I'm trying to find a formula to alow me to work out loading on two anchor points. For example: I have 100Kg on a "yhang knot" and the "y" angle is a 120 deg angle. What weight is on each anchor point? And how do you work it out

I'm trying to find a formula to alow me to work out loading on two anchor points. For example: I have 100Kg on a "yhang knot" and the "y" angle is a 120 deg angle. What weight is on each anchor point? And how do you work it out
Here's an applet.
http://physics.syr.edu/courses/mra/d...mo1/demo1.html
The way to work it out is as follows, assuming it is symmetrical, i.e., equal lengths of rope from the points of attachment to the intersection of the two ropes.
The two ropes share the weight equally, so the vertical component of tension in each rope is half the weight of the load.
The vertical component of the tension in the rope (above the Y) divided by the tension is equal to the sine of the angle the rope makes with the ceiling, which is 30 degrees. Sin 30 degrees = 0.5. Therefore the tension in the rope is twice the vertical component of the tension, which was half the weight of the load. Therefore the tension in each rope section is equal to the weight of the load.
Hi Harold,Originally Posted by Harold14370
Thanks for the info but to be honest I'm having difficulty in understanding it. Is there any chance you can explain how to work it out it more basic terms?
Thanks again
First of all. let's make sure we are talking about the same thing. Two ropes are attached to the ceiling, and a weight is hung from the midpoint. the two ropes meet at an angle of 120 degrees. Correct?
If that is tha case, because it is symmetrical, each rope supports half the weight.
The tension in the rope is directed at an angle to vertical and is in line with the rope. Do you know about vectors? The tension in the rope is a force vector. It can be viewed as having two components, a horizontal component and a vertical component. The magnitude of the force is found by the pythagorean theorem. It is the square root of the sum of the squares of the horizontal and vertical component. Do you follow so far?
Now we know the vertical component. It's what supports the weight. We know the angle that the force is directed. It is in line with the rope, 60 degrees from vertical (half of the 120 degree angle). A rope is floppy so it can only direct a force in the direction it is hanging. So we have a 6030 right triangle and we can use trig to find the answer. The sine is defined as the opposite divided by the hypotenuse. The vertical component, which we know is half the load, is the opposite side of the right triangle, opposite of the 30 degree angle. The hypotenuse is the force (tension), which lies in the direction of the rope. So
sin (30) = 0.5
sin (30) =vertical component/tension = (0.5*100Kg)/T
T=0.5*100Kg /0.5 = 100Kg.
The tension in the rope is the force acting on the anchor point.
T=0.5*100Kg /0.5 = 100Kg.Originally Posted by Harold14370
Thanks again for your time Harold.
I'm an industrial climber and I must often make a YHang very similar to that Java diagram. Some times the angle of the Y is 55 deg some times 40 deg. What I need to work out is how much tension/weight is on each anchor point. It makes sence to me that as the angle of the Y changes so will the weight on each anchor. We try to make the Y equal but sometimes one anchor point is heigher than the other
Hope I'm not causing to much stress
Jason
Well, for different angles you would do it the same way. It's easy as long as everything is symmetrical, which it would be if the ropes are equal in length and hung from the same height.
For example, if the Y angle is 40 degrees, then half that would be 20, and the angle to the ceiling would be 9020=70. Then you would take the sine of 70 degrees, and work it out as above, i.e., sin(70)=0.5(100Kg)/T.
Ok I'm starting to get it, the only thing is, the angle to the ceiling? there is no ceiling. I have my anchor points on the face of a building. What does sin stand for?Originally Posted by Harold14370
Thanks for this
Jason
Then just substitute "horizontal" where I wrote "ceiling."Originally Posted by rigger1734
Sin stands for the trigonometric sine function. If you have a scientific calculator, that is probably what it says on the key.
Originally Posted by Harold14370
Thanks for all of your help but I'm just not getting it.
Thanks for everything
Jason
In order to do the calculations you will need a basic understanding of trigonometry. I think it would be worth studying for someone in your line of work. Maybe you could take a class in a community college.Originally Posted by rigger1734
OK, I've just done some very basic trigonometry on line. I know how to work out the Sin, Cos and Tan.Originally Posted by Harold14370
I also understand the ceiling bit now. I would divide the Sin angle by 2 then take an imaginary line up to horizontal. That would give me a right angle triangle.
NOW What. I know at the point of sin I have 100kg
Maybe a picture would help.
Yep picture is a great help but I thought you needed the length of the opposite and hypot. to work out the sin?Originally Posted by Harold14370
Can you do me another picture showing an angle of 55 deg and a weight of 85Kg?
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