2
$\begingroup$

I am looking at Hibbeler's section on a `Cable Subjected to a Distributed Load' in Engineering Mechanics - Statics, Ed13, as shown in the attached images.

enter image description here

enter image description here

I am getting stuck deriving equation (7-7). I derive the horizontal equilibrium equation just fine;

$$ |\vec{T} + \vec{\Delta T}| \cdot \cos(\theta + \Delta \theta) - |\vec{T}|\cdot \cos(\theta) = 0 $$

I rearrange to describe horizontal tension component as a function of $\Delta x$ (given $\Delta T$ and $\Delta \theta$ are ultimately functions of $\Delta x$);

$$ |\vec{T}|\cdot \cos(\theta) = |\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta) $$

Then I express the differential;

$$ \frac{\Delta(|\vec{T}|\cdot \cos(\theta))}{\Delta x} = \frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x} $$

Then when I go to express the derivative, I come up with a limit which does not exist.

$$ \frac{d(|\vec{T}|\cdot \cos(\theta))}{dx} = \lim_{\Delta x \to 0}\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x} $$ Any pointers as to where my error is? Appreciate any advice offered. Thanks.

$\endgroup$
2
  • $\begingroup$ First of all use don't use vectors but scalars, as your book suggests. Then $T\cos\theta$ must be eliminated before diving by $\Delta x$. $\endgroup$
    – N74
    Apr 22 '17 at 9:07
  • $\begingroup$ Are you able to elaborate on this any further? I am taking the norm of the vector, which is a scalar. I don't see how I can eliminate $T\cos(\theta)$ before dividing by $\Delta x$? Apologies if I am missing something obvious. I tried to apply the sum angle identity but was unable to make anything cancel in the way you suggest. $\endgroup$ Apr 23 '17 at 5:24
2
$\begingroup$

First of all is better to use $T$ as a scalar and write: $$( {T} + {\Delta T})\cdot \cos(\theta + \Delta \theta) - {T}\cdot \cos(\theta) = 0 $$ next expand the cosine: $$( {T} + {\Delta T})\cdot (\cos(\theta)\cos( \Delta \theta)- \sin(\theta)\sin( \Delta \theta) )- {T}\cdot \cos(\theta) = 0 $$ and regroup: $$ {T} \cos(\theta)(\cos( \Delta \theta) -1) + {\Delta T}\cdot \cos(\theta)\cos( \Delta \theta)- (T+ \Delta T) (\sin(\theta)\sin( \Delta \theta) ) = 0$$ now we can divide and take the limit: $$ \lim_{\Delta x \rightarrow 0} {T} \cos(\theta){(\cos( \Delta \theta) -1) \over \Delta \theta} {\Delta \theta \over \Delta x} + \lim_{\Delta x \rightarrow 0} {\Delta T \over \Delta x}\cdot \cos(\theta)\cos( \Delta \theta)- \lim_{\Delta x \rightarrow 0} (T+ \Delta T) \sin(\theta){\sin( \Delta \theta)\over \Delta \theta} {\Delta \theta \over \Delta x}= 0$$ I also introduced a couple of $ {\Delta \theta \over \Delta \theta}$ where needed.

Now, remember that also $\Delta \theta$ and $\Delta T$ go to $0$, so the first limit is $0$, in the second $\cos \Delta \theta$ goes to $1$ and in the third we can neglect $\Delta T$, we remain with: $${d T \over d x}\cos \theta-T \sin \theta {d \theta \over d x}=0$$ that is the expression we were looking for.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.