Matt Springer is a graduate student of physics at Texas A&M university. He is also an occasional writer and tinkerer, and he is probably too curious for his own good.
when you order from Amazon, and a fraction of the purchase price will be sent to me at zero cost to you. Much obliged, and thanks for your patronage.In honor of Physics 201 which I'm teaching this semester, I present a very elementary statics problem.
Here we have a board of uniform composition and weight W. It has length l and the supports are separated by a distance s. What are the two forces (call them A and B) on the boards?
The board isn't moving. It's just sitting there, and so if there's no acceleration there's no net forces. That gives us
Hmm. That's two unknown quantities with one equation. Not enough. Fortunately there's another equation we can use involving the torque. Torque is the angular equivalent of force. Get a wrench and turn a bolt, and you're applying a force at a distance from the bolt. That distance times the force is the torque about the bolt. If there's no angular acceleration (it's not spinning at all here), there's no net torque. There's no bolt either, but fortunately it turns out that you can measure the toque about any convenient point - some might make the problem easier than others. Here's let's pick point A as the point to measure the torque about, and we'll write down each force times its distance from A. They'll have to add to zero since there's no angular acceleration and thus no torque.
That factor of (l/2) represents the torque from the weight of the board. Since the board is uniform we can just pretend all the mass is at the center point. So now we can solve the second equation for B, and use that to find A in the first equation. I get
and
Easy, huh? Everyone has to start somewhere though, and this is a pretty good place to start for statics.
Exit question (easy): If s is too small, the board will fall over. What is the smallest value of s that will allow the board to balance in the picture and how do we know this from the equations?
Exit question (medium): If the board is not uniform but instead has a weight per length (x increasing toward the right) given by
The constants are of course picked so that the total weight is still W. What is the minimum s that allows for stability? Sadly there are no prizes other than fame in the comments, but I'm sure the thrill of success will be reward in itself!
Physical Science
Comments
I think you mean that A=W(1-l/2s).
Posted by: DA | September 2, 2008 11:15 AM
If you reintroduce A = W((1-Wl)/2) into A + B = W, you get B = ((W^2)l)/2 not (Wl)/2.
Pfft, someone beat me to it.
Posted by: Akheloios | September 2, 2008 11:50 AM
Sorry, that's a pretty egregious typo. I've fixed it. Thanks for catching it!
And both challenge questions are still open...
Posted by: Matt Springer | September 2, 2008 12:09 PM
what if there are three supports under the board? what is the third condition?
i think you have to take into account the material properties of the board - the way bends in response to the forces on it. i'm not sure though. what do you think?
Posted by: meichenl | September 2, 2008 12:26 PM
Oh...and the smallest value of s is s=l/2. Not only does that make intuitive sense (once the center of mass moves past support B, it will tip), but the static equation for A and B stop making sense at that point.
Posted by: DA | September 2, 2008 1:08 PM
And by the same center of mass argument, the non-uniform board should tip after s is smaller than the center of mass. Doing the integral COM= /int_{0}^{l} /sigma \cdot x dx shows that happens at 4/5 the length of the board (4*l/5).
Posted by: DA | September 2, 2008 1:17 PM
that was fun
Posted by: Paul Johnson | September 2, 2008 2:34 PM
Commenter DA is right on both counts. Good job!
In particular, when he says the equations stop making sense he's talking about how the force A is providing becomes negative if s
I like this whole challenge question thing. I'll probably start doing it more.
Posted by: Matt Springer | September 2, 2008 5:26 PM
Question: you use the word "torque". This may be a science vs engineering thing, but in pretty much every engineering mechanics class I've been in, the word "moment" would be used instead of torque. They're both defined as r x F. So what the hell is the difference, and if there isn't one, why are there two words for it?
Posted by: C. Chu | September 2, 2008 5:50 PM
They're the same, unless there's some subtle difference I've never heard of. I think it's mainly just one of those weird things with no good explanation. My blind guess is that torque is used to reduce the chance of confusion between "moment" and "moment of inertia".
Posted by: Matt Springer | September 2, 2008 6:17 PM
Yeah i find non-uniform distribution problems to be really lame considering you can just find the center of mass again and then go about your business. This might be interesting if the system was rotating or some crazy jazz like that but thats probably out of the scope of the statics class.
Posted by: Paul Johnson | September 2, 2008 6:17 PM
I don't understand the solution to the second problem. I can't read what DA said...
Posted by: Jack | September 5, 2008 9:11 PM
Technically speaking, those triangles are the engineering symbol for pins, so the board can be whatever length you want it to be (with a constraint form the fact that the pin has a nonzero size). It'll only tip over if W is stronger than the bolts holding the pin in the ground.
As far as I know, in statics you use the terms Moment and Load, and in dynamics you use Torque and Force. So a moment usually causes bending and a Torque causes rotation. It's not really a rigid system for names but it helps some times. I kno wit helps me with formulae, i don't really have to think, it's just Sigma M = 0 and Sigma Tau = I*alpha.
Posted by: nanoAl | September 11, 2008 1:39 AM
Jack, he's just using messy notation for finding the centre of mass. I'm going to use an equally messy one in the hopes that you'll be able to peice something form our two explanations.
its just the integral of x*dw (or in this case x*(sigma)*dx) divided by the integral of dw(which is sigma*dx). so to find it you just multiply the weight density function by x and integrate it(from 0 to L) and then divide it by the integral of the weight function from 0 to L.
this is not a great format for explaining math!
Posted by: nanoAl | September 11, 2008 1:47 AM