The quadratic formula says, given
that
so if we let x \equiv a/L, A \equiv 1, B \equiv -2, and C \equiv 1/2, then
Thanks, Doc and ab; I should have tried that, myself.
This was a particularly thorny problem. I guess I was just a bit frustrated.
I will say, when I started working through this book, every now and then I’d flip toward the back to see what I was in for, and some of it looked absolutely impenetrable. Now that I’ve actually gone through most of the book, problems that used to look impossible now seem doable.
Next up, loads (concentrated and distributed) hanging from cables. I’ll be designing suspension bridges in no time.
Sure thing. BTW, for this particular one, it’s pretty easy to just complete the square manually. Starting again with:
x^2 - 2x + \frac{1}{2} = 0
We can add some terms to get:
x^2 - 2x + 1 - 1 + \frac{1}{2} = 0
Which is then:
(x - 1)^2 = \frac{1}{2}
Or:
x - 1 = \pm \sqrt{\frac{1}{2}}
Finally:
x = 1 \pm \frac{1}{\sqrt{2}}
There’s no difference obviously, but it might be helpful to practice.
Indeed, completing the square is how one derives the Quadratic Formula to begin with.
Yeah, I remember deriving that formula, once, about 45 years ago.
Everybody derives that formula, once. And then, having derived it, just uses the quadratic formula, and most likely then never uses completing the square for anything else, ever again (maybe converting a conic-section equation to a more useful form, but how often does that come up?).
Unless you’re a high school math teacher. Then you re-derive it a few times a year.
I derive it on a regular basis while trying to fall sleep. I have to wait a month or two between uses so I forget exactly how it goes. It’s complicated enough that I can just barely keep all the terms in my head at once.
Anyway, I think it’s always good practice to solve things multiple ways. And it’s good to recognize some common patterns, like that x^2 + 2ax + a^2 = (x + a)^2 or that x^2 - a^2 = (x + a)(x - a).
And you finally got to use it to help a complete stranger on the internet. I love this board.
As luck would have it, that difference of two squares was also needed in the problem I’ve just been posting about.
The quoting function doesn’t seem to work properly for equations.
A complete quote seems to transfer correctly.
But a partial quote does not.
You can always quote the whole post into a scratchpad, copy the parts you want, discard the scratchpad, then paste the good parts into your new post.
The quoting mechanism doesn’t preserve the wrapping $ characters around the expression.
If you put them back it should work again. Takes a bit of messing about.
It is a bit much to suggest it is never used, when it is clearly extremely closely related to the useful law of cosines/polarization identity and the parallelogram identity, not to mention that completing the square is likely to be useful whenever quadratic polynomials come up.
IMHO solving a quadratic equation by completing the square is simpler than using the quadratic formula in cases where the x^2 coefficient is 1 and the x coefficient is an even number. Plus it’s kinda cool how it makes sense visually/geometrically.
Potentially, pretty often if you’re working a lot with conics.
The other situation I can think of off the top of my head where completing the square comes in handy is in integration: completing the square is sometimes useful in wrestling the integrand into a form that’s easy to integrate.
If you’re working with a lot of conics, in rectangular coordinates. With generalized conics, polar coordinates are usually much simpler, and with parabolas specifically, everything you can get from completing the square, you can also get from the quadratic formula.
I’m not denying that completing the square has its place, but for the vast majority of students, it’s mostly just for deriving the quadratic formula.
But yes, absolutely you should be able to recognize quadratics that are already perfect squares, as well as common situations like a difference of squares.
Hello again. I’ve slowed down a bit lately, but still making progress. I almost revived this thread when i got to the section on catenary cables, but I managed to muddle through that on my own. I learned what the HYP and HYP-1 buttons on my calculator do. There’s a whole nother type of trigonometry I never knew about before.
But now I have another question. I’m in the chapter on friction. It mentions that when motion of one surface against another is impending,
friction = coefficient of static friction * normal force. But that’s only when motion is impending. At other times, friction and normal force are independent.
Here’s the problem I’m working on:
The solution I found online assumes that motion is impending at both points of contact, and the relation of friction to normal force applies. Is that valid? I mean, obviously if one end of the rod moves, so will the other. But my thinking goes like this: "Suppose the surface at point A was frictionless. I could solve the problem and find an angle θ1 for which the friction at B is just enough to hold the rod in place. Or I could assume that point B is frictionless, and find the angle for which friction at A is just sufficient to hold the rod in place; call that θ2. So, in the original case with friction at both points, what happens between θ1 and θ2? I’ve exceeded to frictional limit of one of the points of contact, but not the other.
Is it really valid to treat both points as if motion impending, even though they exceed their individual limts at different angles? Is the answer to the problem as written the lesser of θ1 and θ2, or is it smaller than both?
I don’t actually think that problem is solvable, without some additional information. Is the rod resting gently, or is it jammed in there as hard as it can go? The distinction is most obvious in the case where mu > 1, because in that case, a jammed-in rod could stay in place even if you turn the frame upside down, but I think it’s relevant for all nonzero values of mu. The key is that, even if we know that motion is imminent, we don’t know which direction it’s imminent in.
I think it is solvable. The coefficient of static friction is given as 0.25.
If the rod was moved too far to the left, it would reach a position where the static friction isn’t sufficient to hold it in place, and it would slide to the right. Similarly, too far to the right and it would slide back to the left. The goal is the find the angles θ that correspond to those two cases, once where α = 45 °, and again where α = 25°.
Yes, but I maintain that there are multiple states for the rod, which are not uniquely defined by its position, and that different states will have different angles where that happens.
EDIT: To make it clearer, we have five unknowns: Two normal forces, two static friction forces, and an angle. We have four equations: A force balance equation on one axis, a force balance equation on the other axis, a torque balance equation, and one static friction equation, because only one of the ends is on the verge of sliding. Either end can be the one on the verge of sliding, which is how we get the two endpoints of the stable region, but either way, we have more unknowns than equations.
Here’s a book of solutions that I resort to when I get stuck on a problem.
Problem 8.133 covers a general solution to the problem, and 8.134 uses that general solution to solve for the specific case.
In a nutshell, the bar is subject to three forces, gravity (acting at the center of the rod), resultant force at A (combination of normal force and friction), and resultant force at B (also combination of normal and friction). On a three-force body, the forces must be concurrent.
That’s the part I’m trying to figure out. Either both ends of the rod are moving, or both are stationary. In that sense, motion is impending at both A and B at the same time. But in another sense (as I tried to describe above), it feels as though there would be angles for which one end had exceeded its limit, but the other was still holding the rod in place.
The problem is solvable, and because this is a text for a statics class (correct?) the AB member is rigid (no flexure or compression), so there is a singular, deterministic force state for any stable position. The key to solving the problem is to understand that the normal friction coefficients are actually free parameters which can vary independently up to 0.25 with the friction force pointed in either sliding direction, and then find the limiting condition at which the forces are just balanced at the 0.25 limit at one contact. (There are technically four possible limit conditions but I think you’ll find in all cases that two have negative normal forces and thus are aphysical, which reduces down to just two possible conditions.)
It helps to reduce the terms if you orient your coordinate system with the 90° frame and just use 𝛼 to translate the gravity vector into that frame so you don’t have compounding sine/cosine terms. Although I formulated the force relations I didn’t solve the problem for the specified α angles because I don’t think you can make an explicit solution (at least, not without a lot of trigonometric fuckery) and I didn’t care to put in the effort to putting the relations into a numerical solver or iterate it by hand but it should be straightforward once you develop the equations.
Stranger
Ah, but is the limit at just one contact? The solution I linked to above assumes that movement is impending at both points of contact, and therefore F = 0.25 × N at both points. With that assumption, you can calculate the direction of the resultant forces at both points, and solve from there.
That’s what I’m struggling with; is that a valid assumption?
And once again, I do appreciate all the comments in this thread. I do learn a lot here, and I don’t know anyplace else I could ask something like this.