Simplify Equation for T Cosine and Delta X: Step by Step Guide

[Cyrus]

I'm not getting this. Can someone show me how they got from this:

$$- T cos ( \theta ) + (T + \Delta T ) cos ( \theta + \Delta \theta ) =0$$

to below by dividing by $$\Delta x$$

$$\frac { d (T cos\theta)}{dx} = 0$$

Im just not getting it for some reason... The first term would be Tcos(theta)/ dx, which should turn into 1/0, which blows up to infinity?

Thank you,

Cyrus

 

[Tide]

Cyrus,

Clearly, you are treating $\Delta T$ and $\Delta \theta$ as "small" numbers. The lowest order term, $T\cos \theta$, cancels out. Just expand everything in terms of small quantities and your result will be evident.

 

[Cyrus]

like this?


$$-Tcos(\theta) + Tcos(\theta + \Delta \theta ) + \Delta T cos (\theta + \Delta \theta) = 0$$

so I can neglect the delta theta, giving me:

$$-Tcos(\theta) + Tcos(\theta ) + \Delta T cos (\theta) =0$$

which cancels out those two terms, but then I am left with:

$$\Delta T cos (\theta) = 0$$

Now, I am not where I should be...oops?

I'm somehow supposed to get:

$$\Delta T cos (\theta) + T sin(\theta) \Delta \theta = 0$$

 

[Cyrus]

Got it Tide! I'm sorry, but I disagree, or am not seeing, what you said, here's my approach:

$$-T cos (\theta) + (T + \Delta T)cos(\theta + \Delta \theta) = 0$$

which is equivalent to:

$$T cos( \theta + \Delta \theta) - T cos(\theta) + \Delta T cos (\theta + \Delta \theta) = 0$$

which is equivalent to:

$$\frac {[T cos( \theta + \Delta \theta) - T cos(\theta)]\Delta \theta}{\Delta \theta} + \Delta T cos (\theta + \Delta \theta) = 0$$

NOW divide by delta X:

$$\frac {[T cos( \theta + \Delta \theta) - T cos(\theta)]\Delta \theta}{\Delta \theta \Delta x} + \frac {\Delta T cos (\theta + \Delta \theta)}{\Delta x} = 0$$



and take the limit as: $$\Delta x \rightarrow 0; \Delta \theta \rightarrow 0;\Delta T \rightarrow 0$$

So, the $$\frac {cos(\theta + \Delta \theta)-cos(\theta)}{\Delta \theta}$$ term goes to $$sin(\theta)$$, and $$\frac {\Delta \theta}{\Delta x}$$ goes to $$\frac{d \theta } {dx}$$ and $$\frac {\Delta T}{\Delta x}$$ goes to $$\frac {dT}{dx}$$ and $$cos(\theta + \Delta \theta )$$ goes to just good ole, $$cos (\theta)$$

and your left with,

$$T sin(\theta) \frac{ d \theta}{ dx} + \frac {dT}{dx} cos (\theta)}$$

which clearly equals:

$$\frac {d (Tcos\theta)}{dx}$$

Good god, its 4:00 am. I am glad I got this done or I wouldent have fallen asleep. Goodnight.

 

[HallsofIvy]

You can use the fact that $cos(\theta+\Delta\theta)= cos(\theta)cos(\Delta\theta)- sin(\theta)sin(\Delta\theta)$.

Multiply the whole thing out then use the fact that $cos(\Delta\theta)$ goest to 1 and $sin(\Delta\theta)$ goes to 0 as $\Delta\theta$ goes to 0.

 

[maverick6664]

Is it so complicated?
I just think:
$$\theta$$ and $$T$$ are functions of $$x$$ (it means they are not independent to each other. otherwise $$x$$ wouldn't appear here in this context), so let $$S(x)=T\cos(\theta)$$, then the first equation can read $$- S(x) + S(x + \Delta x) = 0$$. Then divide this equation by $$\Delta x$$ and you'll have $$\frac{\Delta S}{\Delta x} = \frac{dS}{dx} = 0$$.

 

[Cyrus]

Hmm,YES, these all seem to be valid methods. Thats very cool. Does my approach seem wrong?

 

[maverick6664]

Your argument #4 looks fine to me. But I just thought it's a bit complicated and not generic. What you wrote in #4 is $$\Delta S = ( \frac{\partial S}{\partial T} \frac{\partial T}{\partial x} + \frac{\partial S}{\partial \theta} \frac{\partial \theta}{\partial x} ) \Delta x$$
It holds even if it isn't $$\cos{\theta}$$, and now that intermediate variables $$\theta$$ and $$T$$ are affected by $$x$$, so we can simply put $$\frac{dS}{dx} = \frac{\partial S}{\partial T} \frac{\partial T}{\partial x} + \frac{\partial S}{\partial \theta} \frac{\partial \theta}{\partial x}$$

and you can ignore these intermediate variables. If you see $$S$$ simply as a function of x, this can be explained without partial differential as in #6.

hope this helps.. and Merry Christmas! (here in Japan it's 2:30am)

 

[benorin]

Are you working the (one dimensional) wave equation for a vibrating string by chance (where T is tension)? As I recall seeing something similar during that proof.

 

[Cyrus]

Its not a wave equation for a string, but close. Its a derviation to find the equation of a string/rope/wire/whatever under a distributed load in the x direction, OR if you replace delta x by delta s, it is the weight distribution of a catenary wire. No vibration, sorry.

 

[kant]

I'm not getting this. Can someone show me how they got from this:
$$- T cos ( \theta ) + (T + \Delta T ) cos ( \theta + \Delta \theta ) =0$$
to below by dividing by $$\Delta x$$
$$\frac { d (T cos\theta)}{dx} = 0$$
Im just not getting it for some reason... The first term would be Tcos(theta)/ dx, which should turn into 1/0, which blows up to infinity?
Thank you,
Cyrus

If ( T+ delT)cos(@+ del@) =o, then Tcos@=0.

The change of Tcos@ is d (Tcos@).

The change of 0 is 0.

divdie d( Tcos@) =0 by delX, and allow delX = dx.

 

[maverick6664]

Its not a wave equation for a string, but close. Its a derviation to find the equation of a string/rope/wire/whatever under a distributed load in the x direction, OR if you replace delta x by delta s, it is the weight distribution of a catenary wire. No vibration, sorry.

Then there should be other equations like

$$(T + \Delta T) \sin(\theta + \Delta \theta) = T \sin(\theta) + \varrho \frac {\Delta x} { \cos(\theta) } g$$

and

$$\Delta y = \Delta x \tan(\theta)$$

along Y axis where $$\varrho$$ is the mass density of the rope.
and

$$\frac {d(T \sin(\theta))} {dx} = \frac {\varrho g} {\cos(\theta)}$$ or just $$\varrho g$$

$$\frac {dy}{dx} = \tan(\theta)$$ or just $$sin(\theta)$$

or something...

 

[Cyrus]

If ( T+ delT)cos(@+ del@) =o, then Tcos@=0.

The change of Tcos@ is d (Tcos@).

The change of 0 is 0.

divdie d( Tcos@) =0 by delX, and allow delX = dx.

...what?

 

[Cyrus]

Then there should be other equations like

$$(T + \Delta T) \sin(\theta + \Delta \theta) = T \sin(\theta) + \varrho \frac {\Delta x} { \cos(\theta) } g$$

and

$$\Delta y = \Delta x \tan(\theta)$$

along Y axis where $$\varrho$$ is the mass density of the rope.
and

$$\frac {d(T \sin(\theta))} {dx} = \frac {\varrho g} {\cos(\theta)}$$ or just $$\varrho g$$

$$\frac {dy}{dx} = \tan(\theta)$$ or just $$sin(\theta)$$

or something...

You're close. No cos in the denominator. Its just the denisty, times gravity, time delta x. Actually its really, w(x)deltaX. w(x) is the more general weight distribution. Your right about the tangent, no sin there.

 

[maverick6664]

You're close. No cos in the denominator. Its just the denisty, times gravity, time delta x. Actually its really, w(x)deltaX. w(x) is the more general weight distribution. Your right about the tangent, no sin there.

oops. you are right..cosine shouldn't be in the denominator

 

[lalbatros]

It seems to me that this is obvious.
T is possibly dependent on $$\theta$$ and therefore:

$$\Delta (T cos(\theta)) = T(\theta + \Delta \theta) * cos(\theta + \Delta \theta) - T(\theta) * cos(\theta)$$

and of course the definition

$$T(\theta + \Delta \theta) = T(\theta) + \Delta T$$

leads to the result