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Balance Forces on a Disc 
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#1
Mar1914, 02:36 PM

P: 9

This seems like it would be pretty simple and I apologize if this is in the wrong area, here goes.
Picture a flat disc. It is supported by a pin or marble in the center. There are no forces acting on it. Now at 0 deg, I apply a force F (all forces are a distance equal from the center), resulting in a force F at 180 deg. Simple enough. Then if I apply an equal force N at 180 deg, the disc becomes flat again, or balanced. If I apply another N force to the 90 deg position, my disc again goes out of balance, tipping up at 270 deg. I think this becomes out of balance equal to when I placed the first force on the disc because the first two forces with the marble, creates a line and a fulcrum. So now I place 4 forces 90 deg apart, and it is balanced. Then add another force at 45 deg. This would cause the disc to want to raise at the 315 deg but I would think that it would not have the same effect as the first force. I am looking for a formula or algorithm that I can input forces around a circle at different angles and calculate the out of balance force and angle. Any information and help would be appreciated. Ron 


#2
Mar1914, 03:04 PM

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#3
Mar1914, 03:25 PM

P: 9

Think of the marble as a single point fulcrum. The disc is sitting on top of it along the X plane. All the forces are directed down (negative Y direction). Applying a force on the edge of the disc will result in the other end of the disc to act in the opposite direction, like a teetertotter. If you push on one end of the teeter totter, you get a force on the other end. But I'm working with a disc. Hopefully this makes things clearer.



#4
Mar1914, 03:27 PM

P: 4,015

Balance Forces on a Disc
http://en.wikipedia.org/wiki/Torque 


#5
Mar1914, 04:41 PM

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I think you should try to restate your original question, making clear whether or not the disc is actually in equilibrium and whether it has mass (moment of inertia). When you say that there is an equal and opposite force on the other side of rim, this will only happen if there is something for that point to push against  and I don't think your model includes this. Without one, all that will happen is that the disc will tip and continue to tip (accelerating). If you are talking about the equilibrium situation then there are many combinations of forces around the disc (downwards, I assume) that will produce equilibrium. Problems about Moments are usually confined to two dimensional bodies but it is quite possible to solve for a 3D situation, like your disc. You just need to split the problem into two axes at 90° across the disc and work out moments about each axis. 


#6
Mar2014, 08:11 AM

P: 9

Ok, lets try this. I have a number of springs in a circle applying force on a disc. The disc is in balance as long as I have the springs equally spaced. I want to try to maintain an equal balance of force all the way around. There would never be a situation where there would only be one spring but there would be cases where I may want to place an odd number of springs around the disc. If I were to use only 1 spring on one side and the spring applied a force of 50 pounds or newtons, the disk would be out of balance (by force applied) by 50 to its direct opposite. Now picture a situation with say 4 springs each applying 50 Force (same distance from center) at places 0 deg, 110 deg, 210 deg, & 285 deg. They are unequally spaced causing an unbalance of force of some amount at some angle.
I tried to create a spreadsheet where I split the disc down an axis and calculated the force applied to one side versus the force applied to the other side. Picture the disc split into left and right sides. But that doesn't work because you can have all your forces on the top of each half and the results show a balanced disc. I speculate I may need to find the forces using polar vectors and sum them somehow. You could also look at this problem as taking a 2" thick steel disc and drilling holes around a bolt circle. The mass removed would cause an imbalance of gravitational force. 


#7
Mar2014, 08:26 AM

P: 4,015

http://en.wikipedia.org/wiki/Torque 


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