# Thread: force imposed trough a triangle

1. ## force imposed trough a triangle

Hi,
Im trying to work out the Force imposed on the top of a triangle for a project Im working on. ( \/ ) If I hung a load on the apex of the V and the tops of the V were supported by a structure or indeed it was a just a triangle, How would I work out the compression force trying to pull the tops of the V together. I can work out the force in the legs by using the sine of the angles. I just cant figure out how to apply it to the top! The formula needs to work for variable sized and angled triangles (or V's).

if you could spare any answers you may have in simple terms (if that is possible it would be much appreciated.)

Thanks

2. I get what you're trying to do now lol. Give me the dimension of the rope or the angle between the rope and the contact points. The mass in Kg you're attaching as well and the dimensions of the triangle.

3. the apex angle is 105 the angle to the left is 30 and to the right is 45 degrees. The side to the left is 5.18 units lets say feet. to the right 7.32 and the top is 10. Load=1000kg This is the example i have been using. I hope this helps and that you understand what the hell I'm going on about lol.

Thanks

4. Is this scheme correct?

Please include a fully detailed scheme next time you have a mecanics problem ok? I'm an engineer, I don't read text.

I need the angles that the rope does with the vertical as well and I should be able to do it:

5. does this help, Im looking for the force pulling the beams together.

thanks

6. Wrong calculation sorry lol.

7. Thanks for the time and effort your putting in to this.

how did you arrive at A=26803 and B=32825.

I will attach a link to download a program which works all this out (windows only). Im just trying to find out the math behind it all. It will give you a better idea of what im getting at. Enter the info of my triangle into the value box's. Your solution seems to differ from the programs. Also I don't think it converts the load to KN. Ive got every thing else figured out its just the force on the beams.

again thanks for your time.

James

8. Originally Posted by continentaljames
Thanks for the time and effort your putting in to this.

how did you arrive at A=26803 and B=32825.

I will attach a link to download a program which works all this out (windows only). Im just trying to find out the math behind it all. It will give you a better idea of what im getting at. Enter the info of my triangle into the value box's. Your solution seems to differ from the programs. Also I don't think it converts the load to KN. Ive got every thing else figured out its just the force on the beams.

again thanks for your time.

James
I made a mistake the first time... I did cos(30) A + Cos(45)B instead of cos(30) A - Cos(45)B. Didn't do much except make the force values 4x bigger LOL -___-. Well **** that method...here's an easier one:

The results match what you get in LD calculator. With the current configuration, 63% of the weight is expressed as compressive force. If you want less compression you must make the rope longer to increase the y component. The shorter the rope, the higher the compression actually.

cheers,

Nuke

9. Hi Nuke,

I think you must be a genius. It took me a while to work that one out but i got it a couple of days ago. you put:

9810/sin(75)=B/sin(60)=A/sin(45)

the only way i could reach the same result was by doing this:

9810/sin(75) x sin(60) = B

9810/sin(75) x sin(45) = A

one more question if I may. If the load is in Kg should the forces be expressed as Kn. Thanks for all your help, we got there in the end.

10. As long as you are on Earth, 1 Kg = 9.81 N that's how you convert. 1 Kn = 1000N = 102 Kg, 1 Kg = 1000g = 9.81 N