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Non-linear force to linear force equation

by jimgram
Tags: equation, force, linear, nonlinear
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jimgram
#1
May10-12, 10:42 AM
P: 92
I need to design a cam based on the following math problem: A non-linear force f(n) is working on a lever that is x + x' long (x+x'=constant). I need an equation for the ratio, x/x', such that the counter-balancing force, f', is constant (non-varying).

By varying the ratio of f(n) to f', that is ratio=x/x', I should be able to find an equation describing the ratio (r) that provides a non-varying output force f'. The distance that f(n) works through is the same as the distance that f' works through, so work in = work out.

Any help will be greatly appreciated. I don't know quite how to approach this problem.

I hope the attached jpeg expalins the problem in enough detail
Attached Thumbnails
Mathcad - force balance.jpg  
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haruspex
#2
May12-12, 07:47 AM
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Not at all sure I've understood the question, but here goes.
By moments, you have f.x = f'.x'.
x+x' = C, constant.
So f.x = f'(C-x)
(f+f')x = f'.C
x = f'.C/(f+f')
Is that it?
(Didn't get the bit about f and f' moving through the same distance. Since f and f' are not equal, if they move the same distanvce there'll be an energy imbalance. Won't the movements be in the inverse ratio?)
jimgram
#3
May12-12, 08:49 AM
P: 92
Your solution works where both f and f' are known. If you know f' and f you can find x, and then f' = f*x/(C-x). But you need x to find f' and you need f' to find x.

I believe I have found a solution: It is required that the distance traveled for f and f' be equal. Work is equal to force times distance, so if distance is equal and f is a function of n, then the work done by f is [itex]\int f(n) dn[/itex]. This yields a constant value for the area under the curve of f(n). So now the ratio(n) (which is x/x') is equal to([itex]\int f(n) dn[/itex])/f(n)*n. Then f'(n) is equal to f(n)*ratio(n), and f'(n) is constant.

The net result is that the ratio varies as f(n) varies to maintain a constant counterbalance force f'.

haruspex
#4
May12-12, 06:45 PM
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Non-linear force to linear force equation

Clearly I have not understood the question - and still don't.
What is n? From your integral it would appear to be the distance through which f has moved at any time. Is that right?
And when you say the two forces move through the same distance, you must mean the same overall distance over a given range, yes? It cannot be true that they have moved the same distance at each instant.
jimgram
#5
May13-12, 09:00 AM
P: 92
A lever that is x+x' long has a force f on one end and a countering force f' on the other end and the fulcrum is at a position determined by x, x'. Say x = x', then the fulcrum is in the center and f must equal f'. This just a simple lever and the distance that f moves will be equal to the distance f' moves.

Now you can vary the relationship between x and x' such that the fulcrum moves to the left or right of center. The ratio x/x' will be equal to the ratio f/f' and the distance each force moves the end of the lever s/s'. Again, simple lever physics.

But lets make f a function (it could be any function) such that f becomes non-linear. Let's say the function (call it 'n') is exponential. In my example, f(n)=(1/n)2. But it could be any function - that part is not important. Just that the force is non-linear. Now, if you have a fixed positon for the fulcrum (that is, x and x' are non-varying) then the countering force f' is simply f(n)*x/x'.

My problem is to find a function so that x/x' becomes non-linear and yields equal work, wf=f*s = wf'=f'*s' and f' is constant and stotal = s'total.

s is not equal to s' at all moments in time since the ratio, x/x', is varying. If area under the curve f(n)(that is, its integral) is made equal to the area under the curve (straight line) for a non-varying f', then work done on both sides will be equal and the total distance traveled on both sides will be equal.
haruspex
#6
May13-12, 04:57 PM
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f becomes a non-linear function of what? Distance travelled? Time? It means nothing to say f is just 'non-linear'.
Perhaps you mean non-constant, so f = f(t). The ratio x/x' = r(t).
Since f' constant, r(t) = f(t)/f'.
The deltas to distance moved must be in the same ratio: ds = r(t).ds'
(Check: work is conserved at all times: f(t).ds = f'.ds')
The remaining constraint is
∫.ds = ∫.ds' over the range of t. I.e.
∫r(t).ds' = ∫.ds'
We can reparamaterise r as a function of s':
∫r(s').ds' = ∫.ds'
I.e. the average value of r(s') as s' varies is 1.
jimgram
#7
May14-12, 10:05 AM
P: 92
I'm sorry - I did make it more confusing by saying that force f was a function of n where n is a series and f(n)=(1/n)^2. The force does vary as a function of time (time is proportional to n).


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