Recent content by IGeekbot

For our balsa wood bridge, 1/8 in square, would a bridge that crosses catilever and arch styles together work well? The steel blocks attached to the test table are acting like abutments, so all of our forces can be directed to them, right?

I have one, and I don't know where you are getting a $200 cost from. All you need is some PVC, PVC glue, the adapters, and hairspray. For an igniter, you can go electric or flint. Each one has its upsides and downsides. And the trick is not how much fuel, but it is how much backpressure...

Can anyone or does anyone know where I can get a good explanation of how to figure out how big an arch should be for a bridge? I am thinking of making one for our competition, and if I can't get that to work right, I will switch to a king post deisgn. Oh, and is it true balsa and other...

Ok, sorry to bring this back up, but I need to get someone to straighten me out. My teacher is telling me that the velocity of Ball B in this problem would be 3.6m/s. He is figuring that ball B is moving southeast at around a 45 degree angle when it makes contact with ball a. This would...

Welll, get out your pens, because i took a picture of the problem, and other than having a wrong number in my typing, which I fixed in my work, I typed it word for word. Apparently, this book has the power to defy physics. I will return later with the answers to question 2, but I need to go do...

I went back and checked. I tpyed what the book says. It seemed odd to my friend and I today as well. We wondered how a ball A could roll along completely horizontally, hit Ball B, have ball B continue on horizontally, and have Ball A deflect to a completely 270 degree direction.

ok, I think I am getting this. Ball B, it will have a velocity of -1 m/s, since it is moving east, and it has a mass of 2kg. Because the original velocity is 6kg*m/s east when the ball makes contact, the two balls afterward will have to have a 6kg*m/s momentum. Since the angle is...

Yeah, its an intro class, and the teacher expects you to pick it up the frist time, and if you ask for help, he still won't really explain it. He just repeats what he says. And I foudn that section, and we have just used it a little bit.

so I need to use the horizontal and vertical numbers? Then, when you get those, they should equal the initial momentum of the ball, which is 6 kg*m/s?

and no, we have never done vectors like that. I think i will read on that more.

yeah, wrong key.

To calculate momentum, it is P=mv. In a closed system you have Pa+Pb=P'a+P'b Then we have the ones for elastic and inelastic equations, but i don't think those are needed here.

The ball would have the original momentum of 3kg x m/s moving west. I got that, but this is where I get lost, because this is all new to me.

ive probably seen them, but that is not ringing a bell. could you please explain what it is?

thats just it, i don't know what to do. Our teachers idea of introducing a new concept is go through it in five minutes to move on to a lab. He justrecites the book so it is no help. I know we have to use the sin and cos of one of the veloctiies, but i am not sure which one. When i get osm...