces in the beams I'm working out the REACTION forces. If a reaction force is in direct opposition to an external force then it meets it "head on" as such and has an equal reaction force at the other end like <-------> hence compression (opposite to what you naturally think as these are reaction forces).
If the reaction force is in same direction as an external force then the reaction force meets half way with it's counterpart in the opposite direction like -----><----- and that's tension as the reaction forces are pulling back on the beam being stretched.

I struggled with this so much but thank you.

The reaction force is always in opposition to the external force regardless of its direction. Your diagrams are correct though.

Quote me if you need a reply, as I don't often visit this forum.

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