Need some help with a trigonometry problem

Because compression makes it smaller.
You know, similar triangles.same angles… if the hypotenuse of one is 2/3rd the length of the hypotenuse of the other, all sides are 2/3ds the length.

A concise way to write what I was saying, is that both counters can be in tension
only if the counters have a similar strength, or stronger, to the other truss members.

Distortion due to stretching of the top truss members (and not counters).

The stretch of the top member is related to the youngs module * CSA (cross sectional area) there. But the stretching of both counters due to this is basically equivalent… 10mm growth of the top member , maybe its 15mm growth on the hypotenuse… worst case its twice… its impossible for it to be high as 2. but it can be shown its unimportant to calculate it precisely.

SR (strength ratio ) = youngs modulus of counter * CSA (cross section area) / youngs modulus of truss * CSA

This sets the amount of change in angle of the truss… distortion due to angular change (due to stretching of counters) which sets the scale of the compression of the compressed counter… Note that with this change in angle, if one counter stretches 2%, the other space for the counter compresses about 2%… it can only be offset by the stretching of both counters by the stretching of the top member (and the bottom member maybe ?) .

So the distortion can only cause stretching in both counters if the compression due to angular change is LESS than the stretch of the horizontal members x 2. But one counter is stretching SR times… so its impossible for the other counter to be in tension if SR is greater than 2… if the counter is half the cross sectional area of the truss members (assuming same youngs modulus for simplicity). We can see that 50 or 500 or 10 wouldn’t really matter, if its a counter, its definite that the counter is far stretchier than the other truss members, one counter must be in compression, force 0.

That may all be true, and it is interesting, but my textbook hasn’t covered any of that. I should be able to solve it with just the techniques the book has taught me.

The purpose of an engineering education (and really, any post-secondary education) isn’t to just be able to solve obvious problems using packaged techniques but to provide the foundation to work through more complex challenges that require creative problem-solving, i.e. how you can reduce a statically-indeterminate problem to one solvable by pure resolution of static forces by checking assumptions. If all a student learns how to do is solving problems by rote application of formulae, they will fail when confronted with the more ambiguously-defined problems in real world engineering challenges.

Stranger

Sure, but things like cross-sectional area, stretching, angular change haven’t even been hinted at in the book. In both solutions (the one for the edition I’m reading, and the one I linked to), it seems like they’re just magically assuming that counter has 0 tension. I’d like to find some reason for it that connects to something I’ve learned so far.

It’s worth pointing out that the solutions I looked at were not prepared by the same people who wrote the textbook.

You don’t really need any of those to puzzle out why if one crossmember is in tension, the other is in compression, and therefore needs to be removed from the calculation. (Since this is a statics text it won’t go into stress and strain or any calculations involving deflection and torsion, and you don’t need that information to resolve this problem.). This is why basic mechanics courses are accompanied by a lab class, to offer the practical insight into how structures react and build a physical intuition that lets you apply the principles to real structures. As an illustration, take a box frame (or a sturdy cardboard box) and make crossmembers using twine or wire rope, and then see what they do when you apply external loads to the corners.

Stranger

If I have a square frame of rigid members, and I connect both pairs of opposite corners with stretched rubber bands, how is it possible to not have both rubber bands in tension? If, in any given shape, one of the rubber bands has zero tension, then what’s stopping the other diagonal from shortening?

They’re not stretched rubber bands. When the frame is square, the “rubber bands” are on the cusp of being slack.

How can we tell that, though? I can’t see any reason those couldn’t be elastic members stretched taut.

This is a text for a statics class, and the underlying assumption is that the members are rigid bodies with no internal preloads on an unloaded structure. These are just pin-connected members with no ability to resist buckling, and hence can only bear tensile loads. In a real world structure you would may to consider the preloaded state (just due to gravity if nothing else) but not in these textbook problems which are really just illustrating the methodology of summing loads and moments on static free bodies.

Stranger

That sounds a lot like complaining that that thing with four legs and an udder can’t possibly be a cow, because it’s not spherical…

Again, it is a problem from a introductory engineering text intended to teach rigid body static mechanics. Once students have mastered the fundamental concepts of applying forces and moments and calculating reactions to the wide array of basic structures and mechanisms, they go onto mechanics of materials where the material and sectional geometry properties are applied to understand deflection, torsion, stress, and strain on structural members, and from there into more complicated methods of calculating static and dynamic loads and stresses on real world structures and mechanisms.

Stranger

Sorta like doing F=ma problems w physics students before we consider the presence of air drag in addition just gravity. And without factoring in that acceleration due to gravity changes as the objects get closer together.

The crawl stage of “crawl, walk, run” includes a lot of “lies to children” regardless of their age or general sophistication.

IIRC highschool math had one more step, and that was to check for d2x/dϕ2 , a positive value showed a minima and a negative value showed a maxima

I plotted the function with TD = 1 and sure enough its sinusoidal with many maxima and minima

https://imgur.com/a/D9NqX9r

I’m back!

Still making progress; I’m on chapter 7 now. As I think I said, I found someone who has worked through the problems and posted the answers online. However, he does take a few shortcuts. On the problem I’m currently working on, he jumps from

\frac{1}{2} ωL^{2}\frac{(L-2a)^2}{4(L-a)^2}=\frac{1}{2}ωa^2

to

(\frac{1}{2}-\frac{a}{L})^2-(\frac{a}{L})^2(1-\frac{a}{L})^2=0

Can anyone help me break that down into more manageable steps?

I found a useful equation editor here. (Unfortunately, when I cut and paste from that I run into the “no posting media items” problem.

While searching for this thread to revive it, I found a similar thread started by @Cubsfan about 20 years ago. Using the same book I am; same edition, even.

\begin{aligned} \frac{1}{2}\omega L^2\frac{(L-2a)^2}{4(L-a)^2} &= \frac{1}{2}\omega a^2 \\ L^2\frac{(L-2a)^2}{4(L-a)^2} &= a^2 \\ \frac{(L-2a)^2}{4(L-a)^2} &= \left(\frac{a}{L}\right)^2 \\ \frac{(\frac{L}{2}-a)^2}{(L-a)^2} &= \left(\frac{a}{L}\right)^2 \\ \frac{L^2(\frac{1}{2}-\frac{a}{L})^2}{L^2(1-\frac{a}{L})^2} &= \left(\frac{a}{L}\right)^2 \\ \frac{(\frac{1}{2}-\frac{a}{L})^2}{(1-\frac{a}{L})^2} &= \left(\frac{a}{L}\right)^2 \\ (\frac{1}{2}-\frac{a}{L})^2 &= \left(\frac{a}{L}\right)^2 (1-\frac{a}{L})^2 \\ (\frac{1}{2}-\frac{a}{L})^2 - \left(\frac{a}{L}\right)^2 (1-\frac{a}{L})^2 &= 0 \end{aligned}

Thanks. Would have taken me a long time to get that (if ever).

I think I like your formatting better than mine. What did you use?

The discourse tool used for this forum supports writing math with TeX, so I typed

$$
\begin{aligned}
\frac{1}{2}\omega L^2 \frac{(L-2a)^2}{4(L-a)^2} &= \frac{1}{2}\omega a^2 \\

\end{aligned}
$$

It’s not the easiest thing at first, but it’s a very powerful way to write math and it’s used pretty pervasively in math, science, and engineering writing. Once you learn it, it’s faster than using a graphical equation editor (although this is about the maximum I would want to try to do on my phone).

Can we try another one?

\begin{aligned} (\frac{a}{L})^2-2\frac{a}{L}+\frac{1}{2}=0 \end{aligned}

to

\begin{aligned} \frac{a}{L}=1\pm \frac{1}{\sqrt{2}} \end{aligned}

That’s just the quadratic formula. Do the substitution x=\frac{a}{L} to get:
x^2 - 2x + \frac{1}{2} = 0

Then plug into:
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

That’s:
\frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \frac{1}{2}}}{2 \cdot 1}
Or just:
1 \pm \frac{1}{\sqrt{2}}