expanding c=-a*sin(α)+b*cos(α)

[joh]

An equation of the following form is given:
[latex]c=-a_1*\sin(\alpha)+a_2*\cos(\alpha)[/latex]

next by expanding the equation above the following equation is obtained:
[latex]c=-b_1*\sin(\beta-\delta)\gamma-b_2(\frac{\partial u}{\partial v})\dot \gamma[/latex]

i have rewritten the first equation to: [latex]c=\sqrt{c_1^2+c_1^2}\cos(\alpha-\epsilon)=\sqrt{(c_1^2+c_2^2)(1-\sin^2(\alpha-\epsilon))}[/latex] where [latex]\epsilon=\tan^{-1}(c_2/c_1)[/latex]

However, I don't think this is the correct path to obtaining the expanded equation and if it is I can't see what the next step should be. How do they obtain the expanded equation?

[daniton]

why are you expanding?

what are you trying to find?

I didn't understand any of what you did may be it's beyond me:-)

[studiot]

I agree that it is rather a big step from your line 1 to line 2, where you have introduced a number of undefined variables.

 

I can observe that the derived function of the sine is the cosine and vice versa to a +/- sign. So are u and v auxiliary functions or what and where did gamma and its (?time) derivative come from?

 

Line 1 looks as though you are working with the solution to a differential equation is this the case?

Edited May 31, 2013 by studiot

[D H]

expanding c=-a*sin(α)+b*cos(α)

 

An equation of the following form is given:

[latex]c=-a_1*\sin(\alpha)+a_2*\cos(\alpha)[/latex]

 

next by expanding the equation above the following equation is obtained:

[latex]c=-b_1*\sin(\beta-\delta)\gamma-b_2(\frac{\partial u}{\partial v})\dot \gamma[/latex]

 

i have rewritten the first equation to: [latex]c=\sqrt{c_1^2+c_1^2}\cos(\alpha-\epsilon)=\sqrt{(c_1^2+c_2^2)(1-\sin^2(\alpha-\epsilon))}[/latex] where [latex]\epsilon=\tan^{-1}(c_2/c_1)[/latex]

 

However, I don't think this is the correct path to obtaining the expanded equation and if it is I can't see what the next step should be. How do they obtain the expanded equation?

 

A big part of the problem is that you are being sloppy. One problem right off the bat: Are you supposed to expand that expression or are you supposed solve or solve for α? I assume it's the latter. How can you expand this? It's already expanded!

 

Another sign of sloppiness is a consistent lack of consistency. The title says the coefficients of sin(α) and cos(α) are -a and b, respectively. Later you use a₁ and a₂, later yet c₁ and c₂. Yet another sign of sloppiness is a bunch of undefined terms. What are beta, gamma, delta, and epsilon supposed to represent? How are they related to your original expression? And what's with that partial? Where did that come from?

 

 

Be consistent, do a better job of telling us (and hence telling yourself) what it is you are trying to do, and define your terms. A lot of your problems will vanish when you do these things.

Edited May 31, 2013 by D H

[joh]

I probably didn't present the problem very well. I did not expand the equation. This is done in a paper. The issue is that the paper does not explain how this happened nor can I think of a way to achieve this (the main problem is that I can't figure out were the partial came from, I did consider using a taylor expansion but got nowhere). In the paper the following information is available:


http://s20.postimg.org/3mqwmdgf1/image.png

Equations (1) and (2) are the force balance on the glass surface shown in the third picture. After these two equations are expanded in equation (3) and (4) they are inserted in the equation of motion of the system.

 

P = force (as shown in figures)

angles as shown in figures
[latex]\mu_0[/latex] = friction coefficient
v = relative velocity between the lip and glass

[latex]\frac{\partial \mu}{\partial v}[/latex]: behavior is described by the stribeck curve

I need to now how the expansion takes place in order to know what b1, b2 and b3 are.

 

P.S.: Apologies for the consistency issues in the first post

Edited May 31, 2013 by joh

[studiot]

I am not familiar with this experiment, but the principal of boundary lubrication is that the weight forces (Py) are taken by the pressure in the fluid and the friction forces (Px) due to the relative movement are taken by the viscoscity of the fluid. If this is the case the the partial will come from this. The b constants will originate from the geometry - separation, contact area etc which appear in theoretical analysis when applying Newton's laws, but do not appear explicitly in your analysis.

 

However there appear to be springs M and D connected to the funnel.

A more detailed description of the experiment would help.

[joh]

Manage to find what b1, b2 and b3 are. Thanks a lot for all your tips.

[studiot]

Manage to find what b1, b2 and b3 are. Thanks a lot for all your tips.

 

 

 

So are you going to enlighten us?